JEE Main + AdvancedClass XIIElectrostaticsPotential & Capacitance

Electrostatic Potential and Capacitance

The electrostatic potential and potential difference, potential due to a point charge, a dipole and a system of charges, equipotential surfaces and their relation to the field, electrostatic potential energy, the electrostatics of conductors, dielectrics, capacitors and capacitance, combinations of capacitors and the energy stored in the electric field

🎯 Overview Chapter hero + roadmap

🔬 Interactive 3D · One view of the whole chapter: the equipotential surfaces around a positive point charge are a set of nested spheres of constant potential $V=kq/r$, and the electric field lines pierce them everywhere at right angles, pointing from high potential to low. Move the test charge and read off how the potential $V$ and the potential energy $U=qV$ change as it crosses from one surface to the next, while no work is done moving it along a surface.

Every problem in electrostatics can be attacked in two complementary ways. The first, developed in the previous chapter, follows the vector electric field $\mathbf{E}$ and the force $q\mathbf{E}$ it exerts on a charge. The second, the subject of this chapter, follows a single scalar quantity — the electrostatic potential $V$ — and the energy that a charge carries by virtue of its position. Because the electrostatic force is conservative, the work done in carrying a charge from one point to another does not depend on the path taken, only on the endpoints; and whenever work is path-independent it can be stored as a potential energy and described by a potential. Trading the three components of a vector field for one scalar function is an enormous simplification, and it is the reason potential is the physicist's tool of choice for all but the most symmetric problems. 🔉⇢

The electrostatic potential difference between two points A and B is defined as the work done by an external agent, against the electrostatic force, in moving a unit positive test charge from A to B without any change in kinetic energy. Equivalently it is minus the work done by the field itself, per unit charge. Its SI unit is the volt (V), equal to one joule per coulomb. The potential at a point is then the potential difference between that point and a chosen reference, conventionally taken to be at infinity where the potential is set to zero. Potential is a property of the field and of position alone; it exists whether or not a test charge is placed there, just as the temperature of a room exists whether or not a thermometer is present. 🔉⇢

For a single point charge $q$, integrating the field $E=kq/r^2$ from infinity in to a distance $r$ gives the potential $V=kq/r$, where $k=1/4\pi\varepsilon_0$. Notice that the potential of a point charge falls off as $1/r$, more gently than the field's $1/r^2$, and that it carries the sign of the charge: positive around a positive charge, negative around a negative one. Because potential is a scalar, the potential due to several charges at a point is simply the algebraic sum of the potentials each would produce there on its own — no components, no angles, no vector diagram. This principle of superposition for potentials is what makes the potential of an arbitrary charge distribution tractable. 🔉⇢

An electric dipole — two equal and opposite charges a small distance apart — produces a potential that reveals a new feature. At a point a distance $r$ from the centre of a short dipole, making an angle $\theta$ with the dipole axis, the potential is $V=k p\cos\theta/r^2$, where $p=q\,(2a)$ is the dipole moment. Two things distinguish this from a point charge: it falls off faster, as $1/r^2$, because the charges partly cancel at a distance; and it depends on direction through $\cos\theta$, vanishing everywhere on the plane that perpendicularly bisects the dipole. The dipole is the prototype of every neutral-but-polar object, from a water molecule to an antenna, and its potential is a recurring examiner favourite. 🔉⇢

A vivid way to picture the potential is through equipotential surfaces: surfaces on which the potential has a single constant value. Because no potential difference exists between any two points on such a surface, no work is done in moving a charge along it, and it follows that the electric field can have no component along an equipotential — the field must be everywhere perpendicular to it. For a point charge the equipotentials are concentric spheres; for a uniform field they are parallel planes. Equipotentials are the electrical analogue of the contour lines on a map, and just as closely spaced contours mark a steep slope, closely spaced equipotentials mark a strong field. 🔉⇢

This geometric picture is made quantitative by the relation between field and potential. Over a small displacement the change in potential is $dV=-E\,dl$, so the field is the negative gradient of the potential, $E=-dV/dr$: its magnitude equals the rate at which the potential falls with distance, and it points in the direction of steepest decrease of $V$. This is why the field points from high potential to low, why a strong field means a rapid change of potential, and why the field is measured equally well in volts per metre as in newtons per coulomb. Being able to move fluently between $E$ and $V$ — differentiate to get the field, integrate to get the potential — is one of the core skills the chapter builds. 🔉⇢

Just as the potential describes the field, the potential energy describes the interaction of charges. The potential energy of a system of point charges is the total work needed to assemble them, brought one at a time from infinity; for two charges it is $U=kq_1q_2/r$, and for more charges it is the sum over every distinct pair. A charge $q$ placed at a point of potential $V$ in an external field carries potential energy $U=qV$, and a dipole of moment $p$ in a uniform field $E$ has orientation energy $U=-\mathbf{p}\!\cdot\!\mathbf{E}=-pE\cos\theta$, least when it is aligned with the field and greatest when anti-aligned. These energies convert directly into kinetic energy by the work-energy theorem, which is how the chapter connects to mechanics in problems on charged particles accelerated through a potential difference. 🔉⇢

Turning from charges in space to charges on matter, the chapter next examines the electrostatics of conductors. A conductor contains mobile charges that rearrange until they feel no net force, and this single condition has sweeping consequences: the electric field is zero everywhere inside a conductor in equilibrium, any excess charge resides entirely on its surface, the field just outside is perpendicular to the surface, and — most usefully — the entire conductor, surface and interior alike, is a single equipotential volume. A hollow conductor shields its interior from external fields, the principle of electrostatic shielding that protects sensitive equipment and keeps you safe inside a car struck by lightning. 🔉⇢

Insulators, or dielectrics, behave quite differently: their charges are bound and cannot flow, but they can be displaced slightly, so that the material develops an induced dipole moment and becomes polarised. The bound charges that appear on the surface of a polarised dielectric set up an internal field that opposes the applied field, reducing the net field inside the material by a factor $K$, the dielectric constant. This modest-sounding effect is the key to the technology of the second half of the chapter, because filling the gap of a capacitor with a dielectric multiplies its ability to store charge. 🔉⇢

A capacitor is any pair of conductors carrying equal and opposite charges, and its capacitance $C=Q/V$ measures how much charge it stores per volt applied — a purely geometric property of the conductors and the medium between them, independent of the charge actually placed on it. For the parallel-plate capacitor, the workhorse of the chapter, $C=\varepsilon_0 A/d$ in vacuum, rising to $C=K\varepsilon_0 A/d$ when a dielectric of constant $K$ fills the gap. Capacitance is measured in farads (F), a very large unit in practice, so real capacitors are rated in microfarads and picofarads. 🔉⇢

Capacitors are combined in series and in parallel to obtain a desired capacitance, and the rules are the mirror image of those for resistors: capacitances in parallel simply add, $C=C_1+C_2+\dots$, because they share the same voltage while their charges add, whereas capacitances in series add as reciprocals, $1/C=1/C_1+1/C_2+\dots$, because they carry the same charge while their voltages add. Charging a capacitor requires work, which is stored as electrostatic potential energy $U=\tfrac12 CV^2=\tfrac12 QV=\tfrac12 Q^2/C$; this energy can be regarded as residing in the electric field itself, with an energy density $u=\tfrac12\varepsilon_0 E^2$ everywhere the field exists. Problems on charge redistribution when charged capacitors are connected together, and on the energy lost in the process, are a staple of the examination. 🔉⇢

For the JEE this chapter is a reliable, high-yield scorer. In JEE Main it contributes one or two questions almost every year, most often on capacitance, the parallel-plate capacitor with a dielectric slab, and capacitor combinations with the stored energy; potential and potential energy of charge systems appear regularly too. In JEE Advanced it feeds multi-concept problems that combine capacitor networks with steady-current circuits, or a dielectric being inserted with the battery either connected or disconnected — a distinction that decides whether it is the charge or the voltage that stays fixed, and the commonest source of error in the whole chapter. Keep the scalar addition of potentials firmly in mind, always ask whether charge or voltage is held constant when a capacitor is altered, reason the sign of every work and energy term from the physics rather than a memorised rule, and work the interactive scenes actively — predict how the capacitance or the stored energy changes before you slide the control — and the chapter is very largely won. 🔉⇢

What you will master 🔉⇢

🧠 Concepts Map of the deep dives

This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.

Electrostatic PotentVPotential Due to an pEquipotential Surfac|E|=- V l▶Potential Energy of U=14_0q_1 q_2r_12Electrostatics of CoE=_0Capacitors, Dielectr+Q▶Combination of Capac1C=_i1C_i▶
🔉⇢
What you are looking at

A map of the whole chapter. Each box is one concept with its governing relation, and the arrows are the recommended order to learn them in.

Click any box to jump straight to that concept’s tab.

🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.

Electrostatic Potential and Potential Difference 🔉⇢

The electrostatic potential $V$ at a point in an electric field is the work done by an external agency (equal and opposite to the electrostatic force) in bringing a unit positive charge, without acceleration, from infinity to that point, so that the potential difference between two points equals the work per unit charge between them, $V_P-V_R=\dfrac{U_P-U_R}{q}$.

Potential Due to an Electric Dipole 🔉⇢

The electrostatic potential due to a point electric dipole of moment $\vec{p}$ at a point with position vector $\vec{r}$ (with $r\gg a$) is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$, so it depends on both the distance $r$ and the angle $\theta$ between $\vec{r}$ and $\vec{p}$, and falls off as $1/r^2$ rather than $1/r$.

Equipotential Surfaces and the Field–Potential Relation 🔉⇢

An equipotential surface is a surface on which the potential has the same constant value at every point; the electric field is everywhere perpendicular to such surfaces and points in the direction of steepest decrease of potential, with magnitude equal to the potential gradient, $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ normal to the surface.

Potential Energy of a System of Charges 🔉⇢

The electrostatic potential energy of a system of charges is the total work done by an external agency in assembling the charges at their locations by bringing them one at a time from infinity; for two charges it is $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, and it is a property of the final configuration alone, independent of the order of assembly.

Electrostatics of Conductors 🔉⇢

In the static situation a conductor arranges its free charges so that the electric field is zero everywhere inside it, the field just outside is everywhere normal to the surface with magnitude $E=\dfrac{\sigma}{\varepsilon_0}$, any excess charge resides only on the surface, and the whole conductor is a single equipotential volume.

Capacitors, Dielectrics and Capacitance 🔉⇢

A capacitor is a pair of conductors carrying charges $+Q$ and $-Q$ at potential difference $V$, characterised by its capacitance $C=\dfrac{Q}{V}$, a purely geometric quantity that for parallel plates in vacuum is $C_0=\dfrac{\varepsilon_0 A}{d}$ and is raised by a factor $K$ (the dielectric constant) to $C=\dfrac{K\varepsilon_0 A}{d}$ when the gap is filled with a dielectric.

Combination of Capacitors and Energy Stored 🔉⇢

Capacitors in series share the same charge and add reciprocals, $\dfrac{1}{C}=\sum_i\dfrac{1}{C_i}$, while capacitors in parallel share the same voltage and add directly, $C=\sum_i C_i$; the energy banked in a charged capacitor is $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$, stored in the field with density $u=\dfrac{1}{2}\varepsilon_0 E^2$.

📖 Contents Full chapter — read straight through

The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.

q = +4.0 uCq = -3.0 uCF = 0.34 Nr = 0.30 mneutral: p = 0.26
🔉⇢
What you are looking at

The chapter banner — the physics of this chapter in a single moving picture, with live symbols and values.

Every diagram below it is interactive: drag the controls and the numbers move with the drawing.

Electrostatic Potential and Potential Difference 🔉⇢

🎯 Potential belongs to the point; energy belongs to the charge you put there. The potential DIFFERENCE between two rings is what moves the charge — flip the test charge and the same point becomes an energy hill instead of a valley.
🔉⇢
V = k·Q / r = — kV  ·  U = q·V = — mJ
V does not depend on q at all — change the test charge and only U moves
What you are looking at
  • Dashed purple rings — equipotentials, each labelled with its own value of V. Every point on one ring is at the same potential, and the gap between two rings is a potential difference.
  • The solid purple curve — V against r. The dashed red curve — the energy U of the test charge you chose.
What to do
  1. Flip the test charge negative. V does not move at all; only U flips.
  2. Set the test charge to zero. U vanishes, V does not — the potential is still there.
  3. Reverse the source Q and watch BOTH curves flip, because now the field itself has reversed.
What it means — V is joules per coulomb and belongs to the position; U is joules and belongs to the charge sitting there. The work to move q from one ring to another is q times the potential DIFFERENCE. A positive charge falls toward LOW potential and a negative charge toward HIGH potential, and both are falling toward lower energy. That single sentence resolves most sign errors in this chapter.
Definition: The electrostatic potential $V$ at a point in an electric field is the work done by an external agency (equal and opposite to the electrostatic force) in bringing a unit positive charge, without acceleration, from infinity to that point, so that the potential difference between two points equals the work per unit charge between them, $V_P-V_R=\dfrac{U_P-U_R}{q}$. 🔉⇢

The starting point for this chapter is an idea you already met in mechanics: the potential energy associated with a conservative force. When an external force does work in taking a body from one point to another against a conservative force such as gravity or a spring force, that work is stored as potential energy of the body; if the external force is then removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy, so that the sum of kinetic and potential energies is conserved. The Coulomb force between two stationary charges is also a conservative force, and this is not surprising, since both the gravitational and the Coulomb laws have an inverse-square dependence on distance and differ mainly in the proportionality constants. Just as we defined the potential energy of a mass in a gravitational field, we can therefore define the electrostatic potential energy of a charge in an electrostatic field. 🔉⇢

Consider the field $\vec{E}$ due to a charge $Q$ placed at the origin, and imagine bringing a small positive test charge $q$ from a point $R$ to a point $P$ against the repulsive force on it. Two remarks fix the idea. First, we take the test charge $q$ to be so small that it does not disturb the original configuration, i.e. it does not move $Q$. Second, we apply an external force $\vec{F}_{ext}$ just large enough to counter the electric force, $\vec{F}_{ext}=-\vec{F}_E$, so there is no net force and no acceleration: the charge is carried with infinitesimally slow, constant speed. In this situation the work done by the external force is the negative of the work done by the electric force, and it is fully stored as potential energy of the charge $q$. 🔉⇢

The work done by external forces in moving the charge $q$ from $R$ to $P$ is written $W_{RP}$, and it increases the potential energy of the charge by exactly that amount. We therefore define the potential energy difference between the two points as $\Delta U=U_P-U_R=W_{RP}$. In words, the electric potential energy difference between two points is the work required to be done by an external force in moving, without accelerating, a charge $q$ from one point to another, for the electric field of any arbitrary charge configuration. The displacement here is in a sense opposite to the electric force, so the work done by the field itself is $-W_{RP}$. 🔉⇢

Two comments about this definition matter for the whole chapter. First, the right side of $\Delta U=W_{RP}$ depends only on the initial and final positions of the charge: the work done by an electrostatic field in moving a charge from one point to another is independent of the path taken. This path-independence is the fundamental characteristic of a conservative force, and the very concept of potential energy would be meaningless if the work depended on the path. Second, because only work appears in the definition, the potential energy is determined only to within an additive constant. Adding any constant $\alpha$ to the energy at every point leaves the difference unchanged, since $(U_P+\alpha)-(U_R+\alpha)=U_P-U_R$. The actual value of potential energy is not physically significant; only the difference is. 🔉⇢

The freedom in the additive constant lets us choose where the energy is zero, and the convenient choice is to set the electrostatic potential energy to be zero at infinity. Taking the point $R$ at infinity then gives $W_{\infty P}=U_P-U_\infty=U_P$. Since $P$ is arbitrary, this provides a clean definition: the potential energy of a charge $q$ at a point, in the presence of the field due to any charge configuration, is the work done by the external force (equal and opposite to the electric force) in bringing the charge $q$ from infinity to that point. 🔉⇢

Now comes the crucial step of removing the dependence on the test charge. The work done on $q$ is obviously proportional to $q$, because the force at any point is $q\vec{E}$, where $\vec{E}$ is the field of the given configuration. It is therefore convenient to divide the work by the amount of charge $q$, so that the resulting quantity is independent of $q$. In other words, the work done per unit test charge is characteristic of the electric field associated with the charge configuration. This leads directly to the idea of the electrostatic potential $V$ due to a given charge configuration, a property of the field alone and not of the charge we happen to be carrying. 🔉⇢

Dividing the potential-energy relation by $q$ gives the potential difference: the work done by an external force in bringing a unit positive charge from $R$ to $P$ is $V_P-V_R=\dfrac{U_P-U_R}{q}$, where $V_P$ and $V_R$ are the electrostatic potentials at $P$ and $R$. As before, it is not the actual value of the potential but the potential difference that is physically significant. If we again choose the potential to be zero at infinity, then the work done by an external force in bringing a unit positive charge from infinity to a point equals the electrostatic potential $V$ at that point. 🔉⇢

Stated fully, the electrostatic potential $V$ at any point in a region with an electrostatic field is the work done in bringing a unit positive charge, without acceleration, from infinity to that point. To obtain the work done per unit test charge with full rigour, one takes an infinitesimal test charge $dq$, finds the work $dW$ in bringing it from infinity to the point, and forms the ratio $dW/dq$; the external force at every point of the path is equal and opposite to the electrostatic force there. Potential is a scalar quantity, so potentials add algebraically, and its SI unit is the volt (V), equal to one joule per coulomb. 🔉⇢

It is worth being careful about the distinction between potential and potential energy, since students often blur them. The potential $V$ is a property of a point in the field, measured in volts; the potential energy $U=qV$ is a property of a particular charge $q$ placed at that point, measured in joules. Two different charges placed at the same point are at the same potential but have different potential energies. This is why potential, being charge-independent, is the more fundamental map of a field: give the potential everywhere and you can immediately find the energy of any charge you like by multiplying by that charge. 🔉⇢

We now apply these definitions to the simplest source, a single point charge $Q$ at the origin, taking $Q$ positive for definiteness. We want the potential at a point $P$ whose position vector from the origin is $\vec{r}$, which means calculating the work done in bringing a unit positive test charge from infinity to $P$. For $Q\gt 0$ this work, done against the repulsive force, is positive. Since the work is independent of the path, we choose the most convenient path of all: a straight radial line from infinity in to $P$. 🔉⇢

At an intermediate point $P'$ a distance $r'$ from the origin, the electrostatic force on a unit positive charge has magnitude $\dfrac{Q}{4\pi\varepsilon_0 r'^2}$, directed radially outward. The work done against this force in moving through a small radial step is $\Delta W=-\dfrac{Q}{4\pi\varepsilon_0 r'^2}\,\Delta r'$, the negative sign arising because for an inward step ($\Delta r'\lt 0$) the work is positive. Integrating this from $r'=\infty$ to $r'=r$ gives the total external work, which by definition is the potential at $P$: $V(r)=\dfrac{Q}{4\pi\varepsilon_0 r}$. This is one of the most important results of the chapter. 🔉⇢

The formula $V(r)=\dfrac{Q}{4\pi\varepsilon_0 r}$ is true for any sign of $Q$, although we derived it for $Q\gt 0$. For $Q\lt 0$ the potential $V$ is negative, meaning the work done per unit positive test charge in bringing it from infinity to the point is negative; equivalently, the electrostatic force does positive work because for a negative source the force on a positive test charge is attractive and points along the displacement. The result is also consistent with our choice that the potential at infinity be zero, since $V\to 0$ as $r\to\infty$. 🔉⇢

A comparison of the potential with the field of a point charge is illuminating. The potential varies as $1/r$, whereas the field varies as $1/r^2$. Thus, as you move away from the charge, the field falls off faster than the potential. A graph of $V$ against $r$ is a hyperbola-like curve decaying as $1/r$, while the field curve decays more steeply as $1/r^2$; both approach zero at large $r$, consistent with the zero-of-potential-at-infinity convention. Remembering that $V\propto 1/r$ but $E\propto 1/r^2$ prevents a great many sign and magnitude errors. 🔉⇢

A quick numerical example fixes the scale. The potential at a point $9\ \text{cm}$ from a charge of $4\times 10^{-7}\ \text{C}$ is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}=(9\times 10^{9})\dfrac{4\times 10^{-7}}{0.09}=4\times 10^{4}\ \text{V}$. The work needed to bring a further charge of $2\times 10^{-9}\ \text{C}$ from infinity to that point is $W=qV=(2\times 10^{-9})(4\times 10^{4})=8\times 10^{-5}\ \text{J}$, and this work is independent of the path along which the charge is brought. 🔉⇢

Real configurations contain many charges, and here the scalar nature of potential pays off. For a system of charges $q_1,q_2,\dots,q_n$ with distances $r_{1P},r_{2P},\dots,r_{nP}$ from a point $P$, each charge contributes a potential $V_i=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_{iP}}$ at $P$, exactly as if the others were absent. Because potential is related to work done by the field, and the field obeys the superposition principle, the potential also obeys superposition: the total potential is the algebraic sum $V=V_1+V_2+\cdots+V_n$. 🔉⇢

Written out, the potential at $P$ due to the whole configuration is $V=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_{1P}}+\dfrac{q_2}{r_{2P}}+\cdots+\dfrac{q_n}{r_{nP}}\right)$. Notice that this is a sum of ordinary numbers with signs, not a vector sum. Compared with adding electric fields, where one must resolve components and worry about directions, adding potentials is far simpler; this is one of the great practical advantages of working with potential rather than field. 🔉⇢

For a continuous charge distribution described by a charge density $\rho(\vec{r})$, we divide the distribution into small volume elements each of size $\Delta v$ carrying charge $\rho\,\Delta v$, find the potential due to each, and sum (strictly, integrate) over all such contributions. In this way the potential of any distribution can be built up. A particularly useful special case is a uniformly charged spherical shell of total charge $q$ and radius $R$: outside the shell the potential is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ for $r\ge R$, as if all the charge were concentrated at the centre. 🔉⇢

Inside that spherical shell the electric field is zero, and since no work is done in moving a charge where there is no field, the potential is constant everywhere inside and equal to its value at the surface, $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}$. This is a favourite examiner's point: the field can vanish in a region while the potential there is a nonzero constant. Field measures how potential changes from place to place, so a region of zero field is simply a region of uniform, unchanging potential, which need not be zero. 🔉⇢

A second standard example asks where, on the line joining two unlike charges, the potential is zero. For charges $3\times 10^{-8}\ \text{C}$ and $-2\times 10^{-8}\ \text{C}$ placed $15\ \text{cm}$ apart, setting the algebraic sum of the two potentials to zero, $\dfrac{3}{x}-\dfrac{2}{15-x}=0$ for a point between them, gives $x=9\ \text{cm}$ from the positive charge; a point on the extended line gives a second zero at $x=45\ \text{cm}$. Because potential is a signed scalar, contributions of opposite sign can cancel, which never happens for the (always positive) magnitude of a field. 🔉⇢

In summary, the electrostatic potential is the field's own map: at each point it records the work per unit positive charge needed to arrive there from infinity, it is a scalar that adds algebraically, and it is defined only up to an additive constant fixed here by taking $V(\infty)=0$. For a point charge it is $\dfrac{Q}{4\pi\varepsilon_0 r}\propto 1/r$, and for any collection of charges it is the algebraic sum of such terms. Potential energy of a charge follows immediately as $U=qV$, and every later idea in the chapter, from equipotential surfaces to capacitance, is built on this single scalar function. 🔉⇢

Derivation 🔉⇢

  1. Aim: find the potential at a point $P$ at distance $r$ from a point charge $Q$ at the origin, i.e. the external work per unit positive charge to bring a unit test charge from infinity to $P$; because the field is conservative, choose the convenient radial path.
  2. At an intermediate point $P'$ a distance $r'$ from $Q$, the electric force on a unit positive charge has magnitude $\dfrac{Q}{4\pi\varepsilon_0 r'^2}$ directed radially outward (for $Q\gt 0$).
  3. The external force is equal and opposite to this, so the external work in an infinitesimal radial step is $\Delta W=-\dfrac{Q}{4\pi\varepsilon_0 r'^2}\,\Delta r'$; the sign makes $\Delta W\gt 0$ for an inward step ($\Delta r'\lt 0$), as expected against repulsion.
  4. Integrate from $r'=\infty$ to $r'=r$: $W=-\displaystyle\int_{\infty}^{r}\dfrac{Q}{4\pi\varepsilon_0 r'^2}\,dr'=\dfrac{Q}{4\pi\varepsilon_0}\left[\dfrac{1}{r'}\right]_{\infty}^{r}=\dfrac{Q}{4\pi\varepsilon_0 r}$.
  5. By definition this external work per unit charge is the potential, so $V(r)=\dfrac{Q}{4\pi\varepsilon_0 r}$, valid for either sign of $Q$ and consistent with $V\to 0$ as $r\to\infty$.
  6. Generalise by superposition: for charges $q_i$ at distances $r_{iP}$ from $P$, each contributes $V_i=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_{iP}}$, and since potential inherits superposition from the field, $V=\displaystyle\sum_i V_i=\dfrac{1}{4\pi\varepsilon_0}\sum_i\dfrac{q_i}{r_{iP}}$.
  7. Relate energy and potential: multiplying the work-per-unit-charge by the actual charge gives $U_P-U_R=q(V_P-V_R)$, and with $V(\infty)=0$, the potential energy of a charge $q$ at a point is $U=qV$.
⚠️ JEE trap: The most common error is to confuse potential with potential energy, and to think that where the electric field is zero the potential must also be zero. Potential $V$ (volts) is a property of a point in the field and adds as a signed scalar; potential energy $U=qV$ (joules) depends on the charge you place there. Inside a charged spherical shell the field is zero, yet the potential is a nonzero constant equal to its surface value, because the field measures only how the potential changes from place to place. A second error is to imagine that the potential of a point charge falls off as $1/r^2$ like its field; in fact $V\propto 1/r$ while $E\propto 1/r^2$, so the field decays faster than the potential. 🔉⇢

Potential Due to an Electric Dipole 🔉⇢

🎯 A dipole's potential depends on the ANGLE and falls off as 1/r² — faster than a single charge. On the axis it peaks; on the equatorial plane it is exactly zero because the two charges are the same distance away.
🔉⇢
V = k·p·cos θ / r² = — kV  at  r = — cm
double r and V drops to a quarter: at 2r it is only — kV — a dipole potential dies as 1/r², not 1/r
What you are looking at
  • The red +q and blue −q — a dipole, with the purple arrow showing the dipole moment p pointing from −q to +q.
  • The teal point P — where you are measuring the potential, at distance r and angle θ from the axis, with the brown arc marking θ.
  • Right — V against θ at your fixed r: a cosine, positive on the +q side, zero at 90°, negative on the −q side.
What to do
  1. Swing θ to 90°. V drops to exactly zero — P is now equidistant from both charges, so their potentials cancel.
  2. Set θ = 0° (on the axis) for the largest positive V, then θ = 180° for the largest negative.
  3. Double r and read the number in the formula bar: V falls to a quarter, because a dipole potential dies as 1/r².
What it means — because potential is a scalar you just ADD the two contributions, and the small separation makes them nearly cancel: what survives falls as 1/r² and carries the angle. The classic error is to expect a 1/r fall-off as for a single charge — a dipole is one step faster.
Definition: The electrostatic potential due to a point electric dipole of moment $\vec{p}$ at a point with position vector $\vec{r}$ (with $r\gg a$) is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$, so it depends on both the distance $r$ and the angle $\theta$ between $\vec{r}$ and $\vec{p}$, and falls off as $1/r^2$ rather than $1/r$. 🔉⇢

An electric dipole consists of two equal and opposite charges $q$ and $-q$ separated by a small distance $2a$. Its total charge is zero, yet it is far from being electrically featureless. The dipole is characterised by a dipole moment vector $\vec{p}$ whose magnitude is $p=q\times 2a$ and which points, by convention, in the direction from the negative charge $-q$ to the positive charge $q$. Because the two charges are separated, the dipole produces both a field and a potential in the surrounding space, and studying its potential is our goal here. 🔉⇢

We already know from the previous chapter that the electric field of a dipole depends not just on the magnitude $r$ of the position vector but also on the angle between $\vec{r}$ and $\vec{p}$, and that at large distances it falls off as $1/r^3$, faster than the $1/r^2$ of a single point charge. We now expect a similar richness in the potential: it too should depend on direction as well as distance, and it too should decay faster with distance than the potential of a single charge. Our task is to determine that potential precisely and to contrast it with the single-charge case. 🔉⇢

Place the origin at the centre of the dipole. The essential tool is that potential obeys the superposition principle. Since the electric field obeys superposition, and the potential is defined through the work done by the field, the potential also follows the superposition principle. Therefore the potential due to the dipole at a point $P$ is simply the algebraic sum of the potentials due to the two charges taken separately: $V=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q}{r_1}-\dfrac{q}{r_2}\right)$, where $r_1$ and $r_2$ are the distances of $P$ from $q$ and from $-q$ respectively. 🔉⇢

This exact expression is correct at every point, but it is awkward because $r_1$ and $r_2$ are different from the distance $r$ measured from the centre. To make progress we use geometry. If $\theta$ is the angle between the position vector $\vec{r}$ and the dipole axis, then applying the law of cosines to the two triangles gives $r_1^2=r^2+a^2-2ar\cos\theta$ and $r_2^2=r^2+a^2+2ar\cos\theta$. These are exact; the approximation enters only when we specialise to points far from the dipole. 🔉⇢

We now take $r$ much greater than $a$, the regime of a point dipole, and retain terms only up to first order in $a/r$. Dropping the $a^2/r^2$ term, $r_1^2\approx r^2\left(1-\dfrac{2a\cos\theta}{r}\right)$ and $r_2^2\approx r^2\left(1+\dfrac{2a\cos\theta}{r}\right)$. Taking the inverse square root and expanding by the binomial theorem to first order gives $\dfrac{1}{r_1}\approx\dfrac{1}{r}\left(1+\dfrac{a}{r}\cos\theta\right)$ and $\dfrac{1}{r_2}\approx\dfrac{1}{r}\left(1-\dfrac{a}{r}\cos\theta\right)$. 🔉⇢

Substituting these into the superposition sum, the leading $1/r$ terms cancel, leaving only the correction that survives because the two charges are of opposite sign: $V=\dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\approx\dfrac{q}{4\pi\varepsilon_0}\dfrac{2a\cos\theta}{r^2}$. Recognising $p=2qa$, this is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$. The cancellation of the $1/r$ term is the mathematical reason the dipole potential is weaker and shorter-ranged than that of a single charge. 🔉⇢

The result can be written compactly using the projection $p\cos\theta=\vec{p}\cdot\hat{r}$, where $\hat{r}$ is the unit vector along the position vector $OP$. Thus the electric potential of a dipole is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$ for $r\gg a$. This form makes the directional dependence explicit and is easy to remember: the numerator is the component of the dipole moment along the line of sight to the point, and the denominator carries the $1/r^2$ fall-off. 🔉⇢

It is important to be clear about the status of the approximation. Equation for $V$ above is only approximately true for distances large compared with the size $2a$ of the dipole, so that higher-order terms in $a/r$ can be neglected. For an idealised point dipole, however, in which $a\to 0$ and $q\to\infty$ with $p=2qa$ held finite, the expression becomes exact at all $r$. Most exam problems either state $r\gg a$ or treat the object as a point dipole, so this formula is the working tool. 🔉⇢

Two special directions deserve attention. On the axis of the dipole ($\theta=0$ for the side of $q$, $\theta=\pi$ for the side of $-q$), $\cos\theta=\pm 1$, so the potential is $V=\pm\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$, with the positive sign towards the $+q$ end and the negative sign towards the $-q$ end. In the equatorial plane, where $\theta=\pi/2$ and $\cos\theta=0$, the potential is exactly zero. This is easy to see physically: every point on the equatorial plane is equidistant from $+q$ and $-q$, so their contributions cancel exactly. 🔉⇢

This brings us to the important contrasting features of the electric potential of a dipole compared with that of a single charge, which are clear from the two formulae. First, the potential due to a dipole depends not just on $r$ but also on the angle $\theta$ between the position vector $\vec{r}$ and the dipole moment $\vec{p}$; the single-charge potential depends only on $r$. The dipole potential is, however, axially symmetric about $\vec{p}$: rotating $\vec{r}$ about the axis while keeping $\theta$ fixed traces a cone of points all at the same potential. 🔉⇢

The second contrasting feature is the rate of fall-off with distance. The dipole potential falls off as $1/r^2$, whereas the potential of a single point charge falls off more slowly, as $1/r$. This makes physical sense: from far away the equal and opposite charges of the dipole nearly cancel, so their net influence is weaker and diminishes faster than that of a lone charge whose full magnitude is always felt. The same near-cancellation is what produced the $1/r$ term dropping out in the derivation. 🔉⇢

It is instructive to line up the hierarchy of fall-offs. A single point charge has potential $\propto 1/r$ and field $\propto 1/r^2$. A dipole has potential $\propto 1/r^2$ and field $\propto 1/r^3$. In each case the field falls off one power of $r$ faster than the potential, which is consistent with the field being the spatial rate of change of the potential. Remembering this ladder, single charge then dipole, each a power steeper, is an efficient way to keep the results straight in problems. 🔉⇢

The angular factor $\cos\theta$ also tells us about the sign of the potential around a dipole. In the hemisphere on the side of the positive charge ($0\le\theta\lt\pi/2$) the potential is positive; in the hemisphere on the side of the negative charge ($\pi/2\lt\theta\le\pi$) it is negative; and on the equatorial plane it is exactly zero, forming a surface that separates the two signs. This pattern is a direct visual signature of the dipole and shows up clearly when equipotential surfaces of a dipole are sketched. 🔉⇢

A subtle point that examiners like to probe is why we could not simply take the potential of a dipole to be zero because its total charge is zero. Total charge governs the very-large-distance monopole term, which here vanishes, but the next term, the dipole term, need not vanish and is precisely what survives. The dipole is the leading nonzero contribution when the net charge is zero, and the systematic expansion in powers of $a/r$ makes this precise: the monopole ($1/r$) piece cancels, and the dipole ($1/r^2$) piece is what remains. 🔉⇢

The derivation also illustrates a technique used throughout physics: the multipole expansion. By writing the exact potential as a sum of potentials of the individual charges and then expanding in the small parameter $a/r$, we organise the answer by powers of $1/r$. The lowest surviving power identifies the character of the source as seen from far away, monopole, dipole, quadrupole, and so on. For the dipole the monopole term is absent and the dipole term dominates, which is exactly the content of our result. 🔉⇢

Because the potential is a scalar and adds algebraically, computing the dipole potential at a general point is far easier than computing the dipole field, which requires vector addition of the two Coulomb fields. Many problems that look forbidding as field problems become straightforward once translated into potential: add the two scalar contributions, expand for $r\gg a$, and read off $\dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$. If the field is subsequently needed, one differentiates the potential, which is generally simpler than adding vectors from scratch. 🔉⇢

It is also worth relating the dipole potential to the potential energy of a charge placed near the dipole. A charge $q_0$ at a point where the dipole produces potential $V$ has potential energy $q_0 V$, so the energy landscape around a dipole mirrors the angular, $1/r^2$ pattern of the potential. This connects the present section to the later discussion of the energy of a dipole in a field and shows again how the single scalar $V$ organises a wide range of results. 🔉⇢

The dipole is not merely a textbook abstraction; it is the dominant electrical model of neutral matter. Many molecules, such as water ($\text{H}_2\text{O}$) and hydrogen chloride ($\text{HCl}$), have their centres of positive and negative charge permanently separated, giving them a built-in dipole moment even with zero net charge. From a distance such a molecule looks electrically like a point dipole, and its influence on a test charge, or on another molecule, is governed by exactly the $\dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$ potential derived here. This is why the dipole potential underlies the behaviour of dielectrics, intermolecular forces, and the response of matter to applied fields, topics that recur later in the chapter. 🔉⇢

It is illuminating to see quantitatively how much weaker the dipole potential is than that of a bare charge. On the axis at distance $r$, the dipole gives $V_{\text{axis}}=\dfrac{p}{4\pi\varepsilon_0 r^2}$, whereas a single charge $q$ gives $V=\dfrac{q}{4\pi\varepsilon_0 r}$. Their ratio is $\dfrac{V_{\text{axis}}}{V}=\dfrac{p}{qr}=\dfrac{2a}{r}$, using $p=2qa$. Since $r\gg a$ by assumption, this ratio is small, confirming that the dipole potential is a small residual left over after the near-cancellation of the two charges. Doubling the distance quarters the dipole potential (because of the $1/r^2$) but only halves the single-charge potential, again showing the faster decay. 🔉⇢

The angular structure repays a closer look through the two hemispheres. Moving a test point around the dipole at fixed $r$, the potential varies as $\cos\theta$: it is largest and positive at $\theta=0$ (the $+q$ end), falls smoothly to zero at $\theta=\pi/2$ (the equatorial plane), then becomes negative and reaches its most negative value at $\theta=\pi$ (the $-q$ end). The equatorial plane is thus a natural zero-potential surface for the dipole, separating the positive-potential region from the negative-potential region. This smooth angular variation is the reason the equipotential surfaces of a dipole are lopsided lobes rather than the concentric spheres of a single charge. 🔉⇢

Once the potential is known, the dipole field can be recovered by differentiation, which is generally easier than adding the two Coulomb field vectors directly. Because the field is the negative gradient of the potential, and $V\propto 1/r^2$ for the dipole, differentiating brings down an extra power of $1/r$, so the field scales as $1/r^3$, exactly the fall-off quoted earlier from the previous chapter. This consistency, potential $\propto 1/r^2$ giving field $\propto 1/r^3$, is a good check and reinforces the general principle that the field always falls off one power of $r$ faster than the potential that generates it. 🔉⇢

To summarise, the potential of a dipole is obtained by superposing the potentials of its two charges and expanding for $r\gg a$, giving $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$. Unlike the potential of a single charge, it depends on direction through $\cos\theta$, is maximal along the axis and zero on the equatorial plane, and decays as $1/r^2$. These features, direction dependence and faster fall-off, are the fingerprints of the dipole and follow directly from the cancellation of the equal and opposite charges' leading contributions. 🔉⇢

A final practical remark concerns evaluating the formula in problems. Always identify $\theta$ carefully as the angle measured from the direction of $\vec{p}$ (from $-q$ to $+q$), not from some arbitrary reference. A sign error in $\cos\theta$ flips the sign of the potential and is one of the commonest slips. When only the magnitude at an axial or equatorial point is required, the shortcuts $V=\dfrac{p}{4\pi\varepsilon_0 r^2}$ on the axis and $V=0$ on the equator save considerable time and reduce the chance of error. 🔉⇢

Derivation 🔉⇢

  1. Write the exact potential at $P$ as the superposition of the two point-charge potentials: $V=\dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$, with $r_1,r_2$ the distances of $P$ from $+q$ and $-q$.
  2. Use the geometry of the two triangles: $r_1^2=r^2+a^2-2ar\cos\theta$ and $r_2^2=r^2+a^2+2ar\cos\theta$, where $\theta$ is the angle between $\vec{r}$ and the dipole axis and $2a$ is the charge separation.
  3. Specialise to $r\gg a$ and keep terms to first order in $a/r$: $r_1^2\approx r^2\left(1-\dfrac{2a\cos\theta}{r}\right)$ and $r_2^2\approx r^2\left(1+\dfrac{2a\cos\theta}{r}\right)$.
  4. Take inverse square roots and expand by the binomial theorem to first order: $\dfrac{1}{r_1}\approx\dfrac{1}{r}\left(1+\dfrac{a}{r}\cos\theta\right)$ and $\dfrac{1}{r_2}\approx\dfrac{1}{r}\left(1-\dfrac{a}{r}\cos\theta\right)$.
  5. Subtract: $\dfrac{1}{r_1}-\dfrac{1}{r_2}\approx\dfrac{2a\cos\theta}{r^2}$, so the $1/r$ monopole terms cancel and only the first-order term survives.
  6. Substitute back and use $p=2qa$: $V\approx\dfrac{q}{4\pi\varepsilon_0}\dfrac{2a\cos\theta}{r^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$.
  7. Write it invariantly using $p\cos\theta=\vec{p}\cdot\hat{r}$: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$ (exact for a point dipole, approximate for $r\gg a$); on the axis $V=\pm\dfrac{p}{4\pi\varepsilon_0 r^2}$ and on the equatorial plane $V=0$.
⚠️ JEE trap: A frequent misconception is that because the dipole's total charge is zero, its potential must be zero everywhere; in fact the monopole ($1/r$) term vanishes but the dipole ($1/r^2$) term survives and is nonzero except on the equatorial plane. Another error is to reuse the single-charge fall-off, treating the dipole potential as $1/r$; the correct dependence is $1/r^2$ for the potential (and $1/r^3$ for the field), because the equal and opposite charges nearly cancel at large distance. Finally, students often forget the angular factor: the dipole potential depends on $\cos\theta$, being maximal on the axis and exactly zero on the equatorial plane, so it is wrong to quote a single distance-only formula for all directions. 🔉⇢

Equipotential Surfaces and the Field–Potential Relation 🔉⇢

🎯 Carry a charge right round one equipotential and the work is exactly zero. Cross to the next ring and it is not — and the field is precisely how steeply the potential falls: E = −dV/dr. Field lines cut equipotentials at 90°, always.
🔉⇢
work along one ring = q·ΔV = 0.00 J  ·  V on this ring = — kV
crossing to the next ring costs = — mJ  ·  local field E = −dV/dr = — kV/m — not zero
What you are looking at
  • Dashed purple circles — equipotentials, labelled with their own voltage.
  • Brown arrows — field lines, which meet every ring at exactly 90° and point from high to low potential.
  • The red arc — the path already travelled by the teal charge, with its running work total.
What to do
  1. Let it go all the way round. The work counter never leaves 0.00 J.
  2. Increase the charge being carried. Still zero — the work does not depend on q, because ΔV is zero.
  3. Move to a different ring and read the cost of CROSSING, and the local field E = −dV/dr the bar reports beside it.
What it means — the force is perpendicular to the motion all the way round, so it does no work; that is why a bird on one power line is safe. Crossing rings costs energy, and the field is exactly minus the slope of the potential: where the rings crowd together, E is large. Field lines and equipotentials can never meet at any angle other than a right one.
Definition: An equipotential surface is a surface on which the potential has the same constant value at every point; the electric field is everywhere perpendicular to such surfaces and points in the direction of steepest decrease of potential, with magnitude equal to the potential gradient, $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ normal to the surface. 🔉⇢

Having built the scalar potential $V$ as a map of a charge configuration, we now ask how to visualise that map and how to recover the electric field from it. The key geometric object is the equipotential surface. An equipotential surface is a surface with a constant value of potential at all points on the surface. Just as a contour line on a topographic map joins points of equal height, an equipotential surface joins all the points in space that share one value of $V$. Assigning a family of such surfaces to a field gives an alternative, and often more intuitive, picture than the field lines alone. 🔉⇢

Full derivation, worked example and interactive 3D on the Equipotential Surfaces and the Field–Potential Relation tab →

Potential Energy of a System of Charges 🔉⇢

🎯 The energy stored in a set of charges is the work to assemble it from infinity — a sum over DISTINCT PAIRS, U = k q₁q₂/r₁₂. Three charges make three pairs; slide them and watch each pair, and the total, move.
🔉⇢
U₁₂ + U₁₃ + U₂₃ = — mJ (total)
U₁₂ = — · U₁₃ = — · U₂₃ = — mJ — each is k·qᵢqⱼ/rᵢⱼ; like pairs are positive, unlike pairs negative
What you are looking at
  • Three charges — q₁ fixed at +3 µC, q₂ at −3 µC, and q₃ you control, joined by the three pair-separations r₁₂, r₁₃, r₂₃.
  • Right — a bar for each pair energy and one for the total. Bars above the line are positive (like pairs), bars below are negative (unlike pairs).
What to do
  1. Slide q₃ from positive to negative. U₁₃ and U₂₃ swap sign together, and the total bar jumps across the line.
  2. Shrink the base separation d. Every pair energy grows in magnitude because each r shrinks.
  3. Lift q₃ far up. Its two pair energies fade toward zero — from far away it barely interacts.
What it means — the energy is a sum over DISTINCT pairs, never counted twice, and it equals the work to bring the charges in from infinity one at a time. A negative total means the configuration is bound: you would have to supply energy to pull it apart. The commonest slip is to forget a pair or to double-count one.
Definition: The electrostatic potential energy of a system of charges is the total work done by an external agency in assembling the charges at their locations by bringing them one at a time from infinity; for two charges it is $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, and it is a property of the final configuration alone, independent of the order of assembly. 🔉⇢

Having a scalar potential in hand lets us assign an energy to an entire arrangement of charges, not just to one charge in someone else's field. The idea is to ask how much work an external agency must do to assemble a given configuration by bringing the charges, one by one, from infinity to their final positions. Because the electrostatic force is conservative, this assembly work is stored as potential energy of the system, and, crucially, it depends only on the final configuration, not on the route or order by which we built it. This section makes that idea precise, first for two charges and then for many. 🔉⇢

Consider first the simple case of two charges $q_1$ and $q_2$ with position vectors $\vec{r}_1$ and $\vec{r}_2$ relative to some origin, and calculate the work done externally in building up this configuration. We imagine both charges initially at infinity and bring them in turn to their given locations. Suppose we bring $q_1$ first from infinity to $\vec{r}_1$. There is no external field present at this stage, so there is no force to work against, and the work done in placing $q_1$ is zero. This first charge does, however, set up a potential in the surrounding space. 🔉⇢

The charge $q_1$ produces at any point $P$ the potential $V_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}}$, where $r_{1P}$ is the distance of $P$ from $q_1$. Now bring the second charge $q_2$ from infinity to its final location $\vec{r}_2$. By the definition of potential, the work done in bringing $q_2$ is $q_2$ times the potential at $\vec{r}_2$ due to $q_1$: work $=q_2 V_1(\vec{r}_2)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, where $r_{12}$ is the distance between the two charges. 🔉⇢

Since the electrostatic force is conservative, this work is stored as the potential energy of the system. Hence the potential energy of a system of two charges is $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$. If we had chosen instead to bring $q_2$ first and $q_1$ afterwards, we would have obtained exactly the same result, because the roles of the charges are symmetric in the formula. This symmetry is the first hint of a general and important truth: the assembly energy does not depend on the order in which the charges are brought in. 🔉⇢

The sign of $U$ carries physical meaning. Equation $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$ holds for any signs of $q_1$ and $q_2$. If $q_1 q_2\gt 0$ (like charges), the potential energy is positive: like charges repel, so a positive amount of external work is needed to push them together from infinity to a finite separation, and that positive work is stored. If $q_1 q_2\lt 0$ (unlike charges), the electrostatic force is attractive; a positive amount of work would be needed to pull the charges apart to infinity, which means the work to bring them together from infinity is negative, so the potential energy is negative. 🔉⇢

This sign convention is worth dwelling on because it recurs throughout physics. A negative potential energy signals a bound configuration: energy must be supplied to separate the charges to infinity (where $U=0$ by convention). A positive potential energy signals a configuration that would fly apart if released, having stored energy that can be liberated. The magnitude $\dfrac{|q_1 q_2|}{4\pi\varepsilon_0 r_{12}}$ measures how strongly the pair is bound or how much energy is stored, growing as the charges are brought closer. 🔉⇢

The two-charge result generalises smoothly to more charges by continuing the assembly, one charge at a time, and adding the work done at each step. Consider three charges $q_1$, $q_2$ and $q_3$ at $\vec{r}_1$, $\vec{r}_2$ and $\vec{r}_3$. Bringing $q_1$ from infinity requires no work. Bringing $q_2$ next requires work $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, exactly as in the two-charge case, since only $q_1$ is present to be worked against. 🔉⇢

Once $q_1$ and $q_2$ are in place they jointly produce a potential $V_{1,2}=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_{1P}}+\dfrac{q_2}{r_{2P}}\right)$ at any point $P$. The work done in bringing $q_3$ from infinity to $\vec{r}_3$ is therefore $q_3$ times this potential evaluated at $\vec{r}_3$: $q_3 V_{1,2}(\vec{r}_3)=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_3}{r_{13}}+\dfrac{q_2 q_3}{r_{23}}\right)$. This is the third step of the assembly. 🔉⇢

Adding the work done in the two nonzero steps gives the total potential energy of the three-charge system: $U=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_2}{r_{12}}+\dfrac{q_1 q_3}{r_{13}}+\dfrac{q_2 q_3}{r_{23}}\right)$. The structure of the answer is transparent: it is the sum over all distinct pairs of charges of the pairwise energy $\dfrac{q_i q_j}{4\pi\varepsilon_0 r_{ij}}$. For $n$ charges the total energy is the sum over all $\tfrac{1}{2}n(n-1)$ distinct pairs, each pair counted exactly once. 🔉⇢

Again the conservative, path-independent nature of the electrostatic force guarantees that this final expression for $U$ is independent of the manner in which the configuration is assembled. Whatever order we choose for bringing the charges in, and whatever paths we use, the total work, and hence the stored energy, is the same. The potential energy is characteristic of the present state of the configuration, and not the way the state is achieved. This is precisely the property that makes 'the energy of a configuration' a well-defined quantity. 🔉⇢

A classic application is four charges placed at the corners of a square. To find the work required to assemble them, one brings the charges in one at a time and sums the pairwise contributions: the first charge costs nothing, the second interacts with the first, the third with the first two, and the fourth with the first three. The total, when summed, gives the electrostatic energy of the arrangement, and one may verify by taking the charges in any other order that the same energy results. This kind of pair-by-pair bookkeeping is the standard method for any discrete configuration. 🔉⇢

A neat feature of such symmetric arrangements is that the potential at special points may vanish. For a square with charges $+q,-q,+q,-q$ on its corners, the potential at the centre is zero, because the contributions of the two positive charges are exactly cancelled by those of the two negative charges (all four are equidistant from the centre). Consequently, no work is required to bring an additional charge from infinity to the centre. This illustrates the earlier lesson that a point can have zero potential even though the field there need not be zero. 🔉⇢

We should carefully distinguish the energy of assembling a set of charges from a related but distinct question: the potential energy of a charge in an external field. In the assembly problem, the charges themselves are the sources of the field they interact with. In the external-field problem, the field is produced by sources external to the charge or charges whose energy we want; those external sources may be unknown, and what is specified is the external potential $V$ or field $\vec{E}$. We assume the charge does not significantly disturb these external sources. 🔉⇢

For a single charge $q$ placed at a point with position vector $\vec{r}$ in an external field, the work done in bringing it from infinity is $qV(\vec{r})$, where $V(\vec{r})$ is the external potential at that point. This work is stored as the potential energy of the charge in the external field, so the potential energy of $q$ at $\vec{r}$ is $U=qV(\vec{r})$. This simple product is the workhorse for problems where a known potential is applied and we ask how much energy a charge acquires by sitting in it. 🔉⇢

This is the origin of a convenient unit of energy. If an electron of charge $e=1.6\times 10^{-19}\ \text{C}$ is accelerated through a potential difference of $1\ \text{volt}$, it gains energy $e\,\Delta V=1.6\times 10^{-19}\ \text{J}$, and this amount of energy is defined as one electron volt, $1\ \text{eV}=1.6\times 10^{-19}\ \text{J}$. Its multiples ($1\ \text{keV}=10^{3}\ \text{eV}$, $1\ \text{MeV}=10^{6}\ \text{eV}$, $1\ \text{GeV}=10^{9}\ \text{eV}$) are the natural energy units of atomic, nuclear and particle physics. 🔉⇢

For a system of two charges $q_1$ and $q_2$ placed in an external field, we combine the two ideas. Bringing $q_1$ to $\vec{r}_1$ costs $q_1 V(\vec{r}_1)$ against the external field. Bringing $q_2$ to $\vec{r}_2$ costs work against both the external field and the field of $q_1$, namely $q_2 V(\vec{r}_2)+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$. Hence the total potential energy of the system is $U=q_1 V(\vec{r}_1)+q_2 V(\vec{r}_2)+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$: two terms for the interaction with the external field plus one for the mutual interaction of the charges. 🔉⇢

It is essential to keep the two kinds of term separate in problems. The interaction terms $q_i V(\vec{r}_i)$ describe how the charges couple to the externally imposed potential and depend on where the charges sit in that potential; the mutual term $\dfrac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}$ describes the charges' interaction with each other and depends only on their separation. If the external field is switched off, only the mutual term survives, returning us to the assembly energy of the previous paragraphs. 🔉⇢

A worked configuration makes the bookkeeping concrete: two charges of $7\ \mu\text{C}$ and $-2\ \mu\text{C}$ placed $18\ \text{cm}$ apart have mutual energy $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r}=(9\times 10^{9})\dfrac{(7\times 10^{-6})(-2\times 10^{-6})}{0.18}=-0.7\ \text{J}$. The negative sign shows the pair is bound (they attract), and the work required to separate them infinitely far is $W=0-U=+0.7\ \text{J}$. Placing the same pair in an additional external field would add the two $q_iV(\vec{r}_i)$ terms without altering this mutual $-0.7\ \text{J}$. 🔉⇢

The lesson that potential energy is path-independent has a practical payoff: we may always compute the energy of a configuration by the most convenient assembly sequence, confident that the answer is the same. For a complicated arrangement, choose an order that makes each successive potential easy to evaluate; the pairwise-sum formula guarantees the total is order-independent. This freedom, a direct consequence of the conservative nature of the Coulomb force, is what makes electrostatic energy calculations tractable. 🔉⇢

The four-charge square is worth carrying through to a numerical answer, because it shows the pairwise sum in action. Take charges $+q,-q,+q,-q$ at the corners $A,B,C,D$ of a square of side $d$. There are six pairs: four sides of length $d$ (each an unlike pair, contributing $-\dfrac{q^2}{4\pi\varepsilon_0 d}$) and two diagonals of length $d\sqrt{2}$ (each a like pair, contributing $+\dfrac{q^2}{4\pi\varepsilon_0 d\sqrt{2}}$). Summing gives the total electrostatic energy $U=\dfrac{q^2}{4\pi\varepsilon_0 d}\left(-4+\dfrac{2}{\sqrt{2}}\right)=-\dfrac{q^2}{4\pi\varepsilon_0 d}\left(4-\sqrt{2}\right)$. The negative result reflects that the attractive side-pairs outweigh the repulsive diagonal-pairs, and one may verify the identical value results whatever order the charges are assembled in. 🔉⇢

The connection to a dipole in an external field is a natural extension of the two-charge external-field formula. A dipole is just the special case $q_1=+q$, $q_2=-q$ separated by $2a$; substituting into $U=q_1 V(\vec{r}_1)+q_2 V(\vec{r}_2)+\dfrac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}$ and using the fact that the potential difference between the two ends equals the work done by the uniform field over the separation, the interaction part reduces to $U=-\vec{p}\cdot\vec{E}=-pE\cos\theta$ (dropping the constant self-energy term). This is the potential energy of a dipole in a uniform field, minimal when $\vec{p}$ is aligned with $\vec{E}$ and maximal when anti-aligned, and it follows directly from the system-energy bookkeeping developed here. 🔉⇢

Finally, energy conservation ties these ideas to motion. If the external constraints holding a charge configuration are released, the stored electrostatic potential energy is converted into kinetic energy as the charges move, with the total energy conserved because the Coulomb force is conservative. A configuration with positive $U$ (like charges pushed together) will fly apart, converting that positive energy into kinetic energy; a configuration with negative $U$ (unlike charges) is bound, and energy must be supplied from outside to disperse it. This is exactly the energy accounting behind particle accelerators, where charges are pushed through potential differences, and behind the release of energy when bound charge configurations rearrange. 🔉⇢

To summarise, the potential energy of a system of charges is the external work needed to assemble it from infinity, and for point charges it is the sum over all distinct pairs of $\dfrac{q_i q_j}{4\pi\varepsilon_0 r_{ij}}$; for two charges this is simply $\dfrac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}$. The sign tells us whether the configuration is bound (negative) or stores energy that would drive it apart (positive). When an external field is also present, add the terms $q_i V(\vec{r}_i)$ for each charge's interaction with that field, keeping them distinct from the mutual interaction terms. Because the force is conservative, the total energy depends only on the final configuration, not on how it was built. 🔉⇢

Derivation 🔉⇢

  1. Assemble two charges from infinity: bring $q_1$ to $\vec{r}_1$ first; with no other charge present there is no field to work against, so the work in this step is zero.
  2. The charge $q_1$ now produces a potential $V_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}}$ at any point $P$ a distance $r_{1P}$ away.
  3. Bring $q_2$ from infinity to $\vec{r}_2$: by the definition of potential the work equals $q_2 V_1(\vec{r}_2)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, which (the force being conservative) is stored as the system's potential energy $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$.
  4. Extend to three charges: bringing $q_3$ to $\vec{r}_3$ against the potential $V_{1,2}$ of the first two costs $q_3 V_{1,2}(\vec{r}_3)=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_3}{r_{13}}+\dfrac{q_2 q_3}{r_{23}}\right)$.
  5. Add the nonzero steps to get $U=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_2}{r_{12}}+\dfrac{q_1 q_3}{r_{13}}+\dfrac{q_2 q_3}{r_{23}}\right)$, i.e. the sum over all distinct pairs of $\dfrac{q_i q_j}{4\pi\varepsilon_0 r_{ij}}$.
  6. Note path-independence: because the electrostatic force is conservative, $U$ is independent of the order and paths of assembly and is characteristic of the final configuration alone.
  7. Include an external field: a single charge has $U=qV(\vec{r})$ in an external potential $V$, and for two charges $U=q_1 V(\vec{r}_1)+q_2 V(\vec{r}_2)+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, separating the external-interaction terms from the mutual term.
⚠️ JEE trap: A common error is to double-count the pairwise energies, for example writing the two-charge energy as $\dfrac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}+\dfrac{q_2 q_1}{4\pi\varepsilon_0 r_{12}}$; each distinct pair must be counted exactly once. A second error concerns the sign of $U$: students often assume potential energy is always positive, but for unlike charges $q_1 q_2\lt 0$ makes $U$ negative, signalling a bound system from which energy must be supplied to separate the charges to infinity. A third error is to confuse the assembly energy of a set of charges (where the charges are their own sources) with the energy $qV$ of a charge in an external field (where the sources are external); when both are present the total energy is $q_1 V(\vec{r}_1)+q_2 V(\vec{r}_2)+\dfrac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}$, and the two kinds of term must be kept separate. 🔉⇢

Electrostatics of Conductors 🔉⇢

🎯 Drop a conductor into a field and its free charges rush to the surfaces until the field INSIDE is exactly zero. The induced surface charge is σ = ε₀E, field lines meet the surface at 90°, and a hollow cavity is completely shielded.
−−−−−−++++++
🔉⇢
E inside a conductor = 0 (always)  ·  σ = ε₀·E₀ = — nC/m²
field just outside = σ/ε₀ = E₀ = — kV/m, and it meets the surface at 90°
What you are looking at
  • Brown arrows — the external field E₀, pushing in from the left and leaving on the right. Notice there are NO arrows inside the metal.
  • The − on the left face and + on the right — induced surface charge, denser when E₀ is stronger, arranged so their own field cancels E₀ inside.
  • The dashed box — a hollow cavity, which stays field-free (switch the body control).
What to do
  1. Raise E₀. More surface charge appears (σ = ε₀E₀) and the arrows brighten, but the inside stays blank.
  2. Switch to the cavity body. The inside of the hollow is still E = 0 — that is electrostatic shielding, the principle behind a Faraday cage.
  3. Look at where the arrows meet the metal: they arrive at 90°, because any sideways component would drive a surface current and cannot persist.
What it means — in a conductor the free charges keep moving until they have killed the field everywhere inside; that is what "conductor" means in electrostatics. The whole body sits at one potential, the surface is an equipotential, and all the leftover charge lives on the outside surface. A car is a safe place in a lightning strike for exactly this reason.
Definition: In the static situation a conductor arranges its free charges so that the electric field is zero everywhere inside it, the field just outside is everywhere normal to the surface with magnitude $E=\dfrac{\sigma}{\varepsilon_0}$, any excess charge resides only on the surface, and the whole conductor is a single equipotential volume. 🔉⇢

Conductors and insulators differ in one decisive respect: a conductor contains mobile charge carriers, while an insulator does not. In a metallic conductor the carriers are electrons. The outer (valence) electrons part away from their atoms and are free to move throughout the metal, though they are not free to leave it. These free electrons behave like a kind of gas: they collide with one another and with the ions and move randomly in all directions, and when an external electric field is applied they drift, on average, against the direction of the field. The positive ions, made up of the nuclei and the tightly bound inner electrons, remain locked in their fixed lattice positions. In electrolytic conductors the carriers are positive and negative ions and the situation is more involved, so throughout this card we restrict attention to metallic solid conductors, for which a small number of clean results govern every electrostatic problem. 🔉⇢

The first and most fundamental result is that in the static situation the electrostatic field is zero everywhere inside the conductor. This can be taken as the very defining property of a conductor in electrostatics. The argument is short and physical: a conductor is full of free charge carriers, and as long as the field inside is not zero these carriers feel a force and keep drifting. They cannot go on drifting forever in a static situation — 'static' means precisely that no current flows. So the free charges rearrange themselves, piling up here and thinning out there, until the field they produce exactly cancels any applied field at every interior point. Only when the interior field has fallen to zero does the drifting stop and equilibrium set in. Hence, inside a conductor in the electrostatic state, the field vanishes. 🔉⇢

The second result concerns the surface. At the surface of a charged conductor the electrostatic field must be normal (perpendicular) to the surface at every point. Suppose it were not — suppose the field at the surface had some component lying along the surface. Then the free charges sitting on the surface would feel a tangential force and would move along the surface. But motion of charge is a current, and in the static situation there is no current. Therefore the tangential component of the field at the surface must be zero, and whatever field exists at the surface points straight out of (or into) it. For a conductor carrying no surface charge, the field is zero even at the surface. 🔉⇢

The third result is that the interior of a conductor can carry no excess charge in the static situation; any net charge given to the conductor must reside entirely on its outer surface. This follows directly from Gauss's law together with the first result. Imagine any small closed surface $S$ drawn entirely inside the conducting material, bounding some volume element. Everywhere on $S$ the field is zero, because $S$ lies inside the conductor. Therefore the total electric flux through $S$ is zero, and by Gauss's law the net charge enclosed by $S$ must be zero. Since $S$ can be shrunk to enclose any point we please, there is no net charge at any interior point. Consequently the excess charge can reside only on the surface in the static situation — a beautifully general conclusion that needs no special symmetry. 🔉⇢

The fourth result ties the conductor to the language of potential: the electrostatic potential is constant throughout the volume of the conductor and takes that same constant value on its surface. In other words, a conductor in equilibrium is a single equipotential body. This follows from the first two results. Because the field is zero inside and has no tangential component on the surface, no work is done in carrying a small test charge from any interior or surface point to any other such point — the line integral of the field along any path within the conductor vanishes. If no work is done in moving between two points, those two points are at the same potential. Hence every point inside the conductor and on its surface sits at one common potential. 🔉⇢

Note carefully what this does and does not say. It says there is no potential difference between points of the conductor; it does not say the potential just outside the surface equals the potential of the surface. If the conductor is charged, there is a nonzero normal field just outside the surface, so as you step off the surface into the surrounding space the potential begins to change. The surface is at one potential; a point a little outside is at a different one. In a system of several conductors of arbitrary size, shape and charge, each conductor is characterised by its own constant value of potential, but this constant may differ from one conductor to another — a fact that underlies the whole idea of capacitance developed in later sections. 🔉⇢

The fifth result gives the magnitude of the field just outside the surface of a charged conductor: $E=\dfrac{\sigma}{\varepsilon_0}\,\hat{n}$, where $\sigma$ is the local surface charge density and $\hat{n}$ is the outward unit normal. This is derived with a Gaussian pill box — a very short cylinder of cross-sectional area $\delta S$ and negligible height, placed so that it straddles the surface, partly inside the conductor and partly outside. Just inside the surface the field is zero, so no flux threads the inner face; along the negligibly short curved side there is essentially no area; only the outer flat face contributes, giving a flux $E\,\delta S$. The charge enclosed is $\sigma\,\delta S$. Gauss's law $E\,\delta S=\dfrac{\sigma\,\delta S}{\varepsilon_0}$ then yields $E=\dfrac{\sigma}{\varepsilon_0}$. 🔉⇢

It is worth pausing on a favourite examination contrast hidden here. An isolated infinite sheet of charge produces a field $\dfrac{\sigma}{2\varepsilon_0}$ on each side, whereas the field just outside a charged conductor is twice as large, $\dfrac{\sigma}{\varepsilon_0}$. There is no contradiction. At the conductor's surface the field from the local patch of charge ($\dfrac{\sigma}{2\varepsilon_0}$) and the field from all the rest of the charge on the conductor ($\dfrac{\sigma}{2\varepsilon_0}$) add outside the conductor to give $\dfrac{\sigma}{\varepsilon_0}$, while inside they cancel to give zero — exactly as the first result demands. The factor of two is not an error; it is the signature of a conductor as opposed to a lone sheet. 🔉⇢

The sixth result is electrostatic shielding. Consider a conductor with an empty cavity inside it, with no charges placed in the cavity. A remarkable theorem says that the electric field inside such a cavity is zero, whatever the size and shape of the cavity, whatever the charge on the conductor, and whatever external field the conductor is immersed in. We have already met a special case: the field inside a charged spherical shell is zero. The general result does not need spherical symmetry. A related statement is that whenever a conductor with a cavity is charged, or has charges induced on it by an outside field, all of that charge resides on the outer surface; none appears on the wall of the empty cavity. 🔉⇢

This shielding is the physics of the Faraday cage. Because the cavity is field-free no matter how strong the outside fields, a hollow conductor can protect sensitive electronic instruments from external electrical disturbances — the interior of a metal enclosure is a quiet, field-free haven. The same idea explains why passengers inside a metal car are relatively safe from a lightning strike: the charge stays on the outer shell and the field inside remains essentially zero. Shielding is one of the most practically important consequences of the simple statement that the field inside a conductor vanishes. 🔉⇢

These abstract results explain many everyday observations, several of which NCERT poses as exercises. A comb run through dry hair becomes charged by friction; it then polarises the molecules of a small piece of paper (an insulator), pulling the induced negative charges slightly closer than the positive ones, so the paper is attracted even though it conducts no current. On a wet or rainy day the friction between hair and comb is reduced, the comb does not charge appreciably, and the attraction disappears. The lesson is that induced polarisation, not free conduction, produces the pull. 🔉⇢

Two safety examples follow the same logic. The tyres of aircraft are made slightly conducting so that static charge, built up by friction during flight and landing, can leak harmlessly to the ground rather than accumulate to a spark. Vehicles carrying inflammable liquids drag a metallic rope or chain that touches the road, providing a conducting path so that frictional charge bleeds continuously into the earth instead of building to a dangerous discharge near flammable vapour. In each case the point is to give excess charge somewhere to go, exploiting the fact that a conductor carries its charge on the outer surface and shares potential with whatever it is connected to. 🔉⇢

The classic bird-on-a-wire puzzle is a pure consequence of the equipotential result. A bird perched on a single bare high-tension line is unharmed because both its feet touch the same conductor; there is no potential difference across the bird and hence no current through it. A person standing on the ground who touches the same line completes a path between the high-potential line and the low-potential earth; the large potential difference drives a current through the body and delivers a fatal shock. Current flows only when there is a difference of potential, and the conductor's own body offers none between two points of itself. 🔉⇢

When a neutral conductor is placed in an external field, the free electrons drift until the induced surface charges create an internal field that exactly opposes the applied field inside the metal, restoring zero interior field. Negative charge accumulates on the face toward which the field points and positive charge on the opposite face; the conductor as a whole stays neutral but becomes polarised, and field lines terminate perpendicularly on its induced surface charges. This induced redistribution — happening almost instantaneously — is the microscopic mechanism behind all the results above, and it is the point of contrast with a dielectric, whose bound charges can only partly, never wholly, cancel the applied field. 🔉⇢

It is illuminating to picture the field lines around a charged conductor. Because the surface is an equipotential and the field must be normal to it, every field line leaves the surface at right angles, exactly as field lines meet an equipotential surface perpendicularly. Where the surface is sharply curved (a spike or point) the surface charge density $\sigma$ is large, so by $E=\dfrac{\sigma}{\varepsilon_0}$ the field just outside is intense — this is why charge tends to concentrate and discharge from sharp points, the principle behind the lightning rod. Where the surface is nearly flat the density and field are smaller. The single relation $E=\dfrac{\sigma}{\varepsilon_0}$ thus links the geometry of a conductor to the strength of the field it produces. 🔉⇢

To summarise the electrostatics of conductors: in equilibrium the field is zero inside; the field just outside is normal to the surface with magnitude $E=\dfrac{\sigma}{\varepsilon_0}$; the excess charge can reside only on the surface in the static situation; the whole conductor (volume and surface) is one equipotential, though its potential generally differs from that of nearby space and from that of other conductors; and any empty cavity is completely shielded, with the field inside it always zero. These six statements, each provable from Gauss's law and the meaning of 'static', are the foundation on which capacitors, and the entire idea of storing charge and energy, are built. 🔉⇢

Derivation 🔉⇢

  1. Zero interior field: a conductor has free carriers, so any nonzero interior field $E$ would exert a force $qE$ and drive a current; the static condition (no current) can hold only after the free charges redistribute so that the field they create cancels the applied field at every interior point, giving $E_{inside}=0$.
  2. Normal surface field: if the surface field had a tangential component $E_\parallel$, surface charges would feel a force $qE_\parallel$ and slide along the surface, which is a current; the static condition forces $E_\parallel=0$, so the field at the surface is purely normal.
  3. No interior charge: enclose any interior point with a small Gaussian surface $S$ lying wholly inside the conductor; since $E=0$ on $S$, the flux is zero, so by Gauss's law $\oint \vec{E}\cdot d\vec{S}=\dfrac{q_{enc}}{\varepsilon_0}=0$ gives $q_{enc}=0$; shrinking $S$ shows any excess charge must lie on the outer surface.
  4. Conductor is equipotential: since $E=0$ inside and $E_\parallel=0$ on the surface, the work $W=-q\int \vec{E}\cdot d\vec{l}=0$ for any path within the conductor; zero work between points means zero potential difference, so $V$ is constant throughout the volume and over the surface.
  5. Surface field magnitude: place a pill box of face area $\delta S$ straddling the surface; the inner face sees $E=0$ and only the outer face contributes flux $E\,\delta S$, while the enclosed charge is $\sigma\,\delta S$; Gauss's law $E\,\delta S=\dfrac{\sigma\,\delta S}{\varepsilon_0}$ gives $E=\dfrac{\sigma}{\varepsilon_0}$, directed along $\hat{n}$.
  6. Reconciliation with the sheet result: the local patch contributes $\dfrac{\sigma}{2\varepsilon_0}$ and the rest of the conductor contributes another $\dfrac{\sigma}{2\varepsilon_0}$; outside these add to $\dfrac{\sigma}{\varepsilon_0}$ and inside they subtract to $0$, consistent with both step 1 and step 5.
  7. Shielding: for an empty cavity, assume a field inside; a field line would have to start and end on cavity-wall charges, but that would require net charge on the wall and a nonzero line integral around a loop closed through the field-free conductor, contradicting $\oint\vec{E}\cdot d\vec{l}=0$; hence the cavity field is zero and all charge stays on the outer surface.
⚠️ JEE trap: A common error is to imagine excess charge spread through the body of a metal, or to think the field inside a charged conductor points from its surplus positive charge inward. In electrostatics neither is true: the excess charge can reside only on the surface, and the interior field is exactly zero. A second frequent slip is to quote the field just outside a conductor as $\dfrac{\sigma}{2\varepsilon_0}$, confusing it with an isolated infinite sheet; the correct value is $E=\dfrac{\sigma}{\varepsilon_0}$, because the conductor's own charge cancels inside and reinforces outside. Finally, students sometimes claim a charged conductor's surface is at the same potential as the space just outside it; the conductor is a single equipotential, but a normal field exists just outside a charged surface, so the potential does change as soon as you step off the surface. 🔉⇢

Capacitors, Dielectrics and Capacitance 🔉⇢

🎯 Capacitance is geometry, not charge. Pile on charge and V rises in exact step so C never moves — but change d, A or the dielectric and C changes at once.
++++++++++−−−−−−−−−−
🔉⇢
C = K·ε₀·A / d = — pF  ·  V = Q / C = — V  ·  U = ½·Q·V = — nJ
slide Q and watch C sit perfectly still — the straight line is Q against V
What you are looking at
  • The two plates, with the gap d and the plate height standing for the area A.
  • Brown arrows — the field between the plates, which thins out as the dielectric constant K rises.
  • The purple line — Q against V. Its slope is the capacitance.
What to do
  1. Sweep the charge Q from 1 to 10 nC. The red dot slides along the line but the LINE does not tilt: C is unchanged.
  2. Halve d and watch C double and the line tilt.
  3. Raise K and watch the field weaken while the charge stays put — that is why V falls and C rises.
What it means — C = Q/V is a ratio fixed entirely by shape and filling. The energy ½QV carries its half because the capacitor charges progressively: the first charge crosses at zero volts and the last at the full voltage.
Definition: A capacitor is a pair of conductors carrying charges $+Q$ and $-Q$ at potential difference $V$, characterised by its capacitance $C=\dfrac{Q}{V}$, a purely geometric quantity that for parallel plates in vacuum is $C_0=\dfrac{\varepsilon_0 A}{d}$ and is raised by a factor $K$ (the dielectric constant) to $C=\dfrac{K\varepsilon_0 A}{d}$ when the gap is filled with a dielectric. 🔉⇢

A capacitor is a system of two conductors separated by an insulator. The two conductors carry charges, say $Q_1$ and $Q_2$, and sit at potentials $V_1$ and $V_2$. In practice the arrangement of interest is the symmetric one: the conductors carry equal and opposite charges $+Q$ and $-Q$, and one speaks of the potential difference $V=V_1-V_2$ between them. We call $Q$ the charge of the capacitor, even though it is really the charge on just one conductor; the total charge of the device is zero. Even a single conductor can act as a capacitor if the second conductor is imagined to be at infinity. The plates are usually charged by connecting them to the two terminals of a battery, which pumps charge from one plate to the other until the plate potentials match the terminal potentials. 🔉⇢

Full derivation, worked example and interactive 3D on the Capacitors, Dielectrics and Capacitance tab →

Combination of Capacitors and Energy Stored 🔉⇢

🎯 Same two capacitors, wired two ways. In PARALLEL the capacitances add; in SERIES the reciprocals add, so the equivalent is smaller than either plate-pair. The energy stored is ½C_eqV² — a parabola in the voltage.
🔉⇢
parallel: C_eq = C₁+C₂   series: C_eq = C₁C₂/(C₁+C₂) = — µF
Q = C_eq·V = — µC  ·  U = ½·C_eq·V² = — µJ
What you are looking at
  • The circuit on the left — a battery (red plate = +) driving two capacitors. Switch the wiring control between parallel (both across the battery) and series (one after the other).
  • Right — the stored energy U = ½C_eqV² as a function of voltage, with your working point marked.
What to do
  1. Toggle parallel and series with the same C₁, C₂. The equivalent jumps from C₁+C₂ down to a value SMALLER than either capacitor.
  2. Make C₂ much larger than C₁ in series. The equivalent hugs the SMALLER one — the small capacitor dominates a series chain.
  3. Double the voltage. The energy quadruples, because U grows as V².
What it means — in parallel every capacitor feels the full voltage and their charges add, so capacitances add. In series they carry the same charge while the voltages add, so it is the reciprocals that add and the equivalent is always less than the smallest member. Either way the energy ends up as ½C_eqV², stored in the field between the plates.
Definition: Capacitors in series share the same charge and add reciprocals, $\dfrac{1}{C}=\sum_i\dfrac{1}{C_i}$, while capacitors in parallel share the same voltage and add directly, $C=\sum_i C_i$; the energy banked in a charged capacitor is $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$, stored in the field with density $u=\dfrac{1}{2}\varepsilon_0 E^2$. 🔉⇢

Several capacitors of capacitance $C_1,C_2,\dots,C_n$ can be combined into a single network with some effective (equivalent) capacitance $C$. The value of $C$ depends entirely on how the individual capacitors are wired together, and two arrangements cover almost every problem: capacitors in series and capacitors in parallel. Mastering these two rules, and knowing how to break a mixed network into series and parallel blocks, is the core skill this section teaches. 🔉⇢

Full derivation, worked example and interactive 3D on the Combination of Capacitors and Energy Stored tab →

Equipotential Surfaces and the Field–Potential Relation 🔉⇢deep concept

Definition: An equipotential surface is a surface on which the potential has the same constant value at every point; the electric field is everywhere perpendicular to such surfaces and points in the direction of steepest decrease of potential, with magnitude equal to the potential gradient, $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ normal to the surface. 🔉⇢

🔬 Interactive 3D · Nested equipotential shells around a point charge with radial field lines piercing them at right angles, illustrating that the field is normal to the equipotentials and $|\vec{E}|=-\dfrac{\delta V}{\delta l}$.

Having built the scalar potential $V$ as a map of a charge configuration, we now ask how to visualise that map and how to recover the electric field from it. The key geometric object is the equipotential surface. An equipotential surface is a surface with a constant value of potential at all points on the surface. Just as a contour line on a topographic map joins points of equal height, an equipotential surface joins all the points in space that share one value of $V$. Assigning a family of such surfaces to a field gives an alternative, and often more intuitive, picture than the field lines alone. 🔉⇢

The simplest case is a single point charge $q$, whose potential is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$. This shows that $V$ is constant if and only if $r$ is constant. The locus of points at a fixed distance $r$ from the charge is a sphere centred on the charge, so the equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge. Surfaces of larger radius correspond to smaller potential (for positive $q$), and the spacing of equally-spaced potential values increases outward because $V$ changes more slowly far from the charge. 🔉⇢

Now recall that the electric field lines for a single charge are radial lines, starting from the charge if $q$ is positive and ending on it if $q$ is negative. Comparing the two pictures, the radial field lines are everywhere perpendicular to the concentric spherical equipotentials. This is not a coincidence of the point charge: the electric field at every point is normal to the equipotential surface passing through that point, and this holds in general for any charge configuration whatsoever. The relationship between field lines and equipotential surfaces is thus one of mutual perpendicularity. 🔉⇢

The proof that the field must be normal to an equipotential surface is short and worth internalising. Suppose, to the contrary, that at some point the field had a nonzero component lying along the surface. Then, to move a unit positive test charge a little way along the surface against that tangential component, an external agency would have to do work. But by the very definition of an equipotential surface there is no potential difference between any two points on it, so no work is required to move a test charge on the surface. These two statements contradict each other unless the tangential component is zero. Hence the field can have no component along the surface; it must be entirely normal to it at every point. 🔉⇢

This perpendicularity gives equipotential surfaces real explanatory power. Because no work is done in moving a charge along an equipotential, such surfaces are the natural 'level sets' of the electrostatic energy landscape for a unit charge. Sketching a few equipotentials immediately reveals the direction of the field everywhere (perpendicular to them) and, through their spacing, the strength of the field, as we shall make quantitative shortly. For this reason equipotential surfaces offer an alternative visual picture, in addition to the picture of electric field lines, around any charge configuration. 🔉⇢

Consider next a uniform electric field $\vec{E}$, say directed along the $x$-axis, the kind of field found between the plates of a large parallel-plate capacitor. Here the field has the same magnitude and direction everywhere, so the surfaces perpendicular to it are planes normal to the $x$-axis, that is, planes parallel to the $y$-$z$ plane. The equipotential surfaces of a uniform field are therefore a set of equally spaced parallel planes, and the equal spacing reflects the constant strength of the field. Moving along any such plane involves no change in potential; moving across them, along $x$, changes the potential steadily. 🔉⇢

More complicated configurations give more interesting surfaces, but the same rules apply. For an electric dipole, the equipotential surfaces are neither spheres nor planes; they bulge out near the positive charge and near the negative charge and are separated by the flat equatorial plane on which $V=0$. For two identical positive charges, the equipotentials far away approach spheres (the pair looks like a single charge of twice the magnitude from a distance), while close in they wrap individually around each charge. In every case the field lines cross these surfaces at right angles. 🔉⇢

We now turn from geometry to the quantitative relation between field and potential, which is the second major result of this section. Consider two closely spaced equipotential surfaces, $A$ and $B$, with potential values $V$ and $V+\delta V$, where $\delta V$ is the change in $V$ measured in the direction of the electric field. Let $P$ be a point on surface $B$, and let $\delta l$ be the perpendicular distance from $P$ to surface $A$. Because the field is normal to the surfaces, moving along this perpendicular is moving along the field direction, which is exactly the situation we can analyse with work done. 🔉⇢

Imagine moving a unit positive charge along the perpendicular from surface $B$ to surface $A$, against the electric field. The work done by the external agency in this displacement is $|\vec{E}|\,\delta l$, since the field magnitude is $|\vec{E}|$ and the displacement $\delta l$ is along the field. By the definition of potential, this external work per unit charge equals the potential difference between the surfaces, $V_A-V_B$. Writing $V_A=V$ and $V_B=V+\delta V$, we have $V_A-V_B=V-(V+\delta V)=-\delta V$. 🔉⇢

Equating the two expressions for the work gives $|\vec{E}|\,\delta l=-\delta V$, so $|\vec{E}|=-\dfrac{\delta V}{\delta l}$. Since we defined $\delta V$ as the change in potential in the direction of the field, and the potential decreases in that direction, $\delta V$ is negative; writing $\delta V=-|\delta V|$, the magnitude of the field is $|\vec{E}|=\dfrac{|\delta V|}{\delta l}$, a positive quantity as it must be. The relation $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ is the differential link between field and potential, valid normal to the equipotential surface. 🔉⇢

Two important conclusions follow from this relation, and they should be stated crisply. First, the electric field is in the direction in which the potential decreases steepest. Field lines therefore run 'downhill' on the potential landscape, from high potential to low, always crossing the equipotentials at right angles. Second, the magnitude of the field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point. In symbols, $E$ equals the rate at which $V$ falls with distance measured perpendicular to the surfaces. 🔉⇢

The relation $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ has an immediate and very useful visual corollary: where equipotential surfaces are crowded close together, the field is strong, and where they are far apart, the field is weak. This is because a fixed step in potential $\delta V$ occurs over a small distance $\delta l$ where the surfaces are dense (large $E$) and over a large distance where they are sparse (small $E$). Reading field strength from the density of equipotentials is exactly analogous to reading the steepness of terrain from the crowding of contour lines on a map. 🔉⇢

It is important to emphasise that the field is determined by how the potential changes, not by its value. A region can be at a high potential and yet have zero field if the potential is uniform there, as inside a charged conductor or a spherical shell; conversely, a region at zero potential can have a strong field if the potential is changing rapidly through zero. The derivative in $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ makes this precise: only spatial variation of $V$ produces a field. This resolves a very common source of confusion between the value of the potential and the strength of the field. 🔉⇢

The direction convention deserves a further word. Because the field points towards decreasing potential, a positive charge released in the field accelerates from high potential to low potential, losing potential energy $qV$ and gaining kinetic energy, exactly as a ball rolls downhill. A negative charge does the opposite, moving towards higher potential. The equipotential surfaces are the 'level ground' of this landscape: motion along them is free of work, motion across them costs or releases energy, and the steepest descent is the field direction. 🔉⇢

The perpendicularity of field and equipotential also explains why the surface of a conductor is an equipotential. In electrostatic equilibrium the field just outside a conductor is normal to its surface (any tangential component would drive the surface charges until it vanished), and a field everywhere normal to a surface is precisely the condition for that surface to be an equipotential. Thus every conductor's surface is an equipotential surface, a fact we will use repeatedly when we study conductors and capacitors in the following sections. 🔉⇢

Equipotential surfaces are also the right tool for thinking about a uniform field between capacitor plates. There the equipotentials are planes parallel to the plates, equally spaced, and the field $E=V/d$ (potential difference divided by plate separation) is a direct application of $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ with a finite step: the potential drops uniformly from one plate to the other across the gap $d$. This picture makes the relation between the applied voltage and the field inside a capacitor almost self-evident and previews the capacitor sections to come. 🔉⇢

There is a useful reciprocity between the two descriptions of a field. Given the charges, we can compute the potential everywhere and then draw the equipotentials; conversely, given a set of equipotential surfaces (for instance, measured in the laboratory), we can reconstruct the field by drawing curves everywhere perpendicular to them and using the spacing to get the magnitude. Neither description is more fundamental than the other; they are two windows on the same physics, and skilled problem-solving moves fluently between them. 🔉⇢

In three dimensions, equipotential surfaces never intersect one another, because a point of intersection would have two different values of potential at once, which is impossible for a single-valued potential. This non-crossing rule is the analogue of the rule that field lines never cross. Together with mutual perpendicularity, it strongly constrains how the two families of curves (field lines and equipotentials) can be drawn, and it is a good check on any sketch you make of a field configuration. 🔉⇢

It is worth restating the field-potential relation in the language of gradients, which is where higher study goes. The relation $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ says that the field is the negative gradient of the potential: the field points in the direction of the greatest rate of decrease of $V$, and its size is that maximum rate. The minus sign encodes 'downhill', and the 'normal to the surface' qualifier encodes that the greatest rate of change is always perpendicular to the level surfaces. This single statement contains both of the conclusions drawn above. 🔉⇢

The topographic-map analogy is worth developing fully, because it makes every result of this section intuitive. On a hill map, contour lines join points of equal height; on an electrostatic map, equipotentials join points of equal potential. The steepest slope on the hill is always perpendicular to the contours, and that is the direction water runs; the field is likewise perpendicular to the equipotentials, and that is the direction a positive charge accelerates. Closely spaced contours mark a steep cliff, closely spaced equipotentials mark a strong field. Walking along a contour is level and effortless, just as moving a charge along an equipotential requires no work. Every property of equipotentials is captured by this single, reliable picture. 🔉⇢

The equipotentials of two identical positive charges reveal a feature absent from the single-charge case: a saddle point. Far from the pair the surfaces close over both charges together, approaching spheres appropriate to a single charge of twice the magnitude. Close in, each charge is wrapped by its own family of surfaces. Between the two charges, on the perpendicular bisector, lies a point where the two fields cancel and the potential has a saddle: it is a minimum along the line joining the charges but a maximum across it. There the field is zero, yet the potential is a nonzero constant along the bisecting plane. Sketching this configuration is an excellent exercise in reconciling field lines and equipotentials. 🔉⇢

The work done in moving a charge between two points can be read directly off the equipotential picture, and this is a powerful problem-solving shortcut. If a charge $q$ is moved from a surface at potential $V_1$ to a surface at potential $V_2$, the work done by an external agency is $W=q(V_2-V_1)$, regardless of the path taken, because the electrostatic force is conservative. In particular, if the start and end points lie on the same equipotential, the net work is zero even if the path wanders far and crosses many other surfaces in between. This is why a charge carried once around a closed loop in an electrostatic field returns with no net work done on it. 🔉⇢

The link between conductors and equipotentials deserves fuller treatment because it drives much of the rest of the chapter. In electrostatic equilibrium the field inside a conductor is zero and the field just outside is normal to the surface, so no work is done in moving a test charge anywhere within the conductor or over its surface: the entire conductor, volume and surface alike, is at one constant potential and is therefore a single equipotential body. When two conductors are at different potentials, as in a capacitor, each is its own equipotential, and the equipotential surfaces in the gap between them interpolate smoothly from the value on one conductor to the value on the other, always meeting each conductor's surface at right angles. 🔉⇢

Equipotentials can be mapped experimentally, which underlines that they are physical, not merely mathematical. Using a sheet of weakly conducting paper or an electrolytic tank connected to a source, one probes with a voltmeter to find sets of points at equal potential and joins them into curves; the field is then reconstructed by drawing lines everywhere perpendicular to these measured equipotentials. This technique was historically important for designing electron-optical devices and vacuum tubes, where the shape of the equipotentials determines how charged particles are focused and steered. The perpendicularity and non-crossing rules we derived are exactly what make such reconstructions unambiguous. 🔉⇢

A closer look at the field-potential relation in components rounds out the picture. If the potential varies in all three directions, the field has a component along each axis equal to the rate of decrease of $V$ in that direction: $E_x=-\dfrac{\delta V}{\delta x}$, $E_y=-\dfrac{\delta V}{\delta y}$, $E_z=-\dfrac{\delta V}{\delta z}$. The full field is the vector sum of these, and its magnitude and direction reproduce the 'steepest descent' statement: the resultant points where $V$ falls fastest, which is perpendicular to the local equipotential. The scalar relation $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ used above is just this same statement taken along the single direction, normal to the surface, in which the change is greatest. 🔉⇢

Finally, the crowding of equipotentials near sharply curved conductors explains a phenomenon of great practical importance. Near a pointed conductor the potential changes very rapidly over a short distance, so the equipotentials bunch tightly and, by $|\vec{E}|=-\dfrac{\delta V}{\delta l}$, the field there is very large. This concentration of field at sharp points is the reason charge leaks off pointed conductors and why lightning rods are made pointed. It is the same relation applied to a curved geometry: wherever equipotentials crowd, whether between capacitor plates or around a sharp tip, the field is intense, and this single idea ties together a wide range of electrostatic behaviour. 🔉⇢

To summarise, an equipotential surface is a surface of constant potential; for a point charge these are concentric spheres, for a uniform field they are parallel planes, and for general configurations they are correspondingly shaped surfaces. The electric field is everywhere normal to these surfaces, points towards decreasing potential, and has magnitude $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ measured normal to them. Crowded equipotentials mean a strong field; widely spaced ones mean a weak field; and along any equipotential no work is done. These results turn the abstract scalar $V$ into a vivid geometric picture and provide the standard route from potential to field. 🔉⇢

Finally, a practical note on using these ideas in problems. When asked for the direction of the field on a given surface, answer 'perpendicular to it, towards lower potential'; when asked to compare field strengths at two places, compare the local spacing of equipotentials; and when a uniform field and a plate separation are given, use $E=V/d$ directly. Recognising a conductor surface as an equipotential often unlocks a problem at once, because it tells you the field just outside is normal to the surface and the interior is field-free at a single constant potential. These habits, grounded in $|\vec{E}|=-\dfrac{\delta V}{\delta l}$, make the field-potential relationship a reliable everyday tool. 🔉⇢

Derivation from first principles 🔉⇢

  1. Define the object: an equipotential surface is the locus of points at a single constant value of potential $V$; for a point charge $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ is constant when $r$ is constant, so the equipotentials are concentric spheres.
  2. Show the field is normal to the surface: if $\vec{E}$ had a tangential component, moving a unit charge along the surface against it would require work, contradicting the fact that no work is needed between two points at the same potential; hence $\vec{E}$ is perpendicular to the equipotential at every point.
  3. Set up two close equipotentials $A$ and $B$ at potentials $V$ and $V+\delta V$, with $\delta l$ the perpendicular distance from a point $P$ on $B$ to $A$, measured along the field direction.
  4. Compute the external work to carry a unit positive charge from $B$ to $A$ against the field: $W=|\vec{E}|\,\delta l$, since the displacement $\delta l$ is along $\vec{E}$.
  5. Set this work equal to the potential difference $V_A-V_B=V-(V+\delta V)=-\delta V$, giving $|\vec{E}|\,\delta l=-\delta V$.
  6. Solve for the field magnitude: $|\vec{E}|=-\dfrac{\delta V}{\delta l}$; since $\delta V\lt 0$ in the field direction, $|\vec{E}|=\dfrac{|\delta V|}{\delta l}\gt 0$.
  7. Read off the two conclusions: (i) $\vec{E}$ points in the direction of steepest decrease of $V$; (ii) $|\vec{E}|$ equals the change in the magnitude of potential per unit displacement normal to the equipotential surface, so crowded surfaces mean a strong field.
⚠️ JEE trap: A widespread misconception is to think that a large potential automatically means a large field, or that where $V=0$ the field must vanish. The relation $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ shows the field depends only on how fast $V$ changes with position, not on its value: inside a charged conductor the potential is a large constant yet the field is zero, while at a point where $V$ passes through zero the field can be strong. A second error is to draw field lines that meet equipotential surfaces at an angle; field lines are always perpendicular to equipotentials, because any tangential field component would do work along a surface of constant potential, which is forbidden. Finally, remember the sign: the field points towards decreasing potential, so a positive charge accelerates from high to low potential. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A parallel-plate arrangement produces a uniform electric field directed along the $x$-axis. Two plane equipotential surfaces are separated by a perpendicular distance $\delta l=2.0\ \text{mm}$ along $x$, and the potential drops by $|\delta V|=10\ \text{V}$ from the first surface to the second in the direction of the field.
TARGET Find the magnitude and direction of the electric field between the surfaces, describe the shape and orientation of the equipotential surfaces, and state what would happen to the field if the two surfaces were drawn twice as far apart for the same potential step.
STRATEGY Apply the field–potential relation $|\vec{E}|=-\dfrac{\delta V}{\delta l}$, taking magnitudes over the small perpendicular step; use the fact that in a uniform field the equipotentials are planes normal to the field and that the field points towards decreasing potential; then use the inverse dependence of $E$ on $\delta l$ at fixed $\delta V$.
EXECUTE $|\vec{E}|=\dfrac{|\delta V|}{\delta l}=\dfrac{10\ \text{V}}{2.0\times 10^{-3}\ \text{m}}=5.0\times 10^{3}\ \text{V m}^{-1}$, directed along $+x$, towards the surface at lower potential. The equipotential surfaces are planes perpendicular to the $x$-axis (parallel to the $y$-$z$ plane), equally spaced because the field is uniform. If the same $|\delta V|=10\ \text{V}$ were spread over $\delta l=4.0\ \text{mm}$, then $|\vec{E}|=\dfrac{10}{4.0\times 10^{-3}}=2.5\times 10^{3}\ \text{V m}^{-1}$, half as large.
REFLECT The field points from the higher-potential plane to the lower-potential one, perpendicular to both, exactly as $|\vec{E}|=-\dfrac{\delta V}{\delta l}$ demands. Doubling the spacing for the same potential step halves the field, which is the quantitative statement of the visual rule that widely spaced equipotentials mean a weak field and crowded ones a strong field. This is the same relation as $E=V/d$ for a capacitor gap, applied over an infinitesimal step.

Source: NCERT XII Ch 2 (§2.6, §2.6.1)

Capacitors, Dielectrics and Capacitance 🔉⇢deep concept

Definition: A capacitor is a pair of conductors carrying charges $+Q$ and $-Q$ at potential difference $V$, characterised by its capacitance $C=\dfrac{Q}{V}$, a purely geometric quantity that for parallel plates in vacuum is $C_0=\dfrac{\varepsilon_0 A}{d}$ and is raised by a factor $K$ (the dielectric constant) to $C=\dfrac{K\varepsilon_0 A}{d}$ when the gap is filled with a dielectric. 🔉⇢

🔬 Interactive 3D · Two charged plates confine a uniform field $E=\dfrac{\sigma}{\varepsilon_0}$ between them, and the geometry alone fixes the capacitance $C=\dfrac{\varepsilon_0 A}{d}$, which rises to $C=\dfrac{K\varepsilon_0 A}{d}$ when a dielectric of constant $K$ is inserted.

A capacitor is a system of two conductors separated by an insulator. The two conductors carry charges, say $Q_1$ and $Q_2$, and sit at potentials $V_1$ and $V_2$. In practice the arrangement of interest is the symmetric one: the conductors carry equal and opposite charges $+Q$ and $-Q$, and one speaks of the potential difference $V=V_1-V_2$ between them. We call $Q$ the charge of the capacitor, even though it is really the charge on just one conductor; the total charge of the device is zero. Even a single conductor can act as a capacitor if the second conductor is imagined to be at infinity. The plates are usually charged by connecting them to the two terminals of a battery, which pumps charge from one plate to the other until the plate potentials match the terminal potentials. 🔉⇢

The electric field in the region between the conductors is proportional to the charge $Q$: if $Q$ is doubled, the field at every point doubles. This follows from the direct proportionality between field and charge in Coulomb's law, together with the superposition principle. The potential difference $V$ is the work done per unit positive charge in carrying a small test charge from conductor 2 to conductor 1 against the field; since $V$ is obtained by integrating the field, and the field is proportional to $Q$, the potential difference is itself proportional to $Q$. Their ratio is therefore a constant, independent of how much charge happens to be on the plates. 🔉⇢

That constant ratio is the capacitance: $C=\dfrac{Q}{V}$. Because both $Q$ and $V$ scale together, $C$ does not depend on the particular charge or voltage — it is fixed by the geometry of the two conductors (their shape, size, and separation) and, as we shall see, by the insulating material between them. The SI unit of capacitance is the farad, $1\ \text{F}=1\ \text{C V}^{-1}$. Capacitance measures how much charge a device banks for each volt of potential difference: a large $C$ stores a large $Q$ at a modest $V$, which is exactly what makes capacitors useful for holding charge and energy. 🔉⇢

There is a practical limit to how much charge a given capacitor can hold. A large potential difference means a strong field around the conductors, and a sufficiently strong field ionises the surrounding air, turning the insulator into a partial conductor so that charge leaks across and the capacitor discharges. The maximum field an insulating medium can sustain without this breakdown is its dielectric strength; for air it is about $3\times10^{6}\ \text{V m}^{-1}$, which over a gap of about a centimetre corresponds to roughly $3\times10^{4}\ \text{V}$. To store much charge without leaking, a capacitor should therefore have a high enough $C$ that the necessary $V$, and hence the field, stays below the breakdown limit. In practice the farad is enormous, and the working units are its submultiples: $1\ \mu\text{F}=10^{-6}\ \text{F}$, $1\ \text{nF}=10^{-9}\ \text{F}$, and $1\ \text{pF}=10^{-12}\ \text{F}$. 🔉⇢

The parallel plate capacitor is the archetype. It consists of two large plane parallel conducting plates, each of area $A$, separated by a small distance $d$, carrying charges $+Q$ and $-Q$. Take the gap to be vacuum for now. Because $d$ is far smaller than the linear size of the plates ($d^2\ll A$), each plate behaves like an infinite plane sheet of uniform surface charge density: plate 1 has $\sigma=\dfrac{Q}{A}$ and plate 2 has $-\sigma$. A single infinite sheet produces a field $\dfrac{\sigma}{2\varepsilon_0}$ on each side, so we simply superpose the fields of the two sheets. 🔉⇢

In the outer region above plate 1, the two sheets give fields of equal magnitude $\dfrac{\sigma}{2\varepsilon_0}$ pointing in opposite senses, so they cancel and the net field is zero. In the outer region below plate 2, the same cancellation occurs and again the field is zero. In the inner region between the plates the two contributions point the same way and add, giving $E=\dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0}=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$. So the field of an ideal parallel plate capacitor is confined between the plates, is uniform there, and points from the positive plate to the negative plate. 🔉⇢

For a uniform field the potential difference is simply the field times the plate separation: $V=Ed=\dfrac{Qd}{\varepsilon_0 A}$. Dividing the charge by this potential difference gives the capacitance of the parallel plate capacitor in vacuum, $C=\dfrac{Q}{V}=\dfrac{\varepsilon_0 A}{d}$, which, as promised, depends only on the geometry — the plate area and their separation. Larger plates or a smaller gap raise the capacitance; a wider gap lowers it. 🔉⇢

A quick numerical check underlines why the farad is so unwieldy. With $A=1\ \text{m}^2$ and $d=1\ \text{mm}$, $C=\dfrac{(8.85\times10^{-12})(1)}{10^{-3}}=8.85\times10^{-9}\ \text{F}$, only about $9\ \text{nF}$ for a square-metre plate. Turned around, to build a $1\ \text{F}$ capacitor with a $1\ \text{cm}$ gap would need a plate area of order $10^{9}\ \text{m}^2$ — a square roughly $30\ \text{km}$ on a side. Real capacitors reach useful values not by giant plates but by using very thin gaps, rolled or stacked layers, and, crucially, dielectric fillings. 🔉⇢

Real plates have finite area, so the idealisation is not perfect near the edges: there the field lines bulge outward instead of running straight across, an effect called fringing of the field, and correspondingly $\sigma$ is not perfectly uniform right at the rim. However, for $d^2\ll A$ these edge effects are negligible in the central region, and the uniform result $E=\dfrac{\sigma}{\varepsilon_0}$ holds well away from the edges. This is why the parallel plate formula is so reliable in practice despite being derived for infinite sheets. 🔉⇢

To understand dielectrics we must contrast them with conductors. A dielectric is a non-conducting substance with no (or a negligible number of) free charge carriers. When a conductor is placed in an external field, its free charges move until the induced-charge field completely cancels the applied field inside, giving zero net field. In a dielectric this free motion is impossible; instead the external field induces dipole moments in the molecules — by stretching them or by re-orienting them — and the collective effect is a set of bound charges on the dielectric's surfaces whose field opposes the applied field. Crucially, this opposing field does not fully cancel the external field; it only reduces it. The extent of the reduction depends on the material. 🔉⇢

At the molecular level, molecules are polar or non-polar. In a non-polar molecule (for example $\text{O}_2$ or $\text{H}_2$) the centres of positive and negative charge coincide, so there is no permanent dipole moment; an external field pulls these centres slightly apart, inducing a dipole moment along the field. In a polar molecule (such as $\text{HCl}$ or water, $\text{H}_2\text{O}$) the centres of positive and negative charge are already separated, giving a permanent dipole moment; with no field these dipoles point randomly because of thermal agitation and average to zero, but an external field tends to align them, again producing a net moment along the field. Either way, a dielectric develops a net dipole moment in an external field — it becomes polarised. 🔉⇢

The dipole moment per unit volume is the polarisation $\vec{P}$. For a linear isotropic dielectric it is proportional to the field, $\vec{P}=\varepsilon_0\chi_e\vec{E}$, where $\chi_e$ is the electric susceptibility of the medium. Inside a uniformly polarised slab the positive end of one molecular dipole sits next to the negative end of its neighbour, so there is no net volume charge; but at the two faces perpendicular to the field the dipole ends are left unneutralised, giving bound surface charge densities $+\sigma_P$ and $-\sigma_P$. The polarised dielectric is thus equivalent to two thin charged sheets whose field opposes the applied field and reduces the net field within the material. These bound charges are not free charges — they cannot be drawn off as current. 🔉⇢

Now insert such a dielectric so that it completely fills the gap of a parallel plate capacitor. The free charge on the plates is still $\pm\sigma$, but the bound surface charges $\mp\sigma_P$ partly cancel it, so the net surface charge density producing the internal field is $\pm(\sigma-\sigma_P)$, and the field becomes $E=\dfrac{\sigma-\sigma_P}{\varepsilon_0}$, smaller than the vacuum value $E_0=\dfrac{\sigma}{\varepsilon_0}$. For a linear dielectric $\sigma_P$ is proportional to $\sigma$, so $(\sigma-\sigma_P)$ is proportional to $\sigma$ and one writes $\sigma-\sigma_P=\dfrac{\sigma}{K}$, defining a material constant $K$ that is clearly greater than $1$. 🔉⇢

With the reduced field, the potential difference across the plates falls to $V=Ed=\dfrac{\sigma d}{\varepsilon_0 K}=\dfrac{Qd}{A\varepsilon_0 K}$, and the capacitance rises to $C=\dfrac{Q}{V}=\dfrac{K\varepsilon_0 A}{d}$. The product $\varepsilon_0 K$ is called the permittivity of the medium, $\varepsilon=\varepsilon_0 K$; for vacuum $K=1$ and $\varepsilon=\varepsilon_0$. The dimensionless ratio $K=\dfrac{\varepsilon}{\varepsilon_0}$ is the dielectric constant of the substance. Comparing the vacuum capacitance $C_0=\dfrac{\varepsilon_0 A}{d}$ with the filled value gives the clean relation $K=\dfrac{C}{C_0}$. 🔉⇢

The dielectric constant of a substance is therefore the factor (greater than one) by which the capacitance grows over its vacuum value when the dielectric is inserted fully between the plates. Although this was derived for a parallel plate capacitor, the relation $K=\dfrac{C}{C_0}$ holds for any capacitor and can be taken as the general definition of the dielectric constant. Physically, the dielectric weakens the field, lowers the voltage needed to hold a given charge, and so lets the device store more charge per volt — this is the second, material way (alongside geometry) of raising capacitance, and it is why practical capacitors are filled with high-$K$ insulators rather than left with a vacuum or air gap. 🔉⇢

A subtle but examinable point is the difference between charging at constant charge and at constant voltage when a dielectric is inserted. If the capacitor is disconnected from the battery before insertion, $Q$ is fixed: inserting the dielectric lowers the field and hence the voltage by a factor $K$, while $C$ rises by $K$ and the stored energy $\dfrac{Q^2}{2C}$ falls. If instead the capacitor stays connected to the battery, $V$ is fixed: $C$ rises by $K$, so the charge $Q=CV$ increases by $K$ as the battery supplies more charge. Deciding which quantity ($Q$ or $V$) is held constant is the key first step in any dielectric-insertion problem. 🔉⇢

It is worth being explicit about how a battery actually charges a capacitor. When the two plates are joined to the terminals of a battery, the battery does work pumping electrons off one plate and onto the other, leaving one plate with $+Q$ and the other with $-Q$. Charge flows until the potential difference between the plates equals the terminal voltage of the battery, at which point the plates are at the terminal potentials and the current stops. A helpful mental picture is a pump raising water to a fixed height: the pump (battery) moves charge until the 'pressure' (potential difference) across the capacitor matches its own, and a device with larger capacitance simply banks more charge for that same pressure, just as a wider tank holds more water at a given height. 🔉⇢

Beyond merely storing charge, a capacitor is a key element of most ac circuits, where it performs important functions described later in the study of alternating currents. Because no steady current can cross the insulating gap, a capacitor blocks direct current while passing alternating current, and it stores and returns energy each cycle rather than dissipating it as a resistor does. This ability to hold charge briefly and release it on demand also makes capacitors indispensable for smoothing rectified voltages, for timing circuits, for tuning radio receivers, and for delivering short, intense bursts of energy — as in a camera flash — that a battery alone could not supply quickly enough. 🔉⇢

Returning to the microscopic origin of $K$, it is useful to distinguish the two mechanisms of polarisation. In non-polar materials the field distorts each molecule, pulling its positive and negative charge centres slightly apart to create an induced (electronic) dipole moment that vanishes the instant the field is removed; this induced effect is nearly independent of temperature. In polar materials the molecules carry permanent dipoles that are normally randomised by thermal agitation; the field only partially aligns them, and because thermal motion fights the alignment, this orientational polarisation weakens as temperature rises. In both cases the polarisation is proportional to the field for modest fields, $\vec{P}=\varepsilon_0\chi_e\vec{E}$, and the susceptibility $\chi_e$ and the dielectric constant are simply related by $K=1+\chi_e$, so a larger susceptibility means a larger $K$ and a greater boost to capacitance. 🔉⇢

The practical construction of real capacitors follows directly from the formula $C=\dfrac{K\varepsilon_0 A}{d}$. Since useful capacitances need either huge area or a tiny gap, manufacturers use very thin dielectric films (paper, plastic, ceramic, or an oxide layer only micrometres thick) sandwiched between metal foils, often rolled up into a compact cylinder to pack a large area into a small volume. A high dielectric constant multiplies the capacitance further, which is why ceramics with large $K$ are prized. The competing constraint is dielectric strength: the film must withstand the working field without breaking down, so the choice of material and thickness is always a compromise between maximising $\dfrac{K}{d}$ and keeping the field safely below the breakdown limit. 🔉⇢

The definition of a capacitor as a pair of conductors is broad enough to include a single isolated conductor, if we imagine its partner plate carried off to infinity and held at zero potential. An isolated conducting sphere of radius $R$ sits at potential $V=\dfrac{Q}{4\pi\varepsilon_0 R}$ when it carries charge $Q$, so its self-capacitance is $C=\dfrac{Q}{V}=4\pi\varepsilon_0 R$ — again purely geometric, depending only on the radius. This shows the concept of capacitance is not tied to two facing plates; any conductor has a capacitance measuring how much charge it banks per volt, and the earth itself, being an enormous conductor, has a correspondingly large capacitance that lets it absorb charge without appreciable change in potential. 🔉⇢

The contrast between a conductor and a dielectric in an external field, first met in the previous section, is worth restating because it explains exactly why $K$ is finite. A conductor placed in a field lets its free charges move until the induced-charge field completely cancels the applied field inside, so the interior field drops to zero — this is the $K\to\infty$, perfect-cancellation limit. A dielectric has no free charges; its bound charges can only be displaced or re-oriented within the molecules, so the induced surface charge is limited and the opposing field it creates only reduces the applied field, never eliminates it. That is why the net field falls from $E_0$ to $\dfrac{E_0}{K}$ with a finite $K$ greater than one, rather than to zero, and why a dielectric-filled capacitor still supports a potential difference across it. 🔉⇢

This finite reduction is precisely what makes dielectrics so valuable in real capacitors. By weakening the internal field for a given free charge, a dielectric lowers the voltage needed to hold that charge, so the same device banks $K$ times as much charge before the field reaches the breakdown value. A good dielectric therefore does double duty: it multiplies the capacitance through the factor $K$, and, by being a better insulator than air, it raises the dielectric strength so that a larger field — and hence a larger stored charge and energy — can be tolerated before leakage sets in. The humble insulating filling is thus central to why practical capacitors can store useful amounts of charge in a small, safe package. 🔉⇢

Finally, the limit $K\to\infty$ of a dielectric slab is instructive because it mimics inserting a conducting slab. A conductor has effectively infinite $K$: the field inside it is zero, so it contributes no potential drop, and a conducting slab of thickness $t$ inserted into a gap $d$ simply shortens the effective gap to $d-t$, giving $C=\dfrac{\varepsilon_0 A}{d-t}$ regardless of where the slab sits. This is exactly the $K\to\infty$ limit of the partial-dielectric result and reinforces the general method: decompose any composite gap into series regions (vacuum, dielectric, or conductor), add the potential drops across each, and recover the capacitance from $C=\dfrac{Q}{V}$. 🔉⇢

Finally, a partially filled gap is a standard exam variant. If a slab of dielectric constant $K$ and thickness less than $d$ is inserted, one treats the gap as a vacuum region and a dielectric region in series, adding the potential drops. For a slab of thickness $\tfrac{3}{4}d$, the field is $E_0$ in the vacuum part and $\dfrac{E_0}{K}$ in the slab, so the total voltage becomes $V=E_0\big(\tfrac{1}{4}d\big)+\dfrac{E_0}{K}\big(\tfrac{3}{4}d\big)=V_0\,\dfrac{K+3}{4K}$ at fixed free charge, and the capacitance rises to $C=\dfrac{4K}{K+3}C_0$. The general recipe — decompose the gap into series regions, add the voltage drops, then use $C=\dfrac{Q}{V}$ — handles every such geometry. 🔉⇢

Derivation from first principles 🔉⇢

  1. Definition of capacitance: the between-plate field is proportional to the plate charge $Q$, and $V=\int \vec{E}\cdot d\vec{l}$ is proportional to that field, so $V\propto Q$ and the ratio $C=\dfrac{Q}{V}$ is a constant fixed by geometry and the medium, independent of $Q$ and $V$.
  2. Field of a parallel plate capacitor: model each plate as an infinite sheet of density $\pm\sigma=\pm\dfrac{Q}{A}$; superposing the two sheet fields ($\dfrac{\sigma}{2\varepsilon_0}$ each) gives zero outside and $E=\dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0}=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$ in the gap.
  3. Vacuum capacitance: for a uniform field $V=Ed=\dfrac{Qd}{\varepsilon_0 A}$, so $C_0=\dfrac{Q}{V}=\dfrac{\varepsilon_0 A}{d}$, a purely geometric result.
  4. Effect of a dielectric on the field: the bound surface charges $\mp\sigma_P$ partly cancel the free charge, leaving net density $\pm(\sigma-\sigma_P)$, so $E=\dfrac{\sigma-\sigma_P}{\varepsilon_0}\lt E_0$; for a linear dielectric $\sigma-\sigma_P=\dfrac{\sigma}{K}$ with $K\gt 1$.
  5. Capacitance with dielectric: $V=Ed=\dfrac{\sigma d}{\varepsilon_0 K}=\dfrac{Qd}{A\varepsilon_0 K}$, hence $C=\dfrac{Q}{V}=\dfrac{K\varepsilon_0 A}{d}=\varepsilon\dfrac{A}{d}$ with permittivity $\varepsilon=\varepsilon_0 K$.
  6. Dielectric constant as a ratio: dividing the filled capacitance by the vacuum value gives $K=\dfrac{C}{C_0}$, the general definition of the dielectric constant, valid for any capacitor shape.
  7. Partially filled gap: with a slab of thickness $\tfrac{3}{4}d$ and constant $K$, add the series voltage drops $V=E_0\big(\tfrac{1}{4}d\big)+\dfrac{E_0}{K}\big(\tfrac{3}{4}d\big)=V_0\dfrac{K+3}{4K}$, giving $C=\dfrac{Q_0}{V}=\dfrac{4K}{K+3}C_0$.
⚠️ JEE trap: The most damaging misconception is that capacitance depends on the charge or voltage placed on the plates — that a capacitor charged to a higher $V$ somehow has a larger $C$. It does not: $C=\dfrac{Q}{V}$ is a fixed ratio set by geometry (area, separation) and the dielectric, so raising $Q$ raises $V$ in exact proportion and leaves $C$ unchanged. A second common error is to confuse constant-charge and constant-voltage insertion of a dielectric: with the battery disconnected $Q$ is fixed and $V$ drops by $K$, while with the battery connected $V$ is fixed and $Q$ rises by $K$. A third is to use $\dfrac{\sigma}{2\varepsilon_0}$ for the field between the plates; the two plates' fields add there to give $E=\dfrac{\sigma}{\varepsilon_0}$, and only outside do they cancel to zero. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A parallel plate capacitor has plate area $A$ and separation $d$ with vacuum between the plates, giving capacitance $C_0=\dfrac{\varepsilon_0 A}{d}$. A slab of dielectric constant $K$, with the same area as the plates but thickness $\tfrac{3}{4}d$, is inserted between the plates while the free charge $Q_0$ on the plates is held fixed.
TARGET Find how the capacitance changes when the slab is inserted.
STRATEGY Treat the gap as two regions in series: a vacuum layer of thickness $\tfrac{1}{4}d$ with field $E_0$ and a dielectric layer of thickness $\tfrac{3}{4}d$ with reduced field $\dfrac{E_0}{K}$. Add the potential drops to get the new $V$, then use $C=\dfrac{Q_0}{V}$.
EXECUTE Let $E_0=\dfrac{V_0}{d}$ be the field with no dielectric. Inside the slab the field is $E=\dfrac{E_0}{K}$. The total potential difference is $V=E_0\big(\tfrac{1}{4}d\big)+\dfrac{E_0}{K}\big(\tfrac{3}{4}d\big)=E_0 d\Big(\dfrac{1}{4}+\dfrac{3}{4K}\Big)=V_0\,\dfrac{K+3}{4K}$. Since $Q_0$ is unchanged, the new capacitance is $C=\dfrac{Q_0}{V}=\dfrac{Q_0}{V_0}\cdot\dfrac{4K}{K+3}=\dfrac{4K}{K+3}\,C_0$.
REFLECT The capacitance increases by the factor $\dfrac{4K}{K+3}$, which is greater than $1$ for any $K\gt 1$ and approaches $4$ as $K\to\infty$ (as if the $\tfrac{3}{4}d$ layer became a conductor, effectively shrinking the gap to $\tfrac{1}{4}d$). Note that the free charge stays fixed while the voltage falls, so $C=\dfrac{Q_0}{V}$ rises — a clean illustration of constant-charge dielectric insertion.

Source: NCERT XII Ch 2 (§2.13)

Combination of Capacitors and Energy Stored 🔉⇢deep concept

Definition: Capacitors in series share the same charge and add reciprocals, $\dfrac{1}{C}=\sum_i\dfrac{1}{C_i}$, while capacitors in parallel share the same voltage and add directly, $C=\sum_i C_i$; the energy banked in a charged capacitor is $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$, stored in the field with density $u=\dfrac{1}{2}\varepsilon_0 E^2$. 🔉⇢

🔬 Interactive 3D · Building charge on a capacitor bit by bit does work $\delta W=\dfrac{Q'}{C}\,\delta Q'$, which integrates to the stored energy $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$, viewable as field energy of density $u=\dfrac{1}{2}\varepsilon_0 E^2$.

Several capacitors of capacitance $C_1,C_2,\dots,C_n$ can be combined into a single network with some effective (equivalent) capacitance $C$. The value of $C$ depends entirely on how the individual capacitors are wired together, and two arrangements cover almost every problem: capacitors in series and capacitors in parallel. Mastering these two rules, and knowing how to break a mixed network into series and parallel blocks, is the core skill this section teaches. 🔉⇢

Consider first two capacitors $C_1$ and $C_2$ in series, with the left plate of $C_1$ and the right plate of $C_2$ connected to the two terminals of a battery and carrying charges $+Q$ and $-Q$. The crucial fact about a series connection is that every capacitor carries the same charge $Q$. Here is why: the right plate of $C_1$ and the left plate of $C_2$ are joined by an isolated conducting stretch that was neutral to begin with. Charge induced on the right plate of $C_1$ must be $-Q$, and by charge conservation on that isolated conductor the left plate of $C_2$ must then be $+Q$. If the charges were unequal, a net charge would sit on the connecting conductor, producing a field inside it that would drive charge to flow until the imbalance vanished. So in series, the charge $\pm Q$ is common to all capacitors. 🔉⇢

The total potential drop $V$ across the series combination is the sum of the drops across the individual capacitors, because potential differences add along a path: $V=V_1+V_2=\dfrac{Q}{C_1}+\dfrac{Q}{C_2}$. Factoring out the common charge gives $V=Q\Big(\dfrac{1}{C_1}+\dfrac{1}{C_2}\Big)$. If we now regard the whole combination as a single effective capacitor holding charge $Q$ at voltage $V$, its capacitance is $C=\dfrac{Q}{V}$, and comparing the two expressions gives $\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$. 🔉⇢

The same reasoning extends immediately to any number of capacitors in series. With $n$ capacitors the voltages still add, $V=V_1+V_2+\cdots+V_n=\dfrac{Q}{C_1}+\dfrac{Q}{C_2}+\cdots+\dfrac{Q}{C_n}$, all sharing the common charge $Q$, and the general series formula follows: $\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots+\dfrac{1}{C_n}$. The reciprocals add, so the effective capacitance of a series string is always smaller than the smallest member — putting capacitors in series is like widening the effective gap, which reduces capacitance. 🔉⇢

Now consider two capacitors in parallel, both connected across the same two nodes so that the same potential difference $V$ appears across each. Here the voltage is common, but the charges need not be: $Q_1=C_1 V$ and $Q_2=C_2 V$ can differ. The equivalent capacitor must hold the total charge drawn from the battery, $Q=Q_1+Q_2$, at the common voltage $V$. Hence $Q=CV=C_1 V+C_2 V$, and cancelling $V$ gives the parallel rule $C=C_1+C_2$. 🔉⇢

For $n$ capacitors in parallel the total charge is $Q=Q_1+Q_2+\cdots+Q_n$, so $CV=C_1 V+C_2 V+\cdots+C_n V$ and therefore $C=C_1+C_2+\cdots+C_n$. Capacitances in parallel simply add, so the effective capacitance of a parallel bank is always larger than the largest member — connecting plates in parallel is like enlarging the total plate area, which increases capacitance. Series adds reciprocals (charge common), parallel adds directly (voltage common): keeping straight which quantity is shared is the whole game. 🔉⇢

A worked network makes the method concrete. Take four $10\ \mu\text{F}$ capacitors with three of them, $C_1,C_2,C_3$, in series and the fourth, $C_4$, in parallel with that series block, connected to a $500\ \text{V}$ supply. The series trio has $\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}=\dfrac{3}{10\ \mu\text{F}}$, so $C'=\dfrac{10}{3}\ \mu\text{F}$. This block sits in parallel with $C_4$, so the network's equivalent capacitance is $C=C'+C_4=\Big(\dfrac{10}{3}+10\Big)\ \mu\text{F}\approx13.3\ \mu\text{F}$. The systematic approach — collapse series blocks first, then combine the parallel pieces — turns any ladder network into a single number. 🔉⇢

The charges then follow. Each of the three series capacitors carries the same charge $Q$, and the sum of their voltages is the supply voltage: $\dfrac{Q}{C_1}+\dfrac{Q}{C_2}+\dfrac{Q}{C_3}=500\ \text{V}$, giving $Q=500\ \text{V}\times\dfrac{10}{3}\ \mu\text{F}\approx1.7\times10^{-3}\ \text{C}$. The parallel capacitor $C_4$ has the full $500\ \text{V}$ across it, so its charge is $Q'=500\ \text{V}\times10\ \mu\text{F}=5.0\times10^{-3}\ \text{C}$. Notice how 'same charge in series, same voltage in parallel' does all the bookkeeping. 🔉⇢

We turn now to the energy stored in a capacitor, which is central to its usefulness. A charged capacitor is a system of two conductors carrying $+Q$ and $-Q$; imagine building this configuration up from two initially uncharged conductors by transferring positive charge from conductor 2 to conductor 1 bit by bit. As soon as any charge has been moved, conductor 1 is at a higher potential than conductor 2, so moving each further increment requires work done by an external agent against the growing potential difference. Summing all these increments gives the total work, which is stored as the electrostatic potential energy of the charged capacitor. 🔉⇢

Consider an intermediate stage where the conductors carry $+Q'$ and $-Q'$, so the potential difference between them is $V'=\dfrac{Q'}{C}$. Transferring a further infinitesimal charge $\delta Q'$ from conductor 2 to conductor 1 requires work $\delta W=V'\,\delta Q'=\dfrac{Q'}{C}\,\delta Q'$. This is the elementary step: the cost of adding the next sliver of charge is that sliver times the voltage the capacitor has already reached. Because the voltage rises steadily as charge accumulates, later increments cost more than earlier ones. 🔉⇢

Integrating this elementary work from $Q'=0$ to $Q'=Q$ gives the total work: $W=\displaystyle\int_0^{Q}\dfrac{Q'}{C}\,dQ'=\dfrac{1}{C}\cdot\dfrac{Q'^2}{2}\Big|_0^{Q}=\dfrac{Q^2}{2C}$. Since the electrostatic force is conservative, this work is stored as potential energy of the system and is recovered when the capacitor discharges. The factor of $\tfrac{1}{2}$ arises precisely because the voltage grew linearly with charge during charging — the average voltage over the process was half the final voltage, not the full final voltage. 🔉⇢

Using $C=\dfrac{Q}{V}$ the stored energy can be written three equivalent ways: $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$. Which form is most convenient depends on what is held fixed. If the capacitor stays connected to a battery, $V$ is fixed and $U=\tfrac{1}{2}CV^2$ makes the energy's dependence on $C$ obvious. If the capacitor is isolated with fixed charge, $U=\dfrac{Q^2}{2C}$ is the natural form. The result is independent of the order or manner in which the charge was assembled, exactly because the force is conservative. 🔉⇢

It is illuminating to see the energy as residing in the electric field between the plates rather than 'on' the charges. For a parallel plate capacitor of plate area $A$ and separation $d$, substitute $C=\dfrac{\varepsilon_0 A}{d}$ and $Q=\sigma A$ into $U=\dfrac{Q^2}{2C}$: $U=\dfrac{(\sigma A)^2}{2}\cdot\dfrac{d}{\varepsilon_0 A}=\dfrac{\sigma^2}{2\varepsilon_0}\,Ad$. Using the relation $E=\dfrac{\sigma}{\varepsilon_0}$ between the surface charge density and the field, $\sigma=\varepsilon_0 E$, this becomes $U=\dfrac{1}{2}\varepsilon_0 E^2\,(Ad)$. 🔉⇢

Now $Ad$ is exactly the volume of the region between the plates, the region where the field exists. Dividing the energy by this volume gives the energy stored per unit volume, the result on energy density of an electric field is $u=\dfrac{1}{2}\varepsilon_0 E^2$. Although we obtained it for a parallel plate capacitor, this expression for the energy density is completely general: wherever an electric field $E$ exists in vacuum, it carries energy of density $\dfrac{1}{2}\varepsilon_0 E^2$, for any configuration of charges whatsoever. The field itself is the seat of the energy. 🔉⇢

A classic and instructive problem exposes a subtlety in energy accounting. A $900\ \text{pF}$ capacitor charged by a $100\ \text{V}$ battery stores $U=\tfrac{1}{2}QV=\tfrac{1}{2}CV^2=\tfrac{1}{2}(900\times10^{-12})(100)^2=4.5\times10^{-6}\ \text{J}$, with charge $Q=CV=9\times10^{-8}\ \text{C}$. Disconnect it from the battery and connect it across an identical uncharged $900\ \text{pF}$ capacitor. In the steady state the two share charge equally, so each holds $\dfrac{Q}{2}$ and, since $C$ is unchanged, the common voltage falls to $\dfrac{V}{2}=50\ \text{V}$. 🔉⇢

The total energy of the two-capacitor system is now $U'=2\times\tfrac{1}{2}\Big(\dfrac{Q}{2}\Big)V'=\tfrac{1}{2}\Big(\dfrac{Q}{2}\Big)V\cdot 2 \cdot \tfrac{1}{2}$; evaluating directly, $U'=\tfrac{1}{4}QV=2.25\times10^{-6}\ \text{J}$, exactly half the original $4.5\times10^{-6}\ \text{J}$. No charge is lost, yet half the energy has vanished. The resolution is physical: during the transient redistribution a current briefly flows from the first capacitor to the second, and the missing energy is dissipated as heat in the connecting wires and as a pulse of electromagnetic radiation. Charge is conserved; stored electrostatic energy is not, whenever a dissipative transient intervenes. 🔉⇢

These ideas of energy and combination connect back to the whole chapter. Building charge on a capacitor is the two-conductor version of assembling a system of point charges: in both cases the stored energy is the work done against the fields already present, and in both cases it is path-independent because electrostatics is conservative. The relation $U=\dfrac{Q^2}{2C}$ shows why a large capacitance banks charge cheaply (low voltage, low energy per unit charge), while the field picture $u=\dfrac{1}{2}\varepsilon_0 E^2$ shows that the energy really lives in the space between the plates. Together with the series and parallel rules, these results let one analyse any capacitor network — its equivalent capacitance, the charge and voltage on every element, and the total energy it stores. 🔉⇢

For a general network the working strategy is to identify blocks that are purely in series or purely in parallel, collapse each block to a single equivalent capacitor, and repeat until one number remains. Two capacitors sharing both nodes are in parallel and add directly; two carrying the same charge in an unbranched chain are in series and add as reciprocals. Symmetry is a powerful shortcut: if two points of a network are always at the same potential, no charge sits on a capacitor bridging them, so it can be ignored, exactly as a balanced Wheatstone bridge lets one drop the central arm. After finding the equivalent capacitance and the total charge from the source, one works backwards through the collapse, using 'same charge in series, same voltage in parallel' to recover the charge and voltage on each original element. 🔉⇢

A particularly common problem connects two capacitors that already carry charge. Suppose capacitors $C_1$ and $C_2$ are charged to voltages $V_1$ and $V_2$ and then joined positive plate to positive plate. Charge is conserved, $Q_1+Q_2=C_1V_1+C_2V_2$, and in the final steady state both capacitors reach a common potential difference. Setting the shared voltage $V$ so that the total charge is redistributed over the total capacitance gives $V=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$. The equal-capacitor case with one uncharged capacitor ($V_2=0$, $C_1=C_2$) reproduces the earlier result $V=\dfrac{V_1}{2}$, and the same formula handles unequal capacitors and arbitrary initial voltages. 🔉⇢

The energy accounting in such redistributions is always revealing. The total energy before connection is $\tfrac{1}{2}C_1V_1^2+\tfrac{1}{2}C_2V_2^2$, and after reaching the common voltage it is $\tfrac{1}{2}(C_1+C_2)V^2$. Substituting the common voltage and simplifying, the energy lost is $\Delta U=\dfrac{1}{2}\dfrac{C_1C_2}{C_1+C_2}(V_1-V_2)^2$, which is strictly positive whenever the initial voltages differ. This loss is unavoidable and independent of the resistance of the connecting wires: a transient current must flow while the voltages equalise, and that current dissipates energy as heat and radiation. Only if $V_1=V_2$ initially — no redistribution needed — is no energy lost. 🔉⇢

The same 'half the energy is lost' theme appears when a capacitor is charged through a resistor from a battery of emf $V$. The battery delivers total charge $Q=CV$ at its fixed terminal voltage, doing work $W_{batt}=QV=CV^2$. But the capacitor ends up storing only $U=\tfrac{1}{2}CV^2$, exactly half of the work done by the battery. The other half is dissipated in the resistor as heat, no matter how small the resistance, because a smaller resistance simply means a larger, briefer current carrying the same total charge. Thus charging any capacitor from a constant-voltage source through a resistive path is at best fifty per cent efficient in energy terms. 🔉⇢

The stored energy also lets one compute the force of attraction between the capacitor plates by the energy method. Imagine slowly increasing the plate separation by $dx$ at fixed charge $Q$; no charge moves, so the battery does no work, and the mechanical work done against the attractive force equals the increase in stored energy. Using $U=\dfrac{Q^2}{2C}=\dfrac{Q^2 x}{2\varepsilon_0 A}$ for a gap $x$, the force is $F=\dfrac{dU}{dx}=\dfrac{Q^2}{2\varepsilon_0 A}$, which can be rewritten $F=\tfrac{1}{2}QE$ — the average of the field acting on the plate charge, the factor $\tfrac{1}{2}$ arising because a plate does not act on itself. Care is needed to specify whether $Q$ or $V$ is held fixed while the plates move, since at constant voltage the battery also exchanges energy with the system. 🔉⇢

The result $u=\dfrac{1}{2}\varepsilon_0 E^2$ deserves emphasis because it changes how we think about where energy resides. Rather than picturing energy as belonging to the charges on the plates, we assign it to the field filling the space between them, at every point in proportion to the square of the local field strength. Integrating this density over all space always recovers the total energy $\dfrac{Q^2}{2C}$, and the picture generalises far beyond capacitors: any electric field, however produced, carries energy of density $\dfrac{1}{2}\varepsilon_0 E^2$. This field-energy viewpoint becomes essential later, when electromagnetic waves are shown to transport energy stored in their own electric (and magnetic) fields through empty space. 🔉⇢

It is finally worth stating the division rules that make series and parallel networks intuitive. In a series pair the common charge is $Q=\dfrac{V}{\frac{1}{C_1}+\frac{1}{C_2}}=\dfrac{C_1C_2}{C_1+C_2}V$, and the voltage divides in inverse proportion to capacitance — the smaller capacitor takes the larger share of the voltage. In a parallel pair the common voltage fixes each charge as $Q_i=C_iV$, so the charge divides in direct proportion to capacitance — the larger capacitor holds the larger share of the charge. Knowing which quantity is shared and how the other divides lets one write down the state of any two-element block by inspection, and repeated application unravels the most elaborate network. 🔉⇢

A short check on the two ways of computing total energy is reassuring. Take the four-capacitor network above, reduced to an equivalent $13.3\ \mu\text{F}$ across $500\ \text{V}$: the total stored energy is $U=\tfrac{1}{2}CV^2=\tfrac{1}{2}(13.3\times10^{-6})(500)^2\approx1.66\ \text{J}$. The same total must emerge from summing $\tfrac{1}{2}Q_iV_i$ over the individual capacitors — the series trio each holding $Q\approx1.7\times10^{-3}\ \text{C}$ at their own smaller voltages, plus $C_4$ holding $5.0\times10^{-3}\ \text{C}$ at the full $500\ \text{V}$. That the network-level and element-level sums agree is a direct consequence of energy conservation and provides a valuable self-check on any network calculation. 🔉⇢

The stored energy is not a mere accounting quantity; it is genuinely available to do work when the capacitor discharges. Connecting a charged capacitor across a load lets the banked energy $\tfrac{1}{2}CV^2$ flow out, powering a circuit for a short time or delivering a rapid, intense burst as in a photographic flash or a defibrillator. Because the energy scales as the square of the voltage, doubling the charging voltage quadruples the stored energy, which is why high-voltage capacitor banks can store and release very large energies from modest capacitances. This release mirrors the charging process in reverse: the field between the plates collapses back toward zero, and the energy that had been stored in that field is handed over, piece by piece, to the circuit, exactly reversing the bit-by-bit work that was done to build the charge up in the first place. 🔉⇢

Bringing the two halves of this card together, the combination rules and the energy formulae are two faces of the same physics. Whenever capacitors are joined, charge and energy must both be tracked: charge is conserved and merely redistributes, but stored electrostatic energy can decrease if a dissipative transient carries charge from a higher to a lower potential. The equivalent capacitance tells you how the network responds as a whole; the per-element charges and voltages tell you the internal state; and the energy relations $U=\dfrac{Q^2}{2C}=\tfrac{1}{2}CV^2=\tfrac{1}{2}QV$ with the field density $u=\tfrac{1}{2}\varepsilon_0 E^2$ tell you how much work the configuration can do and where that capacity is stored. Together they close the study of capacitors that this chapter set out to build. 🔉⇢

One practical closing remark on energy in networks: to find the total stored energy, it is usually cleanest to reduce the network to its equivalent capacitance $C$ and apply $U=\tfrac{1}{2}CV^2$ with the source voltage, or to find each capacitor's own $Q$ and $V$ and sum $\tfrac{1}{2}Q_i V_i$ over all of them — the two methods must agree. Watching where charge is common (series) and where voltage is common (parallel) keeps the energy bookkeeping honest, and remembering that reconnecting capacitors can dissipate energy prevents the common blunder of assuming stored energy is always conserved when charge is merely rearranged. 🔉⇢

Derivation from first principles 🔉⇢

  1. Series, common charge: the isolated conductor joining $C_1$ and $C_2$ was neutral, so if the outer plates carry $\pm Q$ the inner plates must carry $\mp Q$; any imbalance would leave a field in the connector and drive charge until the net charge is zero, so every series capacitor carries the same $Q$.
  2. Series formula: potential drops add, $V=V_1+V_2+\cdots+V_n=\dfrac{Q}{C_1}+\cdots+\dfrac{Q}{C_n}$; writing the block as $C=\dfrac{Q}{V}$ gives $\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots+\dfrac{1}{C_n}$, always less than the smallest member.
  3. Parallel, common voltage: the same $V$ sits across each capacitor, so $Q_i=C_i V$; the equivalent capacitor holds $Q=\sum_i Q_i=V\sum_i C_i$ at voltage $V$, giving $C=\dfrac{Q}{V}=C_1+C_2+\cdots+C_n$, always more than the largest member.
  4. Work to charge: at an intermediate charge $Q'$ the voltage is $V'=\dfrac{Q'}{C}$, so moving a further $\delta Q'$ costs $\delta W=V'\,\delta Q'=\dfrac{Q'}{C}\,\delta Q'$.
  5. Integrate: $W=\displaystyle\int_0^{Q}\dfrac{Q'}{C}\,dQ'=\dfrac{Q^2}{2C}$; using $C=\dfrac{Q}{V}$ this is equivalent to $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$, the factor $\tfrac{1}{2}$ coming from the linearly rising voltage.
  6. Field energy: for parallel plates put $C=\dfrac{\varepsilon_0 A}{d}$ and $Q=\sigma A$ into $U=\dfrac{Q^2}{2C}$ to get $U=\dfrac{\sigma^2}{2\varepsilon_0}Ad$; with $\sigma=\varepsilon_0 E$ this is $U=\dfrac{1}{2}\varepsilon_0 E^2\,(Ad)$.
  7. Energy density: since $Ad$ is the volume occupied by the field, dividing gives the energy per unit volume $u=\dfrac{1}{2}\varepsilon_0 E^2$, a general result valid for the field of any charge configuration, not just a parallel plate capacitor.
⚠️ JEE trap: A frequent slip is to forget the factor of $\tfrac{1}{2}$ and write the stored energy as $QV$ instead of $\tfrac{1}{2}QV$; the half is there because the voltage rises from $0$ to $V$ as the capacitor charges, so the work is done against the average voltage $\tfrac{V}{2}$, not the full $V$. A second, more conceptual error appears in the classic sharing problem: students expect the total electrostatic energy to be conserved when a charged capacitor is connected to an identical uncharged one. It is not — the charge is conserved and shared equally, but exactly half the energy is lost as heat and radiation during the transient current, so the final energy is only $\tfrac{1}{4}QV$. A third error is mixing up the combination rules: in series the charge is common and reciprocals add ($\tfrac{1}{C}=\sum\tfrac{1}{C_i}$), while in parallel the voltage is common and capacitances add directly ($C=\sum C_i$). 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A $900\ \text{pF}$ capacitor is charged by a $100\ \text{V}$ battery. It is then disconnected from the battery and connected across a second, identical, initially uncharged $900\ \text{pF}$ capacitor.
TARGET Find the energy stored initially by the single charged capacitor and the total energy stored by the two-capacitor system after they are connected, and account for any difference.
STRATEGY Use $Q=CV$ and $U=\tfrac{1}{2}CV^2=\tfrac{1}{2}QV$ for the initial state. After connection, apply charge conservation to find the shared charge and common voltage, then compute the total energy and compare with the initial value.
EXECUTE Initial charge: $Q=CV=900\times10^{-12}\times100=9\times10^{-8}\ \text{C}$. Initial energy: $U=\tfrac{1}{2}QV=\tfrac{1}{2}(9\times10^{-8})(100)=4.5\times10^{-6}\ \text{J}$. After connection the two identical capacitors share the charge equally, so each holds $\dfrac{Q}{2}$ and the common voltage is $V'=\dfrac{Q/2}{C}=\dfrac{V}{2}=50\ \text{V}$. Total final energy: $U'=2\times\tfrac{1}{2}C V'^2=2\times\tfrac{1}{2}(900\times10^{-12})(50)^2=2.25\times10^{-6}\ \text{J}$.
REFLECT Although no charge is lost, the final energy $2.25\times10^{-6}\ \text{J}$ is exactly half the initial $4.5\times10^{-6}\ \text{J}$. During the brief transient a current flows from the first capacitor to the second, and the missing half is dissipated as heat in the wires and as electromagnetic radiation. This is the standard reminder that shared charge does not mean conserved electrostatic energy.

Source: NCERT XII Ch 2 (§2.15)

✍️ Worked Examples Polya 5-move · JEE tier

WE1 · Electrostatic Potential and Potential Difference · JEE Advanced 🔉⇢

SITUATION A point charge of $Q=4\times 10^{-7}\ \text{C}$ is fixed in vacuum, and a point $P$ lies $9\ \text{cm}$ away from it.
TARGET (a) Calculate the electrostatic potential at $P$; (b) hence find the work done in bringing a charge $q=2\times 10^{-9}\ \text{C}$ from infinity to $P$, and state whether the answer depends on the path taken.
STRATEGY Use the point-charge potential $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$ with $\dfrac{1}{4\pi\varepsilon_0}=9\times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$, then obtain the work as $W=qV$, since the potential is the work per unit charge to arrive from infinity.
EXECUTE (a) $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}=(9\times 10^{9})\dfrac{4\times 10^{-7}}{0.09}=4\times 10^{4}\ \text{V}$. (b) $W=qV=(2\times 10^{-9})(4\times 10^{4})=8\times 10^{-5}\ \text{J}$.
REFLECT The work is positive because both charges are positive, so we must push the second charge in against repulsion. It does not depend on the path: any infinitesimal path can be resolved into a step along $\vec{r}$ and a step perpendicular to it, and the perpendicular step does no work, so only the radial separation matters. This path-independence is the defining mark of the conservative electrostatic force.

Source: NCERT XII Ch 2 (§2.2, §2.3)

WE2 · Potential Due to an Electric Dipole · JEE Advanced 🔉⇢

SITUATION A short electric dipole has dipole moment magnitude $p$ and is centred at the origin with its axis along a fixed direction. A point $P$ lies at distance $r$ from the centre, with $r\gg a$, and makes an angle $\theta$ with the dipole axis.
TARGET Find the potential at $P$ for a general $\theta$, then evaluate it (i) on the axis at $\theta=0$, (ii) on the equatorial plane at $\theta=\pi/2$, and comment on the ratio of the axial dipole potential to a single point charge's potential at the same distance.
STRATEGY Apply the dipole potential $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$, which follows from superposing the two charge potentials and expanding for $r\gg a$; then substitute the two special angles and compare the distance dependence with the point-charge result $V_{\text{point}}=\dfrac{Q}{4\pi\varepsilon_0 r}$.
EXECUTE General: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$. (i) On the axis, $\cos 0=1$, so $V_{\text{axis}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$. (ii) On the equatorial plane, $\cos(\pi/2)=0$, so $V_{\text{eq}}=0$. Comparing $V_{\text{axis}}\propto 1/r^2$ with $V_{\text{point}}\propto 1/r$, the dipole potential decays one power of $r$ faster.
REFLECT The equatorial potential vanishes because $P$ is equidistant from $+q$ and $-q$, so their scalar contributions cancel exactly; note this is not the same as the field being zero there. The axial potential is nonzero and positive towards the $+q$ end. The extra factor of $1/r$ relative to a point charge is the hallmark of a neutral source whose leading behaviour is dipolar, and it is exactly what the cancellation of the monopole term in the derivation produced.

Source: NCERT XII Ch 2 (§2.4)

WE3 · Potential Energy of a System of Charges · JEE Advanced 🔉⇢

SITUATION Two point charges $q_1=7\ \mu\text{C}$ and $q_2=-2\ \mu\text{C}$ are placed on the $x$-axis at $(-9\ \text{cm},0,0)$ and $(9\ \text{cm},0,0)$ respectively, so their separation is $r_{12}=18\ \text{cm}$, with no external field.
TARGET (a) Determine the electrostatic potential energy of this two-charge system, and (b) find the work required to separate the two charges infinitely far from each other.
STRATEGY Use the two-charge assembly energy $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$ with $\dfrac{1}{4\pi\varepsilon_0}=9\times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$; then, since $U(\infty)=0$ by convention, the separation work is $W=U_{\text{final}}-U_{\text{initial}}=0-U$.
EXECUTE (a) $U=(9\times 10^{9})\dfrac{(7\times 10^{-6})(-2\times 10^{-6})}{0.18}=9\times 10^{9}\times\dfrac{-14\times 10^{-12}}{0.18}=-0.7\ \text{J}$. (b) $W=U_{\text{final}}-U_{\text{initial}}=0-(-0.7)=+0.7\ \text{J}$.
REFLECT The negative potential energy reflects the attraction between the unlike charges: the system is bound, so a positive amount of external work ($0.7\ \text{J}$) must be supplied to pull them apart to infinity, where the energy is defined as zero. Had both charges been of the same sign, $U$ would have been positive, and the charges would tend to fly apart on release, releasing that stored energy. Note the result is independent of which charge we imagine bringing in first.

Source: NCERT XII Ch 2 (§2.7, §2.8)

WE4 · Electrostatics of Conductors · JEE Advanced 🔉⇢

SITUATION A hollow, uncharged, isolated metallic sphere is placed in a uniform external electrostatic field $E_0$. A tiny cavity inside the metal contains a delicate voltmeter and a sensitive circuit.
TARGET State the electric field inside the metal and inside the empty cavity, describe where any induced charge sits, and explain why the instrument in the cavity is protected.
STRATEGY Apply the four core conductor results: zero interior field, surface-only charge, the conductor as an equipotential, and electrostatic shielding of a charge-free cavity.
EXECUTE The free electrons redistribute over the outer surface until the induced surface charges create a field that exactly cancels $E_0$ throughout the metal, so the field in the conducting material is $E=0$. Because the sphere is neutral overall, equal amounts of induced negative and positive charge appear on opposite regions of the outer surface only; none appears on the cavity wall (no interior charge in the static state). The empty cavity therefore has zero field, $E_{cavity}=0$, regardless of how large $E_0$ is. Since $E=0$ within the metal and on the cavity wall, the entire conductor including the cavity boundary sits at one constant potential.
REFLECT The cavity is a field-free, single-potential region — a Faraday cage — so the voltmeter and circuit are shielded from the external field. This is exactly why sensitive electronics are enclosed in metal, and it rests entirely on the general shielding result that a charge-free cavity in any conductor has zero field.

Source: NCERT XII Ch 2 (§2.9)

WE5 · Problem 1 · easy 🔉⇢

SITUATION A point charge $Q=4\times10^{-7}\ \text{C}$ sits in vacuum. A point P lies $9\ \text{cm}$ from it. We then wish to carry a small charge $q=2\times10^{-9}\ \text{C}$ from infinity to P.
TARGET Find (a) the electrostatic potential at P, and (b) the work required to bring the charge $q$ from infinity to P, and state whether that work depends on the path taken.
STRATEGY The potential of an isolated point charge is $V=\dfrac{Q}{4\pi\varepsilon_0 r}$ (with $\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \text{N m}^2\text{C}^{-2}$). By definition, potential is work per unit positive charge from infinity, so the work to bring $q$ is simply $W=qV$. Because the electrostatic force is conservative, $W$ depends only on the endpoints.
EXECUTE (a) $V=\dfrac{Q}{4\pi\varepsilon_0 r}=9\times10^{9}\times\dfrac{4\times10^{-7}}{0.09}=4\times10^{4}\ \text{V}$. (b) $W=qV=(2\times10^{-9})(4\times10^{4})=8\times10^{-5}\ \text{J}$. The work is path-independent because the Coulomb force is conservative.
REFLECT The answer $8\times10^{-5}\ \text{J}$ is positive, consistent with pushing a positive charge toward a positive source. Any curved path decomposes into a radial part (which does the work) and a perpendicular part (which does none), so only the start and end radii matter.

Source: NCERT XII Ch 2, Example 2.1 (derived)

WE6 · Problem 2 · easy 🔉⇢

SITUATION In a region the electrostatic potential at point R is $V_R=+120\ \text{V}$ and at point P is $V_P=+40\ \text{V}$. A charge $q=+5\times10^{-6}\ \text{C}$ is carried slowly from R to P.
TARGET Find the work done by the external agent, and hence the change in the charge's electrostatic potential energy.
STRATEGY Work done by an external force in moving a charge (without acceleration) between two points equals the charge times the potential difference: $W_{RP}=q(V_P-V_R)$. This work equals the change in potential energy $\Delta U=U_P-U_R$.
EXECUTE $W_{RP}=q(V_P-V_R)=5\times10^{-6}\,(40-120)=5\times10^{-6}\times(-80)=-4\times10^{-4}\ \text{J}$. Hence $\Delta U=U_P-U_R=-4\times10^{-4}\ \text{J}$.
REFLECT The negative sign says the charge moves to lower potential energy, which is natural: a positive charge is pushed 'downhill' from high potential ($120\ \text{V}$) to low potential ($40\ \text{V}$), so the field does positive work and the external agent does negative work.

Source: NCERT XII Ch 2 (§2.1–2.2, derived)

WE7 · Problem 3 · medium 🔉⇢

SITUATION Two point charges $q_1=+2\times10^{-6}\ \text{C}$ and $q_2=+3\times10^{-6}\ \text{C}$ are fixed $0.40\ \text{m}$ apart in vacuum. Point M is the midpoint of the line joining them.
TARGET Find the net electrostatic potential at M.
STRATEGY Potential is a scalar, so by the superposition principle the total potential is the algebraic sum $V=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}\right)$. Each charge is $0.20\ \text{m}$ from M.
EXECUTE $V=9\times10^{9}\left(\dfrac{2\times10^{-6}}{0.20}+\dfrac{3\times10^{-6}}{0.20}\right)=9\times10^{9}\times\dfrac{5\times10^{-6}}{0.20}=2.25\times10^{5}\ \text{V}$.
REFLECT Because potentials add as plain numbers (no direction), the calculation is far simpler than adding the two field vectors, which would require components. At the midpoint the fields of two like charges partly cancel while their potentials simply add.

Source: NCERT XII Ch 2 (§2.3, derived)

WE8 · Problem 4 · medium 🔉⇢

SITUATION A charge $+1\times10^{-6}\ \text{C}$ is placed at the origin and a charge $-2\times10^{-6}\ \text{C}$ at $x=0.60\ \text{m}$ on the x-axis.
TARGET Locate the point (or points) on the x-axis between the charges where the net potential is zero.
STRATEGY Set the sum of the two potentials to zero: $\dfrac{q_1}{x}+\dfrac{q_2}{0.60-x}=0$. Since potentials add algebraically, a zero occurs where the positive contribution equals the magnitude of the negative one: $\dfrac{1}{x}=\dfrac{2}{0.60-x}$.
EXECUTE $\dfrac{1}{x}=\dfrac{2}{0.60-x}\Rightarrow 0.60-x=2x\Rightarrow 3x=0.60\Rightarrow x=0.20\ \text{m}$ from the $+1\,\mu\text{C}$ charge. Solving on the far side of the positive charge gives a second zero at $x=-0.60\ \text{m}$.
REFLECT A zero-potential point is not a zero-field point: at $x=0.20\ \text{m}$ the two fields point the same way and add, yet the potentials cancel. Unlike two positive charges (no interior zero), an unlike pair always has such points because the contributions carry opposite signs.

Source: NCERT XII Ch 2 (§2.3, derived)

WE9 · Problem 5 · medium 🔉⇢

SITUATION A short electric dipole has dipole moment $p=2\times10^{-9}\ \text{C m}$. Consider a point at distance $r=0.30\ \text{m}$ from its centre, with $r$ much greater than the dipole size.
TARGET Find the potential at that point when it lies (a) on the dipole axis and (b) on the equatorial (perpendicular bisector) line.
STRATEGY The potential of a short dipole is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$, where $\theta$ is measured from the dipole moment direction. On the axis $\theta=0$ so $\cos\theta=1$; on the equatorial line $\theta=90^\circ$ so $\cos\theta=0$.
EXECUTE (a) Axial: $V=9\times10^{9}\times\dfrac{2\times10^{-9}\times1}{(0.30)^2}=9\times10^{9}\times\dfrac{2\times10^{-9}}{0.09}=200\ \text{V}$. (b) Equatorial: $V=9\times10^{9}\times\dfrac{2\times10^{-9}\times0}{0.09}=0\ \text{V}$.
REFLECT Every point equidistant from the two equal-and-opposite charges is equally close to $+q$ and $-q$, so their potentials cancel exactly on the equatorial line, giving zero. The axial point, closer to one charge, retains a finite potential.

Source: NCERT XII Ch 2 (§2.4, derived)

WE10 · Problem 6 · advanced 🔉⇢

SITUATION For the same short dipole ($p=2\times10^{-9}\ \text{C m}$), compare the potential at $r=0.30\ \text{m}$, $\theta=60^\circ$ with the potential at twice that distance, $r=0.60\ \text{m}$, along the axis ($\theta=0$).
TARGET Compute both potentials and use them to contrast how the dipole potential depends on angle and on distance.
STRATEGY Apply $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ twice. The angular factor is $\cos60^\circ=0.5$; the distance factor scales as $1/r^2$, so doubling $r$ divides the potential by four.
EXECUTE At $r=0.30$, $\theta=60^\circ$: $V_1=9\times10^{9}\dfrac{(2\times10^{-9})(0.5)}{0.09}=100\ \text{V}$. At $r=0.60$, $\theta=0$: $V_2=9\times10^{9}\dfrac{(2\times10^{-9})(1)}{0.36}=50\ \text{V}$. Ratio $V_1/V_2=2$.
REFLECT Two lessons: the dipole potential depends on the angle $\theta$ (not just $r$), unlike a point charge; and it falls as $1/r^2$, faster than a point charge's $1/r$, because the far-field contributions of $+q$ and $-q$ nearly cancel. Doubling $r$ here cut the axial value from $200\ \text{V}$ to $50\ \text{V}$.

Source: NCERT XII Ch 2 (§2.4, derived)

WE11 · Problem 7 · medium 🔉⇢

SITUATION In a region of uniform electric field, two equipotential surfaces are labelled $V_A=100\ \text{V}$ and $V_B=95\ \text{V}$. They are flat parallel planes separated by $2.0\ \text{cm}$ measured along the normal.
TARGET Find the magnitude and direction of the electric field between the surfaces.
STRATEGY For closely spaced equipotentials, $E=-\dfrac{dV}{dr}$, and the field is normal to the surfaces pointing from high to low potential. Magnitude $E=\dfrac{|\Delta V|}{\Delta r}$ using the normal separation.
EXECUTE $E=\dfrac{|\Delta V|}{\Delta r}=\dfrac{100-95}{0.020}=\dfrac{5}{0.020}=250\ \text{V m}^{-1}$, directed normal to the surfaces from the $100\ \text{V}$ plane toward the $95\ \text{V}$ plane.
REFLECT Field lines always cross equipotentials at right angles; if they did not, a component of $E$ would lie along the surface and moving a charge on it would require work, contradicting 'equal potential'. Closely packed equipotentials mean a strong field.

Source: NCERT XII Ch 2 (§2.5–2.6, derived)

WE12 · Problem 8 · advanced 🔉⇢

SITUATION In a region the electrostatic potential is given (in SI units) by $V(x,y,z)=3x^2 y$ volts, with coordinates in metres.
TARGET Find the electric field vector and its magnitude at the point $(1,\,2,\,0)\ \text{m}$.
STRATEGY The field is the negative gradient of potential: $E_x=-\dfrac{\partial V}{\partial x}$, $E_y=-\dfrac{\partial V}{\partial y}$, $E_z=-\dfrac{\partial V}{\partial z}$. Differentiate $V=3x^2y$ with respect to each coordinate, then evaluate at the point.
EXECUTE $E_x=-\dfrac{\partial}{\partial x}(3x^2y)=-6xy$; $E_y=-\dfrac{\partial}{\partial y}(3x^2y)=-3x^2$; $E_z=0$. At $(1,2,0)$: $E_x=-6(1)(2)=-12$, $E_y=-3(1)^2=-3$. So $\vec E=(-12\,\hat i-3\,\hat j)\ \text{V m}^{-1}$, magnitude $|\vec E|=\sqrt{12^2+3^2}=\sqrt{153}\approx12.4\ \text{V m}^{-1}$.
REFLECT The relation $\vec E=-\nabla V$ generalises $E=-dV/dr$ to three dimensions. The minus sign points $\vec E$ toward decreasing potential; note the field varies from point to point here because the potential is non-linear in $x$.

Source: NCERT XII Ch 2 (§2.5, derived)

WE13 · Problem 9 · easy 🔉⇢

SITUATION Two point charges $q_1=7\times10^{-6}\ \text{C}$ and $q_2=-2\times10^{-6}\ \text{C}$ are placed $18\ \text{cm}$ apart in vacuum, with no external field.
TARGET Find the electrostatic potential energy of this two-charge system, and the work needed to separate the charges to infinity.
STRATEGY The interaction energy of a pair of point charges is $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$. The work to pull them infinitely apart is $W=U_\text{final}-U_\text{initial}=0-U=-U$.
EXECUTE $U=9\times10^{9}\times\dfrac{(7\times10^{-6})(-2\times10^{-6})}{0.18}=9\times10^{9}\times\dfrac{-14\times10^{-12}}{0.18}=-0.7\ \text{J}$. Work to separate to infinity: $W=0-(-0.7)=+0.7\ \text{J}$.
REFLECT The negative $U$ signals a bound (attractive) unlike-charge pair, so an external agent must supply $+0.7\ \text{J}$ to tear them apart. Had both charges been positive, $U$ would be positive and they would fly apart on release.

Source: NCERT XII Ch 2, Example 2.5(a,b) (derived)

WE14 · Problem 10 · medium 🔉⇢

SITUATION Three equal point charges $q=1\times10^{-6}\ \text{C}$ are placed at the corners of an equilateral triangle of side $a=0.10\ \text{m}$ in vacuum.
TARGET Find the total work required to assemble this configuration from infinity.
STRATEGY The total electrostatic energy equals the sum over all distinct pairs of $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_iq_j}{r_{ij}}$. There are three identical pairs, each separated by $a$, so $U=3\times\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{a}$.
EXECUTE $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{a}=9\times10^{9}\times\dfrac{(1\times10^{-6})^2}{0.10}=9\times10^{9}\times\dfrac{1\times10^{-12}}{0.10}=0.09\ \text{J}$. Total $U=3\times0.09=0.27\ \text{J}$.
REFLECT Assembly work is path- and order-independent, so we may count each pair exactly once. All three charges are positive, giving positive energy: releasing them lets the triangle explode outward, converting $0.27\ \text{J}$ into kinetic energy.

Source: NCERT XII Ch 2 (§2.7, derived)

WE15 · Problem 11 · medium 🔉⇢

SITUATION An electron (charge magnitude $e=1.6\times10^{-19}\ \text{C}$), initially at rest, is accelerated from a plate to another plate across a potential difference of $200\ \text{V}$ in an external field.
TARGET Find the kinetic energy gained by the electron, expressed both in joules and in electron-volts.
STRATEGY The potential energy of a charge in an external field is $U=qV$, so the energy gained equals $|q|\,\Delta V$. Since $1\ \text{eV}=1.6\times10^{-19}\ \text{J}$ is exactly the energy an electron gains across $1\ \text{V}$, the numeric value in volts equals the energy in eV.
EXECUTE Energy gained $=e\,\Delta V=(1.6\times10^{-19})(200)=3.2\times10^{-17}\ \text{J}$. In electron-volts: $\dfrac{3.2\times10^{-17}}{1.6\times10^{-19}}=200\ \text{eV}$.
REFLECT The electron-volt is defined precisely so that crossing $\Delta V$ volts yields $\Delta V$ eV of energy, which is why the answer is simply '$200\ \text{eV}$'. This unit dominates atomic and nuclear physics because joules are inconveniently large there.

Source: NCERT XII Ch 2 (§2.8.1, derived)

WE16 · Problem 12 · advanced 🔉⇢

SITUATION The two charges $q_1=7\times10^{-6}\ \text{C}$ and $q_2=-2\times10^{-6}\ \text{C}$, still $18\ \text{cm}$ apart, are now placed in an external field described by $E=A/r^2$ with $A=9\times10^{5}\ \text{N C}^{-1}\text{m}^2$, where $r$ is the distance from the field's source origin. Each charge sits at $r=0.09\ \text{m}$ from that origin.
TARGET Find the total electrostatic energy of the configuration in this external field.
STRATEGY Total energy $=q_1V(r_1)+q_2V(r_2)+\dfrac{q_1q_2}{4\pi\varepsilon_0 r_{12}}$. The external potential from a field $E=A/r^2$ is $V=A/r$ (since $-dV/dr=A/r^2$). Add the two 'charge-in-external-field' terms to the mutual interaction energy of $-0.7\ \text{J}$ found earlier.
EXECUTE External potential at each charge: $V=A/r=\dfrac{9\times10^{5}}{0.09}=1\times10^{7}\ \text{V}$. Then $q_1V=7\times10^{-6}\times10^{7}=70\ \text{J}$ and $q_2V=-2\times10^{-6}\times10^{7}=-20\ \text{J}$. Total $=70-20+(-0.7)=49.3\ \text{J}$.
REFLECT The mutual interaction term ($-0.7\ \text{J}$) is unchanged by the external field, exactly as in NCERT Example 2.5(c): the external field only adds the independent $q_iV(r_i)$ terms. The large $49.3\ \text{J}$ is dominated by the interaction with the strong external source.

Source: NCERT XII Ch 2, Example 2.5(c) (derived)

WE17 · Problem 13 · medium 🔉⇢

SITUATION An electric dipole of moment $p=2\times10^{-9}\ \text{C m}$ is placed in a uniform external field $E=1\times10^{5}\ \text{N C}^{-1}$. The angle $\theta$ between $\vec p$ and $\vec E$ can be varied.
TARGET Find the potential energy of the dipole when it is aligned with the field ($\theta=0$), perpendicular to it ($\theta=90^\circ$), and anti-aligned ($\theta=180^\circ$).
STRATEGY The potential energy of a dipole in a uniform field is $U(\theta)=-\vec p\cdot\vec E=-pE\cos\theta$, taking the reference $U=0$ at $\theta=90^\circ$. Substitute the three angles.
EXECUTE $pE=(2\times10^{-9})(1\times10^{5})=2\times10^{-4}\ \text{J}$. Then: $U(0)=-pE\cos0=-2\times10^{-4}\ \text{J}$; $U(90^\circ)=-pE\cos90^\circ=0$; $U(180^\circ)=-pE\cos180^\circ=+2\times10^{-4}\ \text{J}$.
REFLECT Energy is minimum ($-2\times10^{-4}\ \text{J}$) when the dipole aligns with the field, so that is the stable equilibrium; the anti-aligned state is the maximum-energy unstable equilibrium. This is why dipoles (like polar molecules) tend to rotate into alignment with an applied field.

Source: NCERT XII Ch 2 (§2.8.3, derived)

WE18 · Problem 14 · advanced 🔉⇢

SITUATION The same dipole ($p=2\times10^{-9}\ \text{C m}$) in the field $E=1\times10^{5}\ \text{N C}^{-1}$ is to be rotated in the plane containing $\vec E$.
TARGET Find (a) the maximum torque the field exerts on the dipole, and (b) the external work needed to rotate it from the aligned position $\theta=0$ to the fully reversed position $\theta=180^\circ$.
STRATEGY Torque is $\tau=pE\sin\theta$, maximal at $\theta=90^\circ$. Work done by an external agent rotating the dipole equals the change in potential energy, $W=U(\theta_2)-U(\theta_1)=pE(\cos\theta_1-\cos\theta_2)$.
EXECUTE (a) $\tau_{\max}=pE\sin90^\circ=(2\times10^{-9})(1\times10^{5})=2\times10^{-4}\ \text{N m}$. (b) $W=pE(\cos0-\cos180^\circ)=2\times10^{-4}\,(1-(-1))=2pE=4\times10^{-4}\ \text{J}$.
REFLECT Reversing the dipole costs $2pE$, twice the depth of the potential well, because we climb from the minimum at $\theta=0$ up to the maximum at $\theta=180^\circ$. The torque vanishes at both $0$ and $180^\circ$ (equilibria) and peaks broadside at $90^\circ$.

Source: NCERT XII Ch 2 (§2.8.3, derived)

WE19 · Problem 15 · easy 🔉⇢

SITUATION A solid conducting sphere of radius $R=0.10\ \text{m}$ carries a total charge $Q=1\times10^{-8}\ \text{C}$ in electrostatic equilibrium in vacuum.
TARGET Find the electric field and the potential at the sphere's centre, and the potential at its surface.
STRATEGY For a charged conductor in equilibrium the field inside is zero and the whole conductor is an equipotential. Outside it behaves like a point charge, so the surface potential is $V=\dfrac{Q}{4\pi\varepsilon_0 R}$; the interior potential equals this same value (constant), and the interior field is zero.
EXECUTE Field at centre: $E_\text{inside}=0$. Surface potential $V=9\times10^{9}\times\dfrac{1\times10^{-8}}{0.10}=900\ \text{V}$. Since the conductor is equipotential, the potential at the centre is also $900\ \text{V}$.
REFLECT Zero field but non-zero potential inside is not a contradiction: $E=-dV/dr$, and a constant $V$ has zero gradient. All the excess charge resides on the surface, and the interior is shielded, the basis of electrostatic shielding.

Source: NCERT XII Ch 2 (§2.9, derived)

WE20 · Problem 16 · medium 🔉⇢

SITUATION A parallel-plate capacitor has a uniform field $E=1\times10^{6}\ \text{V m}^{-1}$ in the vacuum gap between its plates. Each plate has area $A=0.02\ \text{m}^2$ and the gap is $d=1.0\ \text{mm}$.
TARGET Find the energy density of the field and the total electrostatic energy stored in the gap.
STRATEGY Energy per unit volume of an electric field is $u=\tfrac12\varepsilon_0 E^2$ (a general result). The total stored energy is this density times the field volume $A\,d$: $U=u\,(A d)$. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$.
EXECUTE $u=\tfrac12\varepsilon_0 E^2=\tfrac12(8.85\times10^{-12})(1\times10^{6})^2=\tfrac12(8.85\times10^{-12})(10^{12})=4.43\ \text{J m}^{-3}$. Volume $=A d=0.02\times1.0\times10^{-3}=2.0\times10^{-5}\ \text{m}^3$. Total $U=4.43\times2.0\times10^{-5}\approx8.85\times10^{-5}\ \text{J}$.
REFLECT Viewing the energy as residing in the field itself, at density $\tfrac12\varepsilon_0E^2$, is more general than $\tfrac12CV^2$: it applies to any field configuration, not just capacitors, and underlies the energy carried by electromagnetic waves.

Source: NCERT XII Ch 2 (§2.15, derived)

WE21 · Problem 17 · easy 🔉⇢

SITUATION A capacitor acquires a charge of $1\times10^{-6}\ \text{C}$ on each plate when connected across a potential difference of $200\ \text{V}$.
TARGET Find its capacitance, and the charge it would hold if the voltage were raised to $500\ \text{V}$.
STRATEGY Capacitance is defined by $C=\dfrac{Q}{V}$ and is a fixed geometric property, independent of $Q$ and $V$. Compute $C$, then use $Q'=CV'$ at the new voltage.
EXECUTE $C=\dfrac{Q}{V}=\dfrac{1\times10^{-6}}{200}=5\times10^{-9}\ \text{F}=5\ \text{nF}$. At $V'=500\ \text{V}$: $Q'=CV'=(5\times10^{-9})(500)=2.5\times10^{-6}\ \text{C}$.
REFLECT Charge scales linearly with voltage because $C$ is constant, set only by plate geometry and the dielectric. The farad is huge: $5\ \text{nF}$ is a typical practical value, illustrating why sub-multiples like nF and pF are the common units.

Source: NCERT XII Ch 2 (§2.11, derived)

WE22 · Problem 18 · medium 🔉⇢

SITUATION A parallel-plate capacitor in vacuum has plates of area $A=1.0\ \text{m}^2$ separated by $d=1.0\ \text{mm}$.
TARGET Find its capacitance, and the potential difference produced if a charge $Q=1\times10^{-8}\ \text{C}$ is placed on it.
STRATEGY The parallel-plate result is $C=\dfrac{\varepsilon_0 A}{d}$, which depends only on geometry. Then use $V=Q/C$ (equivalently $V=Ed$ with $E=Q/\varepsilon_0 A$).
EXECUTE $C=\dfrac{\varepsilon_0 A}{d}=\dfrac{(8.85\times10^{-12})(1.0)}{1.0\times10^{-3}}=8.85\times10^{-9}\ \text{F}=8.85\ \text{nF}$. Then $V=\dfrac{Q}{C}=\dfrac{1\times10^{-8}}{8.85\times10^{-9}}\approx1.13\ \text{V}$.
REFLECT To make $C=1\ \text{F}$ at $d=1\ \text{mm}$ would need plates of area $\sim10^{8}\ \text{m}^2$, kilometres across, confirming that the farad is an enormous unit. Capacitance rises with area and falls with separation.

Source: NCERT XII Ch 2, Eqs. 2.43–2.44 (derived)

WE23 · Problem 19 · medium 🔉⇢

SITUATION The vacuum capacitor of the previous problem ($C_0=8.85\ \text{nF}$) is completely filled with a dielectric of constant $K=5$. It remains connected to a battery holding it at $V=100\ \text{V}$.
TARGET Find the new capacitance and the new charge on the plates, and compare with the vacuum case.
STRATEGY Inserting a dielectric multiplies capacitance by $K$: $C=K C_0=\dfrac{K\varepsilon_0 A}{d}$. With the battery keeping $V$ fixed, the charge is $Q=CV$, so it also rises by the factor $K$.
EXECUTE $C=KC_0=5\times8.85\ \text{nF}=44.25\ \text{nF}$. Charge $Q=CV=(44.25\times10^{-9})(100)=4.43\times10^{-6}\ \text{C}$. Without the dielectric it would have been $Q_0=C_0V=8.85\times10^{-7}\ \text{C}$, five times smaller.
REFLECT Because $K\gt1$, the dielectric always increases capacitance: its polarisation partly cancels the field, so more charge is needed to reach the same $100\ \text{V}$. Note that if the battery were disconnected first, $Q$ would stay fixed and instead $V$ would drop by the factor $K$.

Source: NCERT XII Ch 2 (§2.13, derived)

WE24 · Problem 20 · advanced 🔉⇢

SITUATION A slab of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but a thickness of only $\tfrac34 d$, where $d$ is the plate separation. The slab is inserted between the plates while the free charge $Q_0$ on the plates is held fixed.
TARGET Derive the factor by which the capacitance changes.
STRATEGY With fixed free charge, the field in the vacuum part is $E_0=V_0/d$ and inside the slab it is reduced to $E_0/K$. Add the potential drops across the vacuum thickness $\tfrac14 d$ and the slab thickness $\tfrac34 d$ to get the new $V$, then use $C=Q_0/V$.
EXECUTE $V=E_0\left(\tfrac14 d\right)+\dfrac{E_0}{K}\left(\tfrac34 d\right)=E_0 d\left(\dfrac14+\dfrac{3}{4K}\right)=V_0\,\dfrac{K+3}{4K}$. Hence $C=\dfrac{Q_0}{V}=\dfrac{4K}{K+3}\,C_0$. For example, with $K=4$: $C=\dfrac{16}{7}C_0\approx2.3\,C_0$.
REFLECT The slab need not fill the whole gap to boost capacitance; a partial slab gives $C=\dfrac{4K}{K+3}C_0$, always greater than $C_0$ since $K\gt1$. As $K\to\infty$ (a conductor slab of thickness $\tfrac34d$) the factor approaches $4$, matching the intuition that the effective gap shrinks to $\tfrac14 d$.

Source: NCERT XII Ch 2, Example 2.8 (derived)

WE25 · Problem 21 · easy 🔉⇢

SITUATION Two capacitors $C_1=6\ \mu\text{F}$ and $C_2=3\ \mu\text{F}$ are connected in series across a $60\ \text{V}$ battery.
TARGET Find the equivalent capacitance, the common charge on each capacitor, and the voltage across each.
STRATEGY In series the charge on each capacitor is the same and reciprocals add: $\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$. Then $Q=CV$, and each individual voltage is $V_i=Q/C_i$.
EXECUTE $\dfrac{1}{C}=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}\Rightarrow C=2\ \mu\text{F}$. Charge $Q=CV=(2\times10^{-6})(60)=1.2\times10^{-4}\ \text{C}$. Voltages: $V_1=Q/C_1=\dfrac{1.2\times10^{-4}}{6\times10^{-6}}=20\ \text{V}$, $V_2=Q/C_2=\dfrac{1.2\times10^{-4}}{3\times10^{-6}}=40\ \text{V}$.
REFLECT Series capacitance ($2\ \mu\text{F}$) is smaller than either member, opposite to resistors in series. The smaller capacitor ($C_2$) takes the larger share of voltage ($40\ \text{V}$), and $V_1+V_2=60\ \text{V}$ checks the total.

Source: NCERT XII Ch 2 (§2.14.1, derived)

WE26 · Problem 22 · medium 🔉⇢

SITUATION Two capacitors $C_1=2\ \mu\text{F}$ and $C_2=3\ \mu\text{F}$ are connected in parallel across a $100\ \text{V}$ supply.
TARGET Find the equivalent capacitance, the charge on each capacitor, and the total charge drawn.
STRATEGY In parallel the voltage across each capacitor is the same and capacitances add: $C=C_1+C_2$. Each charge is $Q_i=C_iV$, and the total is $Q=CV=Q_1+Q_2$.
EXECUTE $C=C_1+C_2=2+3=5\ \mu\text{F}$. $Q_1=C_1V=(2\times10^{-6})(100)=2\times10^{-4}\ \text{C}$; $Q_2=C_2V=(3\times10^{-6})(100)=3\times10^{-4}\ \text{C}$. Total $Q=CV=(5\times10^{-6})(100)=5\times10^{-4}\ \text{C}=Q_1+Q_2$.
REFLECT Parallel capacitance is the sum, exceeding either member, because the plate areas effectively combine. The larger capacitor stores proportionally more charge at the same shared voltage, and the individual charges add to the total, confirming charge conservation.

Source: NCERT XII Ch 2 (§2.14.2, derived)

WE27 · Problem 23 · advanced 🔉⇢

SITUATION Four capacitors, each of $10\ \mu\text{F}$, form a network across a $500\ \text{V}$ supply: $C_1$, $C_2$, $C_3$ are in series with one another, and that series branch is in parallel with $C_4$.
TARGET Find (a) the equivalent capacitance of the network and (b) the charge on the series capacitors and on $C_4$.
STRATEGY First combine $C_1,C_2,C_3$ in series to get $C'$, then add $C_4$ in parallel: $C=C'+C_4$. The full $500\ \text{V}$ appears across both the series branch and $C_4$. The series branch carries a single charge $Q=C'\times500$; $C_4$ carries $Q'=C_4\times500$.
EXECUTE Series: $\dfrac{1}{C'}=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{3}{10}\Rightarrow C'=\dfrac{10}{3}\ \mu\text{F}$. Equivalent: $C=\dfrac{10}{3}+10=\dfrac{40}{3}\approx13.3\ \mu\text{F}$. Charge on each of $C_1,C_2,C_3$: $Q=C'V=\dfrac{10}{3}\times10^{-6}\times500\approx1.7\times10^{-3}\ \text{C}$. Charge on $C_4$: $Q'=C_4V=(10\times10^{-6})(500)=5.0\times10^{-3}\ \text{C}$.
REFLECT The three series capacitors share the same charge $1.7\ \text{mC}$ (series rule), while $C_4$, seeing the full $500\ \text{V}$ alone, holds a much larger $5.0\ \text{mC}$. Reducing three capacitors in series to $10/3\ \mu\text{F}$ shows how series stacking weakens capacitance.

Source: NCERT XII Ch 2, Example 2.9 (derived)

WE28 · Problem 24 · medium 🔉⇢

SITUATION A $900\ \text{pF}$ capacitor is charged by a $100\ \text{V}$ battery.
TARGET Find the charge stored and the electrostatic energy stored in the capacitor.
STRATEGY Charge is $Q=CV$. The stored energy has three equivalent forms $U=\tfrac12 CV^2=\tfrac12 QV=\dfrac{Q^2}{2C}$; use $\tfrac12 CV^2$ directly from the given $C$ and $V$.
EXECUTE $Q=CV=(900\times10^{-12})(100)=9\times10^{-8}\ \text{C}$. Energy $U=\tfrac12 CV^2=\tfrac12(900\times10^{-12})(100)^2=\tfrac12(900\times10^{-12})(10^{4})=4.5\times10^{-6}\ \text{J}$.
REFLECT The factor $\tfrac12$ arises because the plate voltage grows from $0$ to $V$ as charge accumulates, so the average work per unit charge is $\tfrac12 V$, not $V$. Checking with $\tfrac12 QV=\tfrac12(9\times10^{-8})(100)=4.5\times10^{-6}\ \text{J}$ agrees.

Source: NCERT XII Ch 2, Example 2.10(a) (derived)

WE29 · Problem 25 · advanced 🔉⇢

SITUATION The $900\ \text{pF}$ capacitor, charged to $100\ \text{V}$ (so $Q=9\times10^{-8}\ \text{C}$, energy $4.5\ \mu\text{J}$), is disconnected from the battery and then connected to a second, identical, uncharged $900\ \text{pF}$ capacitor.
TARGET Find the common final voltage, the final total energy stored, and the energy that appears to be lost.
STRATEGY Charge is conserved and redistributes until both capacitors reach a common potential $V'$. With equal capacitances the charge splits equally, so $Q'=Q/2$ on each and $V'=V/2$. The final energy is $U'=\tfrac12(2C)V'^2$; compare it with the initial $4.5\ \mu\text{J}$.
EXECUTE Common voltage $V'=V/2=50\ \text{V}$; each capacitor holds $Q'=Q/2=4.5\times10^{-8}\ \text{C}$. Final energy $U'=\tfrac12(2C)V'^2=\tfrac12(1800\times10^{-12})(50)^2=\tfrac12(1800\times10^{-12})(2500)=2.25\times10^{-6}\ \text{J}$. Energy 'lost' $=4.5-2.25=2.25\ \mu\text{J}$, i.e. exactly half.
REFLECT No charge is lost, yet half the energy disappears: during the transient a current flows between the capacitors and dissipates energy as heat and electromagnetic radiation in the connecting wires. This charge-sharing loss is unavoidable and independent of the wire resistance in the ideal limit.

Source: NCERT XII Ch 2, Example 2.10(b) (derived)

On the concept tabs

These worked examples are taught in full alongside their interactive scene:

📐 Formula Sheet Printable · every formula cited

Potential and Potential Difference

QuantityFormulaWhat it means / when to useSource
Potential difference (definition) 🔉⇢$V_B-V_A=\dfrac{W_{AB}}{q}=-\displaystyle\int_A^B \mathbf{E}\!\cdot\!d\mathbf{l}$Work done per unit charge against the field in moving a test charge from A to B. SI unit: volt (V) $=$ J C$^{-1}$. Potential is a scalar.NCERT XII Ch 2 (§2.2)
Potential due to a point charge 🔉⇢$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}=\dfrac{kq}{r}$Reference $V=0$ at infinity. Falls off as $1/r$ and carries the sign of $q$. $k=1/4\pi\varepsilon_0=9\times10^9$ N m$^2$ C$^{-2}$.NCERT XII Ch 2 (§2.3)
Potential of a system of charges 🔉⇢$V=\dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_i \dfrac{q_i}{r_i}$Scalar (algebraic) superposition — add the potentials, not vectors. No angles involved, unlike the field.NCERT XII Ch 2 (§2.5)
Potential due to a short dipole 🔉⇢$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$$p=q(2a)$ is the dipole moment; $\theta$ measured from the dipole axis. Falls as $1/r^2$; zero on the perpendicular bisector ($\theta=90^\circ$).NCERT XII Ch 2 (§2.4)

Field–Potential Relation and Potential Energy

QuantityFormulaWhat it means / when to useSource
Field from potential 🔉⇢$E=-\dfrac{dV}{dr}\quad(\mathbf{E}=-\nabla V)$The field is the negative gradient of the potential; it points toward decreasing $V$ and is perpendicular to equipotential surfaces. Unit V m$^{-1}$ = N C$^{-1}$.NCERT XII Ch 2 (§2.6.1)
PE of two point charges 🔉⇢$U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$Work to assemble the pair from infinity. For a system, sum over every distinct pair. Positive for like charges, negative for unlike.NCERT XII Ch 2 (§2.7)
PE of a charge in an external field 🔉⇢$U=qV(\mathbf{r})$$V$ is the potential of the external field at the charge's location. Converts to kinetic energy via the work-energy theorem.NCERT XII Ch 2 (§2.8)
PE of a dipole in a uniform field 🔉⇢$U=-\mathbf{p}\!\cdot\!\mathbf{E}=-pE\cos\theta$Minimum ($-pE$) when aligned ($\theta=0$), maximum ($+pE$) when anti-aligned. Torque $\tau=pE\sin\theta$.NCERT XII Ch 2 (§2.8.3)

Conductors, Dielectrics and Capacitance

QuantityFormulaWhat it means / when to useSource
Field just outside a charged conductor 🔉⇢$E=\dfrac{\sigma}{\varepsilon_0}\,\hat{\mathbf{n}}$$\sigma$ = local surface charge density. Field is normal to the surface; zero inside the conductor, which is an equipotential volume.NCERT XII Ch 2 (§2.9)
Capacitance (definition) 🔉⇢$C=\dfrac{Q}{V}$Charge stored per unit potential difference. Geometric property. SI unit farad (F) = C V$^{-1}$; practical units $\mu$F, pF.NCERT XII Ch 2 (§2.11)
Parallel-plate capacitor 🔉⇢$C=\dfrac{\varepsilon_0 A}{d}\;\xrightarrow{\text{dielectric}}\;\dfrac{K\varepsilon_0 A}{d}$$A$ = plate area, $d$ = separation. A dielectric of constant $K$ filling the gap multiplies the capacitance by $K$.NCERT XII Ch 2 (§2.12, §2.13)
Dielectric constant 🔉⇢$K=\dfrac{C}{C_0}=\dfrac{E_0}{E}$Ratio of capacitance with to without the dielectric; equivalently the factor by which the dielectric reduces the field inside it. $K\ge 1$.NCERT XII Ch 2 (§2.13)

Combinations and Energy Stored

QuantityFormulaWhat it means / when to useSource
Capacitors in parallel 🔉⇢$C_{\text{eq}}=C_1+C_2+\dots$Same voltage across each; charges add. Equivalent capacitance is larger than the largest member.NCERT XII Ch 2 (§2.14.2)
Capacitors in series 🔉⇢$\dfrac{1}{C_{\text{eq}}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dots$Same charge on each; voltages add. Equivalent capacitance is smaller than the smallest member.NCERT XII Ch 2 (§2.14.1)
Energy stored in a capacitor 🔉⇢$U=\tfrac12 CV^2=\tfrac12 QV=\dfrac{Q^2}{2C}$Work done in charging. Use $\tfrac12 CV^2$ at fixed $V$, $Q^2/2C$ at fixed $Q$ (the safe form when a dielectric is inserted with the battery disconnected).NCERT XII Ch 2 (§2.15)
Energy density of the field 🔉⇢$u=\tfrac12\varepsilon_0 E^2$Energy per unit volume stored in the electric field itself; integrating it over the field region recovers the capacitor's stored energy.NCERT XII Ch 2 (§2.15)

📜 Previous-Year Questions Authentic NTA · 62 questions

Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.

JEE Main 2021 · Paper 1 · July 27 Shift 1 · Q15 (official key (printed in paper)) Answer: (C)

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

  • (A) ${V \over {K + 2}}$
  • (B) ${V \over K}$
  • (C) ${{3V} \over {K + 2}}$
  • (D) ${{3V} \over K}$
JEE Main 2021 · Paper 1 · August 26 Shift 1 · Q19 (official key (printed in paper)) Answer: (C)

The material filled between the plates of a parallel plate capacitor has resistivity 200 $\Omega$m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permittivity of material is 50)

  • (A) 9.0 $\mu$A
  • (B) 9.0 mA
  • (C) 0.9 mA
  • (D) 0.9 $\mu$A
JEE Main 2021 · Paper 1 · March 16 Shift 1 · Q2 (official key (printed in paper)) Answer: (C)

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is ${3 \over 4}$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance ($C_{0}$) is given by the following relation :

  • (A) $C' = {{3 + K} \over {4K}}{C_0}$
  • (B) $C' = {{4 + K} \over {3}}{C_0}$
  • (C) $C' = {{4K} \over {K + 3}}{C_0}$
  • (D) $C' = {{4} \over {3 + K}}{C_0}$
JEE Main 2023 · Paper 1 · January 31 Shift 2 · Q17 (official key (printed in paper)) Answer: (A)

Considering a group of positive charges, which of the following statements is correct ?

  • (A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point
  • (B) Net potential of the system at a point can be zero but net electric field can't be zero at that point.
  • (C) Both the net potential and the net electric field cannot be zero at a point.
  • (D) Both the net potential and the net field can be zero at a point.
JEE Main 2023 · Paper 1 · January 25 Shift 2 · Q21 (official key (printed in paper)) Answer: 6 (numerical value)

A capacitor has capacitance 5$\mu$F when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $\frac{d}{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ $\mu$F.

JEE Main 2023 · Paper 1 · January 25 Shift 1 · Q7 (official key (printed in paper)) Answer: (C)

A parallel plate capacitor has plate area 40 cm$^2$ and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :

  • (A) $\mathrm{10\varepsilon_0~F}$
  • (B) $\mathrm{24\varepsilon_0~F}$
  • (C) $\mathrm{\frac{3}{10}\varepsilon_0~F}$
  • (D) $\mathrm{\frac{10}{3}\varepsilon_0~F}$
IIT-JEE 2008 · Paper 2 · Q30 (official key) Answer: A

A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K = 2$. The level of liquid is $\dfrac{d}{3}$ initially. The capacitor is connected in series with a resistance $R$ and a battery. Suppose the liquid level decreases at a constant speed $V$, the time constant as a function of time $t$ is

  • (A) $\dfrac{6\varepsilon_0 R}{5d+3Vt}$
  • (B) $\dfrac{(15d+9Vt)\varepsilon_0 R}{2d^2-3dVt-9V^2t^2}$
  • (C) $\dfrac{6\varepsilon_0 R}{5d-3Vt}$
  • (D) $\dfrac{(15d-9Vt)\varepsilon_0 R}{2d^2+3dVt-9V^2t^2}$
IIT-JEE 2008 · Paper 2 · Q34 (official key) Answer: A

STATEMENT-1: For practical purposes, the earth is used as a reference at zero potential in electrical circuits. and STATEMENT-2: The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\dfrac{Q}{4\pi\varepsilon_0 R}$.

  • (A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
  • (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
  • (C) STATEMENT-1 is True, STATEMENT-2 is False
  • (D) STATEMENT-1 is False, STATEMENT-2 is True
IIT-JEE 2011 · Paper 1 · Q30 (official key) Answer: D

A 2 $\mu$F capacitor is charged by a source $V$ with the switch S in position 1. The percentage of its stored energy dissipated after the switch S is turned to position 2, which disconnects the source and connects the charged 2 $\mu$F capacitor across an uncharged 8 $\mu$F capacitor, is

  • (A) 0%
  • (B) 20%
  • (C) 75%
  • (D) 80%
JEE Advanced 2014 · Paper 1 · Q4 (official key) Answer: A, D

A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1/3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. The electric field in the dielectric is $E_1$ and that in the other portion is $E_2$. Choose the correct option/options, ignoring edge effects.

  • (A) $\dfrac{E_1}{E_2} = 1$
  • (B) $\dfrac{E_1}{E_2} = \dfrac{1}{K}$
  • (C) $\dfrac{Q_1}{Q_2} = \dfrac{3}{K}$
  • (D) $\dfrac{C}{C_1} = \dfrac{2+K}{K}$
JEE Advanced 2015 · Paper 2 · Q14 (official key) Answer: D

A parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C_1$ in air. When two dielectrics of different relative permittivities ($\varepsilon_1 = 2$ and $\varepsilon_2 = 4$) are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\dfrac{C_2}{C_1}$ is [From the figure: the plate area is divided into two equal halves of area $S/2$. Over one half the dielectric $\varepsilon_1$ fills the whole gap $d$. Over the other half, $\varepsilon_1$ fills a thickness $d/2$ and $\varepsilon_2$ fills the remaining thickness $d/2$, the two being in series across that half.]

  • (A) $6/5$
  • (B) $5/3$
  • (C) $7/5$
  • (D) $7/3$
JEE Advanced 2017 · Paper 2 · Q15 (official key) Answer: A

Consider a simple $RC$ circuit consisting of a source of constant voltage $V$, a switch $S$, a resistance $R$ and a capacitor $C$ in series. Process 1: In the circuit the switch $S$ is closed at $t = 0$ and the capacitor is fully charged to voltage $V_0$ (i.e., charging continues for time $T \gg RC$). In the process some dissipation $(E_D)$ occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $E_C$. Process 2: In a different process the voltage is first set to $\dfrac{V_0}{3}$ and maintained for a charging time $T \gg RC$. Then the voltage is raised to $\dfrac{2V_0}{3}$ without discharging the capacitor and again maintained for a time $T \gg RC$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final voltage $V_0$ as in Process 1. In Process 1, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are related by:

  • (A) $E_C = E_D$
  • (B) $E_C = E_D\ln 2$
  • (C) $E_C = \dfrac{1}{2}E_D$
  • (D) $E_C = 2E_D$
JEE Advanced 2017 · Paper 2 · Q16 (official key) Answer: C

Consider a simple $RC$ circuit consisting of a source of constant voltage $V$, a switch $S$, a resistance $R$ and a capacitor $C$ in series. Process 1: In the circuit the switch $S$ is closed at $t = 0$ and the capacitor is fully charged to voltage $V_0$ (i.e., charging continues for time $T \gg RC$). In the process some dissipation $(E_D)$ occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $E_C$. Process 2: In a different process the voltage is first set to $\dfrac{V_0}{3}$ and maintained for a charging time $T \gg RC$. Then the voltage is raised to $\dfrac{2V_0}{3}$ without discharging the capacitor and again maintained for a time $T \gg RC$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final voltage $V_0$ as in Process 1. In Process 2, total energy dissipated across the resistance $E_D$ is:

  • (A) $E_D = \dfrac{1}{2}CV_0^2$
  • (B) $E_D = 3\left(\dfrac{1}{2}CV_0^2\right)$
  • (C) $E_D = \dfrac{1}{3}\left(\dfrac{1}{2}CV_0^2\right)$
  • (D) $E_D = 3\,CV_0^2$
JEE Advanced 2018 · Paper 1 · Q11 (official key) Answer: 1.50

Three identical capacitors $C_1$, $C_2$ and $C_3$ have a capacitance of $1.0\ \mu\mathrm{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\epsilon_r$. The cell electromotive force (emf) $V_0 = 8\ \mathrm{V}$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $C_3$ is found to be $5\ \mu\mathrm{C}$. The value of $\epsilon_r = $ __________. [Figure: the cell $V_0$ in series with switch $S_1$ forms one branch between the top and bottom rails; a second branch contains switch $S_2$ in series with $C_1$ and $C_2$ (with $C_1$ and $C_2$ themselves in series); a third branch contains $C_3$ alone. All three branches are connected in parallel between the same two rails.]

JEE Advanced 2019 · Paper 1 · Q16 (official key) Answer: 1 (numerical value)

A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta = \frac{d}{N}$. The dielectric constant of the $m^{th}$ layer is $K_m = K\left(1 + \frac{m}{N}\right)$. For a very large $N$ ($> 10^{3}$), the capacitance $C$ is $\alpha\left(\frac{K\epsilon_0 A}{d\ln 2}\right)$. The value of $\alpha$ will be ____. [$\epsilon_0$ is the permittivity of free space]

JEE Advanced 2019 · Paper 1 · Q2 (official key) Answer: C

A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_0$. A hole with a small area $\alpha 4\pi R^{2}$ ($\alpha \ll 1$) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

  • (A) The potential at the center of the shell is reduced by $2\alpha V_0$
  • (B) The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha V_0}{2R}$
  • (C) The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2}R$ from center towards the hole will be $\frac{1-\alpha}{1-2\alpha}$
  • (D) The magnitude of electric field at a point, located on a line passing through the hole and shell's center, on a distance $2R$ from the center of the spherical shell will be reduced by $\frac{\alpha V_0}{2R}$
JEE Advanced 2021 · Paper 1 · Q10 (official key) Answer: 3.00

Two point charges $-Q$ and $+Q/\sqrt{3}$ are placed in the xy-plane at the origin $(0, 0)$ and a point $(2, 0)$, respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V = 0$ in the xy-plane with its center at $(b, 0)$. All lengths are measured in meters. The value of $b$ is ___ meter.

JEE Advanced 2021 · Paper 1 · Q9 (official key) Answer: 1.73

Two point charges $-Q$ and $+Q/\sqrt{3}$ are placed in the xy-plane at the origin $(0, 0)$ and a point $(2, 0)$, respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V = 0$ in the xy-plane with its center at $(b, 0)$. All lengths are measured in meters. The value of $R$ is ___ meter.

JEE Advanced 2022 · Paper 2 · Q12 (official key) Answer: A, C, D

A disk of radius $R$ with uniform positive charge density $\sigma$ is placed on the $xy$ plane with its center at the origin. The Coulomb potential along the $z$-axis is $V(z) = \dfrac{\sigma}{2\epsilon_0}\left(\sqrt{R^2 + z^2} - z\right)$. A particle of positive charge $q$ is placed initially at rest at a point on the $z$ axis with $z = z_0$ and $z_0 > 0$. In addition to the Coulomb force, the particle experiences a vertical force $\vec{F} = -c\,\hat{k}$ with $c > 0$. Let $\beta = \dfrac{2c\epsilon_0}{q\sigma}$. Which of the following statement(s) is(are) correct?

  • (A) For $\beta = \dfrac{1}{4}$ and $z_0 = \dfrac{25}{7}R$, the particle reaches the origin.
  • (B) For $\beta = \dfrac{1}{4}$ and $z_0 = \dfrac{3}{7}R$, the particle reaches the origin.
  • (C) For $\beta = \dfrac{1}{4}$ and $z_0 = \dfrac{R}{\sqrt{3}}$, the particle returns back to $z = z_0$.
  • (D) For $\beta > 1$ and $z_0 > 0$, the particle always reaches the origin.
JEE Advanced 2022 · Paper 1 · Q9 (official key) Answer: B

A medium having dielectric constant $K > 1$ fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$, as shown in Figure (a). Now, both the plates are moved by a distance of $\dfrac{d}{2}$ from their original positions, as shown in Figure (b). [In Figure (b) the dielectric slab of thickness $d$ stays in place while each plate moves outward by $\dfrac{d}{2}$, so that the plate separation becomes $2d$, with an air gap of $\dfrac{d}{2}$ on each side of the slab; the battery remains connected.] In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct?

  • (A) The electric field inside the dielectric material is reduced by a factor of $2K$.
  • (B) The capacitance is decreased by a factor of $\dfrac{1}{K+1}$.
  • (C) The voltage between the capacitor plates is increased by a factor of $(K+1)$.
  • (D) The work done in the process DOES NOT depend on the presence of the dielectric material.
JEE Advanced 2023 · Paper 2 · Q1 (official key) Answer: B

An electric dipole is formed by two charges $+q$ and $-q$ located in the $xy$-plane at $(0,2)$ mm and $(0,-2)$ mm, respectively. The electric potential at point P $(100,100)$ mm due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1,2)$ mm and $(1,-2)$ mm, respectively. What is the value of electric potential at P due to the new dipole?

  • (A) $V_0/4$
  • (B) $V_0/2$
  • (C) $V_0/\sqrt{2}$
  • (D) $3V_0/4$
JEE Advanced 2023 · Paper 1 · Q5 (official key) Answer: B

A container has a base of $50\ \text{cm} \times 5\ \text{cm}$ and height $50$ cm. It has two parallel electrically conducting walls each of area $50\ \text{cm} \times 50\ \text{cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250\ \text{cm}^3\,\text{s}^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: Permittivity of free space $\epsilon_0 = 9 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]

  • (A) $27$ pF
  • (B) $63$ pF
  • (C) $81$ pF
  • (D) $135$ pF
JEE Advanced 2024 · Paper 1 · Q15 (official key) Answer: C

Four identical thin, square metal sheets, $S_1$, $S_2$, $S_3$ and $S_4$, each of side $a$ are kept parallel to each other with equal distance $d\ (\ll a)$ between them, as shown in the figure. Let $C_0 = \varepsilon_0 a^2/d$, where $\varepsilon_0$ is the permittivity of free space. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. List-I: (P) The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected, is (Q) The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$, is (R) The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$, is (S) The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$, is List-II: (1) $3C_0$ (2) $C_0/2$ (3) $C_0/3$ (4) $2C_0/3$ (5) $2C_0$

  • (A) P $\to$ 3; Q $\to$ 2; R $\to$ 4; S $\to$ 5
  • (B) P $\to$ 2; Q $\to$ 3; R $\to$ 2; S $\to$ 1
  • (C) P $\to$ 3; Q $\to$ 2; R $\to$ 4; S $\to$ 1
  • (D) P $\to$ 3; Q $\to$ 2; R $\to$ 2; S $\to$ 5
JEE Advanced 2025 · Paper 2 · Q5 (official key) Answer: A, B, C

A positive point charge of $10^{-8}\ \mathrm{C}$ is kept at a distance of $20\ \mathrm{cm}$ from the center of a neutral conducting sphere of radius $10\ \mathrm{cm}$. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of $10\ \mathrm{cm}$ further away from the center of the sphere along the radial direction. Taking $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \mathrm{Nm^2/C^2}$ (where $\epsilon_0$ is the permittivity of free space), which of the following statements is/are correct:

  • (A) Before the grounding, the electrostatic potential of the sphere is $450\ \mathrm{V}$.
  • (B) Charge flowing from the sphere to the ground because of grounding is $5 \times 10^{-9}\ \mathrm{C}$.
  • (C) After the grounding is removed, the charge on the sphere is $-5 \times 10^{-9}\ \mathrm{C}$.
  • (D) The final electrostatic potential of the sphere is $300\ \mathrm{V}$.
JEE Advanced 2026 · Paper 2 · Q16 (official key) Answer: 1.97

Question Stem for Question Nos. 15 and 16: A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area horizontal metal plates ($M_1$ and $M_2$) which extend far beyond the vertical walls in all directions. The container is partitioned into two equal chambers with a thin insulating vertical wall. The partition wall contains a small hole of cross-sectional area $\sqrt{10}$ cm$^2$ near its bottom edge. Initially the hole is closed and the left chamber of the container is completely filled with a liquid of dielectric constant $\epsilon_r = 15$ and the right chamber is empty ($\epsilon_r = 1$). At time $t = 0$, the hole is opened and the liquid flows from the left chamber to the right chamber. In both the chambers, the space above the liquid has $\epsilon_r = 1$ and is maintained at atmospheric pressure. The schematic of the container at a time $t > 0$ is shown in the figure. [Given: acceleration due to gravity is 10 ms$^{-2}$.] The difference in the capacitance (in F) between the metal plates at $t = 0$ and that at $t = 500$ s is $(8 - n)\epsilon_0$, where $\epsilon_0$ is the permittivity of free space. The value of $n$ is:

JEE Main 2026 · Paper 1 · April 2 Shift 2 · Q39 (official key) Answer: A

Two metal plates (A, B) are kept horizontally with separation of $ \left( \frac{12}{\pi} \right) $ cm , with plate A on the top. An atomizer jet sprays oil (density 1.5 g/$cm^{3}$ ) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC . The potentials $V_{A}$ and $V_{B}$ are required on plates A and B respectively in order to ensure the droplets do not descend. The values of $V_{A}$ and $V_{B}$ are ______. (Neglect the air resistance to the droplets and take g = 10 m/$s^{2}$ )

  • (A) 100 V and 580 V
  • (B) 580 V and 100 V
  • (C) 60 V and 400 V
  • (D) 0 V and -200 V
JEE Main 2026 · Paper 1 · April 4 Shift 1 · Q40 (official key) Answer: A

A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed $v$. If $x$ is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to $x^\alpha$, where $\alpha$ is $\_\_\_\_$ .

  • (A) -2
  • (B) 1
  • (C) -1
  • (D) 2
JEE Advanced 2026 · Paper 2 · Q7 (official key) Answer: A, B, C

Two charges $Q_1 = q$ and $Q_2 = mq$ are placed at the points $P_1(a, b)$ and $P_2(ma, mb)$, respectively, in the $XY$ plane, where $a, b \neq 0$ and $m \neq 0, 1$. If $V_1$ is the potential at a point in the $XY$ plane due to charge $Q_1$ and $V_2$ is the potential at that point due to charge $Q_2$. Correct statement(s) for the points at which $|V_1| = |V_2|$ is/are:

  • (A) For $m = -1$, locus of these points is $ax + by = 0$.
  • (B) For $m = 2$, the locus of these points is a circle of radius $\tfrac{2}{3}\sqrt{a^2 + b^2}$ centered at $\left(\tfrac{2}{3}a, \tfrac{2}{3}b\right)$
  • (C) For $m = -2$, the locus of these points is a circle of radius $2\sqrt{a^2 + b^2}$ centered at $(2a, 2b)$
  • (D) For $m = -3$, locus of these points is $3bx + 3ay = 0$.
JEE Main 2020 · Paper 1 · September 3 Shift 1 · Q10 (published compilation) Answer: C⚑ verify

Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius ${2 \over 3}R$ and ${1 \over 3}R$ have 12 $\mu$C and –3 $\mu$C charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on $S_{1}$ and $S_{2}$ are respectively :

  • (A) 4.5 $\mu$C on both
  • (B) +4.5 $\mu$C and –4.5 $\mu$C
  • (C) 6 $\mu$C and 3 $\mu$C
  • (D) 3 $\mu$C and 6 $\mu$C
JEE Main 2020 · Paper 1 · September 5 Shift 2 · Q14 (published compilation) Answer: D⚑ verify

A parallel plate capacitor has plate of length 'l', width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

  • (A) ${l \over 4}$
  • (B) ${l \over 2}$
  • (C) ${{2l} \over 3}$
  • (D) ${l \over 3}$
JEE Main 2020 · Paper 1 · September 3 Shift 2 · Q15 (published compilation) Answer: C⚑ verify

Concentric metallic hollow spheres of radii R and 4R hold charges $Q_{1}$ and $Q_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is :

  • (A) ${{3{Q_2}} \over {4\pi {\varepsilon _0}R}}$
  • (B) ${{3{Q_1}} \over {4\pi {\varepsilon _0}R}}$
  • (C) ${{3{Q_1}} \over {16\pi {\varepsilon _0}R}}$
  • (D) ${{{Q_2}} \over {4\pi {\varepsilon _0}R}}$
JEE Main 2020 · Paper 1 · September 2 Shift 2 · Q16 (published compilation) Answer: B⚑ verify

A 10 $\mu$F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

  • (A) 20 $\mu$F
  • (B) 15 $\mu$F
  • (C) 10 $\mu$F
  • (D) 30 $\mu$F
JEE Main 2020 · Paper 1 · January 7 Shift 2 · Q21 (published compilation) Answer: 6 (numerical value)⚑ verify

A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is conneced to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) _____

JEE Main 2020 · Paper 1 · September 5 Shift 2 · Q4 (published compilation) Answer: A⚑ verify

Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E at the centre of the circle are respectively. (Take V = 0 at infinity)

  • (A) V = 0; E = 0
  • (B) $V = {{10q} \over {4\pi {\varepsilon _0}R}}$; $E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}$
  • (C) $V = {{10q} \over {4\pi {\varepsilon _0}R}}$; E = 0
  • (D) V = 0; $E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}$
JEE Main 2020 · Paper 1 · September 5 Shift 1 · Q5 (published compilation) Answer: B⚑ verify

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

  • (A) Zero
  • (B) ${3 \over 2}C{V^2}$
  • (C) ${9 \over 2}C{V^2}$
  • (D) ${{25} \over 6}C{V^2}$
JEE Main 2020 · Paper 1 · January 8 Shift 1 · Q6 (published compilation) Answer: C⚑ verify

Effective capacitance of parallel combination of two capacitors $C_{1}$ and $C_{2}$ is 10 $\mu$F. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor $C_{2}$ is 4 times that of $C_{1}$. If these capacitors are connected in series, their effective capacitance will be :

  • (A) 4.2 $\mu$F
  • (B) 8.4 $\mu$F
  • (C) 1.6 $\mu$F
  • (D) 3.2 $\mu$F
JEE Main 2021 · Paper 1 · February 25 Shift 2 · Q16 (published compilation) Answer: A⚑ verify

An electron with kinetic energy $K_{1}$ enters between parallel plates of a capacitor at an angle '$\alpha$' with the plates. It leaves the plates at angle '$\beta$' with kinetic energy $K_{2}$. Then the ratio of kinetic energies $K_{1}$ : $K_{2}$ will be :

  • (A) ${{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$
  • (B) ${{\cos \beta } \over {\cos \alpha }}$
  • (C) ${{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}$
  • (D) ${{\cos \beta } \over {\sin \alpha }}$
JEE Main 2021 · Paper 1 · February 24 Shift 1 · Q4 (published compilation) Answer: D⚑ verify

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :

  • (A) 4 : 1
  • (B) 1 : 2
  • (C) 2 : 1
  • (D) 1 : 4
JEE Main 2022 · Paper 1 · July 28 Shift 1 · Q10 (published compilation) Answer: A⚑ verify

Two capacitors, each having capacitance $40 \,\mu \mathrm{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $\mathrm{K}$ such that the equivalence capacitance of the system became $24 \,\mu \mathrm{F}$. The value of $\mathrm{K}$ will be :

  • (A) 1.5
  • (B) 2.5
  • (C) 1.2
  • (D) 3
JEE Main 2022 · Paper 1 · June 26 Shift 2 · Q10 (published compilation) Answer: B⚑ verify

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 $\mu$C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

  • (A) 1 : 4
  • (B) 4 : 1
  • (C) 1 : 8
  • (D) 8 : 1
JEE Main 2022 · Paper 1 · July 26 Shift 1 · Q11 (published compilation) Answer: A⚑ verify

The total charge on the system of capacitors $C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}$ and $\mathrm{C}_{4}=3 \mu \mathrm{F}$ connected in parallel is : (Assume a battery of $20 \mathrm{~V}$ is connected to the combination)

  • (A) $200 \,\mu \mathrm{C}$
  • (B) 200 C
  • (C) $10 \,\mu \mathrm{C}$
  • (D) 10 C
JEE Main 2022 · Paper 1 · July 29 Shift 1 · Q11 (published compilation) Answer: A⚑ verify

Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options given below.

  • (A) Both Statement I and Statement II are correct
  • (B) Both Statement I and Statement II are incorrect
  • (C) Statement I is correct but Statement II is incorrect
  • (D) Statement I is incorrect but Statement II is correct
JEE Main 2022 · Paper 1 · June 27 Shift 1 · Q14 (published compilation) Answer: A⚑ verify

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

  • (A) 5 N
  • (B) 10 N
  • (C) 20 N
  • (D) Zero
JEE Main 2022 · Paper 1 · June 29 Shift 1 · Q17 (published compilation) Answer: A⚑ verify

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

  • (A) increase by 50%
  • (B) decrease by 15%
  • (C) increase by 25%
  • (D) increase by 33%
JEE Main 2022 · Paper 1 · July 29 Shift 2 · Q3 (published compilation) Answer: C⚑ verify

Two identical thin metal plates has charge $q_{1}$ and $q_{2}$ respectively such that $q_{1}>q_{2}$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

  • (A) $\frac{\left(q_{1}+q_{2}\right)}{C}$
  • (B) $\frac{\left(q_{1}-q_{2}\right)}{C}$
  • (C) $\frac{\left(q_{1}-q_{2}\right)}{2 C}$
  • (D) $\frac{2\left(q_{1}-q_{2}\right)}{C}$
JEE Main 2022 · Paper 1 · July 25 Shift 2 · Q6 (published compilation) Answer: A⚑ verify

Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes n times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $\left( {{{{R_2}} \over {{R_1}}}} \right)$ is :

  • (A) ${n \over {n - 1}}$
  • (B) ${{2n} \over {2n + 1}}$
  • (C) ${{n + 1} \over n}$
  • (D) ${{2n + 1} \over n}$
JEE Main 2022 · Paper 1 · June 29 Shift 2 · Q7 (published compilation) Answer: D⚑ verify

A capacitor is discharging through a resistor R. Consider in time $t_{1}$, the energy stored in the capacitor reduces to half of its initial value and in time $t_{2}$, the charge stored reduces to one eighth of its initial value. The ratio $t_{1}$/$t_{2}$ will be

  • (A) 1/2
  • (B) 1/3
  • (C) 1/4
  • (D) 1/6
JEE Main 2022 · Paper 1 · June 25 Shift 2 · Q8 (published compilation) Answer: A⚑ verify

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness ${d \over 2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

  • (A) 2 : 1
  • (B) 1 : 2
  • (C) 1 : 4
  • (D) 4 : 1
JEE Main 2022 · Paper 1 · July 28 Shift 2 · Q9 (published compilation) Answer: A⚑ verify

A slab of dielectric constant $\mathrm{K}$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathrm{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given $\mathrm{C}_{0}$ = capacitance of capacitor with air as medium between plates.)

  • (A) $\frac{4 K C_{0}}{3+K}$
  • (B) $\frac{3 K C_{0}}{3+K}$
  • (C) $\frac{3+K}{4 K C_{0}}$
  • (D) $\frac{K}{4+K}$
JEE Main 2023 · Paper 1 · January 30 Shift 1 · Q22 (published compilation) Answer: 225 (numerical value)⚑ verify

A capacitor of capacitance $900 \mu \mathrm{F}$ is charged by a $100 \mathrm{~V}$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-} { }^{2} \mathrm{~J}$. The value of $x$ is _____________.

JEE Main 2023 · Paper 1 · January 24 Shift 2 · Q24 (published compilation) Answer: 105 (numerical value)⚑ verify

A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes $\frac{x}{4}$ pF. The value of $x$ is ____________.

JEE Main 2023 · Paper 1 · January 29 Shift 2 · Q24 (published compilation) Answer: 12 (numerical value)⚑ verify

For a charged spherical ball, electrostatic potential inside the ball varies with $r$ as $\mathrm{V}=2ar^2+b$. Here, $a$ and $b$ are constant and r is the distance from the center. The volume charge density inside the ball is $-\lambda a\varepsilon$. The value of $\lambda$ is ____________. $\varepsilon$ = permittivity of the medium

JEE Main 2023 · Paper 1 · January 31 Shift 2 · Q28 (published compilation) Answer: 55 (numerical value)⚑ verify

Two parallel plate capacitors $C_{1}$ and $C_{2}$ each having capacitance of $10 \mu \mathrm{F}$ are individually charged by a 100 V D.C. source. Capacitor $C_{1}$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $\mathrm{C}_{2}$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_{1}$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ________ V. (Assuming Dielectric constant $=10$ )

JEE Main 2023 · Paper 1 · April 10 Shift 2 · Q39 (published compilation) Answer: B⚑ verify

The distance between two plates of a capacitor is $\mathrm{d}$ and its capacitance is $\mathrm{C}_{1}$, when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $\mathrm{C}_{2}$. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$ is

  • (A) 1 : 1
  • (B) 3 : 1
  • (C) 2 : 1
  • (D) 4 : 1
JEE Main 2023 · Paper 1 · April 11 Shift 1 · Q40 (published compilation) Answer: A⚑ verify

A parallel plate capacitor of capacitance $2 \mathrm{~F}$ is charged to a potential $\mathrm{V}$, The energy stored in the capacitor is $E_{1}$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $\mathrm{E}_{2}$. The ratio $\mathrm{E}_{2} / \mathrm{E}_{1}$ is :

  • (A) 1 : 2
  • (B) 2 : 3
  • (C) 2 : 1
  • (D) 1 : 4
JEE Main 2023 · Paper 1 · April 11 Shift 2 · Q50 (published compilation) Answer: B⚑ verify

A capacitor of capacitance $\mathrm{C}$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :

  • (A) Zero
  • (B) $\frac{C V}{\varepsilon_{0}}$
  • (C) $\frac{C V}{2 \varepsilon_{0}}$
  • (D) $\frac{2 C V}{\varepsilon_{0}}$
JEE Main 2023 · Paper 1 · April 8 Shift 2 · Q51 (published compilation) Answer: 6 (numerical value)⚑ verify

A $600 ~\mathrm{pF}$ capacitor is charged by $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 ~\mathrm{pF}$ capacitor. Electrostatic energy lost in the process is ____________ $\mu \mathrm{J}$

JEE Main 2026 · Paper 1 · April 2 Shift 1 · Q34 (published compilation) Answer: D⚑ verify

Two charged conducting spheres $S_1$ and $S_2$ of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on $S_1$ and $S_2$ spheres are $E_{S1}$ and $E_{S2}$ respectively. The value of $\dfrac{E_{S1}}{E_{S2}}$ is ________.

  • (A) $\dfrac{3}{2}$
  • (B) $\dfrac{2}{3}$
  • (C) $\dfrac{4}{9}$
  • (D) $\dfrac{9}{4}$
JEE Main 2026 · Paper 1 · April 6 Shift 2 · Q37 (published compilation) Answer: A⚑ verify

The electric potential as a function of $x, y$ is given by $V=5\left(x^2-y^2\right) V$. The electric field at a point $(2,3) \mathrm{m}$ is $\_\_\_\_ \mathrm{V} / \mathrm{m}$.

  • (A) $(-20 \hat{i}+30 \hat{j})$
  • (B) $(20 \hat{i}-30 \hat{j})$
  • (C) $(20 \hat{i}+45 \hat{j})$
  • (D) $(-4 \hat{i}+6 \hat{j})$
JEE Main 2026 · Paper 1 · April 6 Shift 2 · Q40 (published compilation) Answer: B⚑ verify

A sphere of capacitance 100 pF is charged to a potential of 100 V . Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is $\alpha \times 10^{-7} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . (combined capacitance of spheres is 200 pF )

  • (A) 5
  • (B) $\frac{5}{2}$
  • (C) $\frac{7}{2}$
  • (D) $\frac{9}{2}$
JEE Main 2026 · Paper 1 · April 8 Shift 2 · Q46 (published compilation) Answer: 2 (numerical value)⚑ verify

A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of $1 \mu \mathrm{~F}$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13} \Omega \mathrm{~m}$ and resistance $17.7 \times 10^{14} \Omega$. Relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is $\_\_\_\_$ . (Take permittivity of free space $=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$ )

JEE Main 2026 · Paper 1 · April 6 Shift 1 · Q49 (published compilation) Answer: 186 (numerical value)⚑ verify

A three coulomb charge moves from the point $(0,-2,-5)$ to the point $(5,1,2)$ in an electric field expressed as $\vec{E}=2 x \hat{\mathrm{i}}+3 \mathrm{y}^2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}$. The work done in moving the charge is $\_\_\_\_$ J.

🎯 Question Bank 100 MCQs · graded

Distribution — advanced: 15 · easy: 25 · hard: 24 · medium: 36. Every question carries a source trace; each ends in an SME-verify solution.

Q1 Electric potential at a point is defined as easy
Step solution + source
Potential $V$ equals the work done by an external agent, against the electric force, in carrying a unit positive test charge from infinity to the point without acceleration. It is a scalar defined as $V=\dfrac{W}{q}$ and measured in volts, whereas force per unit charge is the field, not the potential. 🔉⇢

Source: NCERT XII Ch 2

Q2 The SI unit of electric potential is the easy
Step solution + source
Potential is energy per unit charge, so its unit is the joule per coulomb, given the special name volt, $1\,\text{V}=1\,\text{J/C}$. Newton per coulomb is the field, coulomb per volt is the farad, and joule per second is the watt, so only the volt measures potential. 🔉⇢

Source: NCERT XII Ch 2

Q3 The electric potential at distance $r$ from an isolated point charge $q$ in vacuum is easy
Step solution + source
Integrating the field $E=\dfrac{kq}{r^2}$ from infinity to $r$ gives the potential $V=\dfrac{kq}{r}$, which falls off as one over distance. The inverse-square form is the field, not the potential, so the correct expression is $\dfrac{kq}{r}$ with $k=\dfrac{1}{4\pi\varepsilon_0}$. 🔉⇢

Source: NCERT XII Ch 2

Q4 Electric potential is a easy
Step solution + source
Potential is defined through work and energy per unit charge, which are scalars, so it has magnitude and sign but no direction. Only the electric field is a vector. Because potential is scalar, potentials from several charges add algebraically, which is far simpler than adding field vectors. 🔉⇢

Source: NCERT XII Ch 2

Q5 The potential due to a charge of $-2\,\mu C$ at a distance of $0.5\,\text{m}$ in vacuum is about medium
Step solution + source
Using $V=\dfrac{kq}{r}=\dfrac{9\times10^{9}\times(-2\times10^{-6})}{0.5}$ gives $-3.6\times10^{4}\,\text{V}$. The sign of the potential follows the sign of the source charge, so a negative charge produces a negative potential everywhere around it, here about 36 kilovolts negative. 🔉⇢

Source: authored

Q6 The potential difference $V_A-V_B$ equals medium
Step solution + source
By definition $V_A-V_B=\dfrac{W_{B\to A}}{q}$, the external work per unit positive charge carried from B to A against the field. It is not a difference of fields and is generally nonzero unless A and B lie on the same equipotential. The work is path-independent because the electrostatic force is conservative. 🔉⇢

Source: NCERT XII Ch 2

Q7 Two points in a field are at $20\,\text{V}$ and $5\,\text{V}$. The work done in moving $+3\,\mu C$ from the $5\,\text{V}$ point to the $20\,\text{V}$ point is medium
Step solution + source
The work equals $q(V_f-V_i)=3\times10^{-6}\times(20-5)=45\times10^{-6}\,\text{J}$. Since the positive charge moves to a higher potential, an external agent does positive work against the field, so the value is $+45\,\mu J$ and the charge's potential energy increases. 🔉⇢

Source: authored

Q8 The electric potential at infinity is conventionally taken as easy
Step solution + source
The reference level for electrostatic potential is chosen at infinity, where the influence of any finite charge distribution vanishes, so $V_\infty=0$. All finite-distance potentials are measured relative to this zero, which makes the point-charge potential $V=\dfrac{kq}{r}$ tend correctly to zero as $r$ grows large. 🔉⇢

Source: NCERT XII Ch 2

Q9 Moving along the direction of the electric field, the electric potential medium
Step solution + source
The relation $E=-\dfrac{dV}{dr}$ shows the field points from higher to lower potential, so moving along the field means moving toward smaller $V$. Hence potential always decreases in the direction of the field, and positive charges accelerate this way, converting electrical potential energy into kinetic energy. 🔉⇢

Source: NCERT XII Ch 2

Q10 An electron (charge $-e$) is accelerated through a potential difference of $100\,\text{V}$. Its kinetic energy gain is hard
Step solution + source
The kinetic energy gained equals $|q|\,\Delta V=e\times100\,\text{V}=100\,\text{eV}$. An electron accelerates toward the higher-potential plate, and the electron-volt is defined as the energy an electron gains across one volt, so regardless of the sign of the charge the magnitude gained is $|q|\Delta V$. 🔉⇢

Source: authored

Q11 The potential $2\,\text{m}$ from a point charge is $30\,\text{V}$. The charge is approximately hard
Step solution + source
From $V=\dfrac{kq}{r}$ we get $q=\dfrac{Vr}{k}=\dfrac{30\times2}{9\times10^{9}}\approx6.7\times10^{-9}\,\text{C}$. The positive potential indicates a positive charge, and substituting the distance and the Coulomb constant yields about 6.7 nanocoulombs. 🔉⇢

Source: authored

Q12 The electric potential at the midpoint between two equal positive charges is medium
Step solution + source
Potential is a scalar that adds algebraically, so the two positive charges contribute equal positive potentials at the midpoint, giving a nonzero positive total $V=2\times\dfrac{kq}{r}$. The electric field there is zero by symmetry, but a zero field does not force the potential to vanish; the two are different quantities. 🔉⇢

Source: authored

Q13 Two charges $+q$ and $-q$ are separated by distance $2a$. At the midpoint the potential is hard
Step solution + source
At the midpoint each charge is a distance $a$ away, and the potentials $+\dfrac{kq}{a}$ and $-\dfrac{kq}{a}$ cancel exactly, giving zero. The field there is not zero, since both charges push a test charge the same way, illustrating that scalar potential and vector field must be judged separately. 🔉⇢

Source: authored

Q14 One electron-volt equals easy
Step solution + source
An electron-volt is the energy gained by a charge equal to one electronic charge when accelerated through one volt, so $1\,\text{eV}=e\times1\,\text{V}=1.6\times10^{-19}\,\text{J}$. It is a unit of energy, not charge, and is convenient at atomic scales where the joule is impractically large. 🔉⇢

Source: NCERT XII Ch 2

Q15 A charged conductor is said to be at high potential. This means medium
Step solution + source
High potential means each unit of positive charge placed there has high electrostatic potential energy, since $V=\dfrac{U}{q}$. A small conductor can reach high potential with little charge, so large potential does not imply large charge. The field inside any conductor is zero, and the sign of charge is not fixed by high potential alone. 🔉⇢

Source: authored

Q16 The work done in moving a charge around any closed loop in an electrostatic field is hard
Step solution + source
The electrostatic field is conservative, so the line integral of the field around any closed path is zero, meaning the net work per unit charge over a closed loop vanishes. Consequently potential is single-valued and path-independent, which is exactly why a scalar potential can be defined at all. 🔉⇢

Source: NCERT XII Ch 2

Q17 The electric potential due to a short dipole at a point on its axis at distance $r$ is easy
Step solution + source
For a short dipole the axial potential is $V=\dfrac{kp}{r^2}$, where $p$ is the dipole moment. It falls faster than a point charge's potential because the opposite charges partially cancel. On the axis $\theta=0$ so $\cos\theta=1$, giving the maximum potential for a given $r$. 🔉⇢

Source: NCERT XII Ch 2

Q18 The general potential due to a short dipole at distance $r$ making angle $\theta$ with the axis is easy
Step solution + source
The dipole potential is $V=\dfrac{kp\cos\theta}{r^2}$, where $\theta$ is measured from the dipole axis. The $\cos\theta$ factor makes the potential maximum on the axis and zero on the perpendicular bisector, and the inverse-square dependence is characteristic of a dipole rather than a single charge. 🔉⇢

Source: NCERT XII Ch 2

Q19 The potential due to a short dipole on its equatorial (perpendicular-bisector) plane is medium
Step solution + source
On the equatorial plane $\theta=90^\circ$ so $\cos\theta=0$, making $V=\dfrac{kp\cos\theta}{r^2}=0$. Physically every equatorial point is equidistant from the equal and opposite charges, so their scalar potentials cancel exactly. The electric field there is not zero, but the potential is, showing field and potential are independent. 🔉⇢

Source: NCERT XII Ch 2

Q20 The potential of a short dipole falls off with distance as medium
Step solution + source
For a short dipole $V\propto\dfrac{1}{r^2}$, faster than the $1/r$ of a point charge because the two opposite charges nearly cancel at large distances. The dipole field, obtained from $E=-\dfrac{dV}{dr}$, then falls as $1/r^3$, one power faster than the potential, as expected. 🔉⇢

Source: NCERT XII Ch 2

Q21 A dipole has moment $p=2\times10^{-9}\,\text{C\,m}$. The axial potential at $0.1\,\text{m}$ is about hard
Step solution + source
On the axis $V=\dfrac{kp}{r^2}=\dfrac{9\times10^{9}\times2\times10^{-9}}{(0.1)^2}=\dfrac{18}{0.01}=1.8\times10^{3}\,\text{V}$. The $\cos\theta=1$ on the axis gives the full value, so substituting the dipole moment and distance yields about 1800 volts. 🔉⇢

Source: authored

Q22 Compared with the potential of a single point charge, the dipole potential medium
Step solution + source
A point-charge potential goes as $1/r$, while a dipole's goes as $1/r^2$, so the dipole potential decreases more rapidly because the fields of the two opposite charges tend to cancel far away. The dipole potential also depends on angle through $\cos\theta$, unlike an isotropic point charge. 🔉⇢

Source: NCERT XII Ch 2

Q23 For a dipole, the locus of points where the potential is zero is hard
Step solution + source
Setting $V=\dfrac{kp\cos\theta}{r^2}=0$ requires $\cos\theta=0$, i.e. $\theta=90^\circ$, which is the equatorial plane through the dipole centre. Every point on this plane is equidistant from the two charges, so their potentials cancel, whereas along the axis the potential is largest in magnitude. 🔉⇢

Source: authored

Q24 The dipole moment vector $\vec p$ is directed medium
Step solution + source
By convention the dipole moment $\vec p=q\,\vec d$ points from the negative charge toward the positive charge, with magnitude $p=q\times2a$. This convention makes the axial potential positive on the positive-charge side, consistent with $V=\dfrac{kp\cos\theta}{r^2}$ where $\theta$ is measured from this direction. 🔉⇢

Source: NCERT XII Ch 2

Q25 The exact potential of a dipole (charges $\pm q$ separated by $2a$) at axial distance $r$ from its centre is advanced
Step solution + source
On the axis the potentials add as $V=kq\left(\dfrac{1}{r-a}-\dfrac{1}{r+a}\right)=\dfrac{kq(2a)}{r^2-a^2}$. For $r\gg a$ this reduces to $\dfrac{k(2qa)}{r^2}=\dfrac{kp}{r^2}$, the short-dipole result, so keeping the full expression matters when the point is not far compared with the dipole size. 🔉⇢

Source: authored

Q26 For a dipole the potential can be written compactly as $V=\dfrac{k\,\vec p\cdot\hat r}{r^2}$. This shows the potential is advanced
Step solution + source
The dot product $\vec p\cdot\hat r=p\cos\theta$ picks out the component of the dipole moment along the line to the field point, so $V=\dfrac{kp\cos\theta}{r^2}$. Being a dot product it yields a scalar, and it can be positive or negative depending on the angle $\theta$. 🔉⇢

Source: authored

Q27 At a fixed distance $r$, the magnitude of the dipole potential is greatest when $\theta$ equals hard
Step solution + source
Since $V=\dfrac{kp\cos\theta}{r^2}$, the magnitude is largest when $\cos\theta=\pm1$, i.e. along the axis. At $\theta=0^\circ$ the potential is maximum positive, at $180^\circ$ it is the most negative, and at $90^\circ$ it is zero, so the greatest positive value occurs at $0^\circ$. 🔉⇢

Source: authored

Q28 The dipole moment of two charges $\pm4\,\mu C$ separated by $5\,\text{mm}$ is medium
Step solution + source
The dipole moment is $p=q\times d=4\times10^{-6}\times5\times10^{-3}=2\times10^{-8}\,\text{C\,m}$. It uses the magnitude of either charge times their separation, pointing from negative to positive, so substituting the values gives 20 nanocoulomb-metres. 🔉⇢

Source: authored

Q29 An equipotential surface is one on which easy
Step solution + source
On an equipotential surface every point is at the same potential, so no work is done moving a charge between any two of its points, since $W=q\Delta V=0$. Such surfaces map the field conveniently; the field need not be large nor the charge uniform, only $V$ constant across the surface. 🔉⇢

Source: NCERT XII Ch 2

Q30 The electric field is always ___ to an equipotential surface. easy
Step solution + source
If the field had a component along the surface, it would do work moving a charge between points of equal potential, contradicting $\Delta V=0$. Hence the field must be entirely normal to the equipotential surface everywhere, which lets us sketch field lines directly from a set of equipotential surfaces. 🔉⇢

Source: NCERT XII Ch 2

Q31 The equipotential surfaces of an isolated point charge are easy
Step solution + source
For a point charge $V=\dfrac{kq}{r}$ depends only on $r$, so surfaces of constant $V$ are spheres centred on the charge. The radial field lines pierce each sphere at right angles, consistent with the field being perpendicular to equipotentials, and closer spheres represent larger potential magnitudes. 🔉⇢

Source: NCERT XII Ch 2

Q32 The relation between field and potential in one dimension is medium
Step solution + source
The field equals the negative gradient of potential, $E=-\dfrac{dV}{dr}$, so it points toward decreasing potential. The negative sign encodes that positive charges move from high to low potential. Given $V(r)$, differentiating and negating recovers the field, and the unit volt per metre equals newton per coulomb. 🔉⇢

Source: NCERT XII Ch 2

Q33 Closely spaced equipotential surfaces (for equal potential steps) indicate medium
Step solution + source
Since $E=-\dfrac{dV}{dr}$, a large field corresponds to a rapid change of potential over a short distance, i.e. closely packed equipotentials for equal potential steps. Widely spaced surfaces mean the potential changes slowly, so the field is weak; the spacing is a direct visual measure of field strength. 🔉⇢

Source: NCERT XII Ch 2

Q34 For a uniform electric field, the equipotential surfaces are medium
Step solution + source
A uniform field has constant magnitude and direction, so surfaces of constant potential are flat planes oriented perpendicular to the field. Moving along the field, $V$ decreases linearly, giving equally spaced planes for equal potential intervals, as in the interior of a parallel-plate capacitor. 🔉⇢

Source: NCERT XII Ch 2

Q35 In a region $V=5x$ (volts, $x$ in metres). The electric field is hard
Step solution + source
Using $E_x=-\dfrac{dV}{dx}=-\dfrac{d}{dx}(5x)=-5\,\text{V/m}$, the field is uniform and points in the negative $x$-direction. The magnitude $5\,\text{V/m}$ is constant because $V$ varies linearly with $x$, and the negative sign shows the field points toward lower potential. 🔉⇢

Source: authored

Q36 No work is done in moving a charge on an equipotential surface because easy
Step solution + source
Work equals $q\Delta V$, and on an equipotential $\Delta V=0$ between any two points, so the work is zero regardless of the path taken on the surface. This holds even though the field itself may be strong; it is the vanishing potential difference, not a vanishing field, that makes the work zero. 🔉⇢

Source: NCERT XII Ch 2

Q37 Two equipotential surfaces at $10\,\text{V}$ and $6\,\text{V}$ are separated by $2\,\text{cm}$ along the field. The average field is hard
Step solution + source
The field magnitude is $E=\dfrac{|\Delta V|}{d}=\dfrac{10-6}{0.02}=\dfrac{4}{0.02}=200\,\text{V/m}$. The potential drops 4 volts over 2 centimetres, and dividing by the separation in metres gives 200 volts per metre, directed from the higher to the lower equipotential. 🔉⇢

Source: authored

Q38 Equipotential surfaces of different potential values can never intersect because advanced
Step solution + source
If two equipotentials of different values crossed, the intersection point would simultaneously have two different potentials, impossible since potential is a single-valued scalar function of position. Equivalently the field direction, normal to each surface, would be ambiguous there, so distinct equipotential surfaces are always non-intersecting. 🔉⇢

Source: authored

Q39 In three dimensions the field relates to potential by advanced
Step solution + source
The field is the negative gradient of the scalar potential, $\vec E=-\nabla V=-\left(\dfrac{\partial V}{\partial x}\hat x+\dfrac{\partial V}{\partial y}\hat y+\dfrac{\partial V}{\partial z}\hat z\right)$. This vector points in the direction of steepest decrease of $V$, perpendicular to equipotential surfaces. 🔉⇢

Source: NCERT XII Ch 2

Q40 Just outside a conductor, an equipotential surface runs medium
Step solution + source
A conductor's surface is itself an equipotential, and the equipotential surfaces just outside hug its shape, running parallel to it. The field, being normal to equipotentials, therefore emerges perpendicular to the conductor surface, which is why field lines meet conductor surfaces at right angles. 🔉⇢

Source: NCERT XII Ch 2

Q41 If the potential is constant throughout a region, the electric field there is hard
Step solution + source
If $V$ is constant, all its spatial derivatives vanish, so $\vec E=-\nabla V=0$ throughout the region, as inside a conductor in electrostatic equilibrium. Constant potential and zero field go together, since the field measures how fast potential changes in space. 🔉⇢

Source: authored

Q42 The unit volt per metre for the electric field is equivalent to medium
Step solution + source
From $E=-\dfrac{dV}{dr}$ the unit is volt per metre; from $E=F/q$ it is newton per coulomb. These are identical because $1\,\text{V}=1\,\text{J/C}=1\,\text{N\,m/C}$, so $1\,\text{V/m}=1\,\text{N/C}$, confirming the field-potential relation is dimensionally consistent. 🔉⇢

Source: NCERT XII Ch 2

Q43 The potential energy of a system of two point charges $q_1,q_2$ separated by $r$ is easy
Step solution + source
The interaction energy is $U=\dfrac{kq_1q_2}{r}$, the work needed to bring the charges from infinity to separation $r$. It is positive for like charges (work against repulsion) and negative for unlike charges, and varies as $1/r$, matching the potential of one charge times the other. 🔉⇢

Source: NCERT XII Ch 2

Q44 As two like charges are brought closer together, the potential energy of the system easy
Step solution + source
Like charges repel, so external work must be done against the repulsion to reduce their separation, storing energy in the system. Since $U=\dfrac{kq_1q_2}{r}$ with $q_1q_2\gt0$, decreasing $r$ increases the positive potential energy, which is released as kinetic energy if the charges are freed. 🔉⇢

Source: NCERT XII Ch 2

Q45 For two unlike charges, the electrostatic potential energy is easy
Step solution + source
With opposite signs $q_1q_2\lt0$, so $U=\dfrac{kq_1q_2}{r}$ is negative. This reflects that the charges attract and energy is released as they come together from infinity. A negative potential energy signals a bound configuration that needs external work to separate. 🔉⇢

Source: NCERT XII Ch 2

Q46 The total potential energy of three point charges is found by medium
Step solution + source
Electrostatic energy is additive over pairs, so $U=\dfrac{kq_1q_2}{r_{12}}+\dfrac{kq_1q_3}{r_{13}}+\dfrac{kq_2q_3}{r_{23}}$. Each distinct pair contributes once; there are three pairs for three charges. This follows from building the configuration by bringing in charges one at a time. 🔉⇢

Source: NCERT XII Ch 2

Q47 The work done in assembling a charge configuration from infinity equals medium
Step solution + source
The work done by an external agent, against electrostatic forces, in bringing charges from infinity to their final positions is stored as the configuration's potential energy $U$. By energy conservation this work equals $U$ exactly when charges start and end at rest; for attractive systems it can be negative. 🔉⇢

Source: NCERT XII Ch 2

Q48 The potential energy of a charge $q$ placed at a point where the potential is $V$ is medium
Step solution + source
Potential energy is $U=qV$, since potential is defined as energy per unit charge. Placing charge $q$ where the potential is $V$ gives it energy $qV$ relative to infinity, linking the single-charge energy to the potential produced by all other charges, with sign set by $q$ and $V$. 🔉⇢

Source: NCERT XII Ch 2

Q49 Two charges $+3\,\mu C$ and $+3\,\mu C$ are $3\,\text{m}$ apart. Their potential energy is about hard
Step solution + source
Using $U=\dfrac{kq_1q_2}{r}=\dfrac{9\times10^{9}\times(3\times10^{-6})^2}{3}=\dfrac{9\times10^{9}\times9\times10^{-12}}{3}=2.7\times10^{-2}\,\text{J}$. Both charges are positive, so the energy is positive, indicating work done against repulsion, about 27 millijoules. 🔉⇢

Source: authored

Q50 A charge $+2\,\mu C$ is moved from a point at $100\,\text{V}$ to a point at $250\,\text{V}$. The change in its potential energy is hard
Step solution + source
The change is $\Delta U=q\,\Delta V=2\times10^{-6}\times(250-100)=3\times10^{-4}\,\text{J}$. Because a positive charge is taken to higher potential, its potential energy increases, so $\Delta U$ is positive and external work of $+300\,\mu J$ is required. 🔉⇢

Source: authored

Q51 The potential energy of a dipole of moment $p$ in a uniform field $E$ at angle $\theta$ is medium
Step solution + source
The orientation energy of a dipole is $U=-\vec p\cdot\vec E=-pE\cos\theta$. It is minimum ($-pE$) when the dipole aligns with the field ($\theta=0$) and maximum ($+pE$) when anti-aligned, which is why a dipole tends to rotate into alignment, lowering its energy. 🔉⇢

Source: NCERT XII Ch 2

Q52 A dipole in a uniform field is in stable equilibrium when it is hard
Step solution + source
Since $U=-pE\cos\theta$, the energy is minimum at $\theta=0$, the condition for stable equilibrium: a small displacement produces a restoring torque. At $\theta=180^\circ$ the energy is maximum, giving unstable equilibrium, so a dipole rests stably with its moment along the field. 🔉⇢

Source: NCERT XII Ch 2

Q53 Three equal charges $q$ sit at the corners of an equilateral triangle of side $a$. The total potential energy is advanced
Step solution + source
There are three identical pairs, each separated by $a$, so $U=3\times\dfrac{kq^2}{a}=\dfrac{3kq^2}{a}$. Every pair contributes the same positive energy because all charges and separations are equal, reflecting the work done against mutual repulsion to assemble the symmetric configuration. 🔉⇢

Source: authored

Q54 The work required to bring a charge from infinity to a point near a fixed charge is negative when advanced
Step solution + source
The assembly work equals $U=\dfrac{kq_1q_2}{r}$. For opposite signs $q_1q_2\lt0$, so $U$ is negative, meaning the field does positive work and an external agent does negative work to bring them together slowly. Like-sign charges instead require positive external work against repulsion. 🔉⇢

Source: authored

Q55 The maximum torque on a dipole of moment $p$ in a uniform field $E$ has magnitude advanced
Step solution + source
The torque is $\tau=pE\sin\theta$, greatest when $\theta=90^\circ$, giving $\tau_{max}=pE$. This is where the dipole is perpendicular to the field, precisely where its potential energy changes fastest with angle. The torque vanishes at $\theta=0$ and $180^\circ$, the equilibrium orientations. 🔉⇢

Source: authored

Q56 The energy released when a dipole rotates from perpendicular ($90^\circ$) to aligned ($0^\circ$) in a field $E$ is medium
Step solution + source
Using $U=-pE\cos\theta$, the energy at $90^\circ$ is $0$ and at $0^\circ$ is $-pE$, so the decrease is $pE$. This energy is released as rotational kinetic energy as the dipole swings into alignment, and the result depends only on the energy difference, not the sign convention. 🔉⇢

Source: authored

Q57 In electrostatic equilibrium, the electric field inside the material of a conductor is easy
Step solution + source
Free charges in a conductor redistribute until the interior field vanishes; otherwise they would keep moving, contradicting equilibrium. So $E=0$ everywhere inside the conducting material, any excess charge resides on the surface, and this zero interior field is the basis of electrostatic shielding. 🔉⇢

Source: NCERT XII Ch 2

Q58 Excess charge given to an isolated conductor resides easy
Step solution + source
Because the interior field is zero, Gauss's law applied to any interior surface encloses no net charge, so all excess charge must lie on the outer surface. On an irregular conductor it distributes so the surface stays equipotential, concentrating where the curvature is greatest. 🔉⇢

Source: NCERT XII Ch 2

Q59 Just outside a charged conductor's surface, the electric field is medium
Step solution + source
The field just outside a conductor is $E=\dfrac{\sigma}{\varepsilon_0}$, normal to the surface, where $\sigma$ is the local surface charge density. It is normal because any tangential component would drive surface currents, and it is twice the field of an isolated sheet because charge sits on only one side. 🔉⇢

Source: NCERT XII Ch 2

Q60 A conductor in electrostatic equilibrium is medium
Step solution + source
Since the interior field is zero and the surface field has no tangential component, no work is needed to move a charge anywhere within or on the conductor, so the whole conductor sits at one potential. Its potential need not be zero; it simply has a single constant value throughout. 🔉⇢

Source: NCERT XII Ch 2

Q61 Electrostatic shielding means the field inside an empty cavity of a hollow conductor is medium
Step solution + source
The cavity of a hollow conductor is shielded: external charges and fields induce surface charges that keep the cavity field zero. This is why sensitive equipment is enclosed in metal boxes, and the result holds for any external arrangement as long as no charge is placed inside the cavity. 🔉⇢

Source: NCERT XII Ch 2

Q62 The surface charge density on a charged conductor is greatest where the surface is hard
Step solution + source
Charge accumulates where curvature is highest, so sharp points carry the largest surface charge density and hence the strongest local field, $E=\dfrac{\sigma}{\varepsilon_0}$. This is why lightning rods are pointed: the intense field ionises nearby air and lets charge leak off, the action of points. 🔉⇢

Source: NCERT XII Ch 2

Q63 A charge $+Q$ is placed inside the cavity of a neutral conducting shell. The charge induced on the inner wall is hard
Step solution + source
The interior of the metal must be field-free, so by Gauss's law a Gaussian surface within the conductor encloses zero net charge. Thus $-Q$ is induced on the inner cavity wall to cancel the enclosed $+Q$, and to keep the shell neutral, $+Q$ then appears on its outer surface. 🔉⇢

Source: authored

Q64 The potential inside a charged hollow spherical conductor is medium
Step solution + source
Inside a hollow charged conductor the field is zero, so the potential does not change and equals its surface value $\dfrac{kQ}{R}$ throughout the interior. Only outside does the potential fall as $\dfrac{kQ}{r}$, and a constant interior potential with zero field is consistent with $E=-\dfrac{dV}{dr}$. 🔉⇢

Source: NCERT XII Ch 2

Q65 The electrostatic pressure (force per unit area) on the surface of a charged conductor is advanced
Step solution + source
Each surface element feels the field of all other charges, $\dfrac{\sigma}{2\varepsilon_0}$, giving an outward pressure $P=\sigma\times\dfrac{\sigma}{2\varepsilon_0}=\dfrac{\sigma^2}{2\varepsilon_0}$. This equals the field energy density $\dfrac12\varepsilon_0E^2$ just outside and always acts outward, tending to expand the charged surface. 🔉⇢

Source: authored

Q66 Two isolated conducting spheres of radii $R$ and $2R$ carry equal charge $Q$. The ratio of their surface charge densities (small : large) is hard
Step solution + source
Surface density is $\sigma=\dfrac{Q}{4\pi R^2}$, so with equal $Q$, $\sigma\propto\dfrac{1}{R^2}$. Thus $\dfrac{\sigma_{small}}{\sigma_{large}}=\dfrac{(2R)^2}{R^2}=4$, i.e. $4:1$. The smaller sphere has higher density, consistent with charge crowding where curvature is greater. 🔉⇢

Source: authored

Q67 Two conducting spheres of radii $R_1$ and $R_2$ are joined by a long wire. The ratio of their surface charge densities is advanced
Step solution + source
Connected spheres share a common potential, so $\dfrac{kQ_1}{R_1}=\dfrac{kQ_2}{R_2}$, giving $Q\propto R$. Then $\sigma=\dfrac{Q}{4\pi R^2}\propto\dfrac{R}{R^2}=\dfrac{1}{R}$, so $\dfrac{\sigma_1}{\sigma_2}=\dfrac{R_2}{R_1}$; the smaller sphere carries the higher density and stronger field. 🔉⇢

Source: authored

Q68 When a neutral conductor is placed in an external field, charges are induced such that medium
Step solution + source
Free electrons shift until the field they produce inside exactly cancels the applied field, restoring zero interior field at equilibrium. This creates induced positive and negative charges on opposite faces, but the conductor's net charge stays zero, and removing the external field lets the induced charges relax back. 🔉⇢

Source: NCERT XII Ch 2

Q69 A charge $q$ is placed in the previously empty cavity of a conductor. The field within the conducting material remains hard
Step solution + source
Even with a charge in the cavity, the conducting material stays field-free at equilibrium; induced charges $-q$ on the inner wall and $+q$ on the outer surface arrange so the metal's interior field is still zero. The external field then looks as though $q$ sat at the conductor's centre. 🔉⇢

Source: NCERT XII Ch 2

Q70 A metal enclosure used to shield electronic equipment from external electric fields is called a easy
Step solution + source
A Faraday cage is a conducting enclosure that keeps its interior field-free because induced surface charges cancel any external field inside the cavity. This electrostatic shielding protects sensitive electronics and passengers in cars struck by lightning, and it works for any external field provided no charge is placed inside. 🔉⇢

Source: NCERT XII Ch 2

Q71 Capacitance is defined as easy
Step solution + source
Capacitance measures a conductor's ability to store charge, $C=\dfrac{Q}{V}$, the charge needed to raise its potential by one volt. It depends only on geometry and the surrounding medium, not on the actual charge or voltage, and its SI unit is the farad, one coulomb per volt. 🔉⇢

Source: NCERT XII Ch 2

Q72 The SI unit of capacitance is the easy
Step solution + source
The farad is defined as one coulomb per volt, $1\,\text{F}=1\,\text{C/V}$. It is a large unit, so practical capacitors are rated in microfarads or picofarads. The volt measures potential, the coulomb measures charge, and the henry measures inductance, so only the farad measures capacitance. 🔉⇢

Source: NCERT XII Ch 2

Q73 The capacitance of a parallel-plate capacitor with plate area $A$ and separation $d$ in vacuum is easy
Step solution + source
For a parallel-plate capacitor $C=\dfrac{\varepsilon_0 A}{d}$: capacitance grows with plate area and falls with separation. Larger plates hold more charge at the same voltage, while a smaller gap raises the field and stored charge. The formula assumes a uniform field and neglects edge effects. 🔉⇢

Source: NCERT XII Ch 2

Q74 Inserting a dielectric of constant $K$ that completely fills a capacitor changes its capacitance to medium
Step solution + source
A dielectric of constant $K$ increases capacitance by the factor $K$, giving $C'=KC=\dfrac{K\varepsilon_0 A}{d}$. The polarised dielectric reduces the net field for a given charge, lowering the voltage and thus raising $C=\dfrac{Q}{V}$, so the capacitor stores more charge at the same voltage. 🔉⇢

Source: NCERT XII Ch 2

Q75 A dielectric increases capacitance because it medium
Step solution + source
The dielectric's molecules polarise, creating an internal field opposing the applied one, so the net field and hence the voltage drop for a fixed charge. Since $C=\dfrac{Q}{V}$, a smaller $V$ at the same $Q$ means larger $C$ by the factor $K$; the separation and charge are unchanged by inserting the slab. 🔉⇢

Source: NCERT XII Ch 2

Q76 The dielectric constant of vacuum is medium
Step solution + source
By definition the dielectric constant (relative permittivity) of free space is exactly 1, since $K=\dfrac{\varepsilon}{\varepsilon_0}$ and for vacuum $\varepsilon=\varepsilon_0$. Air is very close to 1, while glass or water have $K$ well above 1. The number $8.85\times10^{-12}$ is $\varepsilon_0$ itself in farads per metre, not $K$. 🔉⇢

Source: NCERT XII Ch 2

Q77 A parallel-plate capacitor has $C=10\,\mu F$. If the plate separation is halved, the new capacitance is hard
Step solution + source
Since $C=\dfrac{\varepsilon_0 A}{d}$, capacitance is inversely proportional to separation, so halving $d$ doubles $C$ to $20\,\mu F$. The smaller gap raises the field for a given charge, storing more charge per volt; area and medium are unchanged, so only the factor of two matters. 🔉⇢

Source: authored

Q78 A $5\,\mu F$ capacitor is charged to $200\,\text{V}$. The charge stored is hard
Step solution + source
Using $Q=CV=5\times10^{-6}\times200=1\times10^{-3}\,\text{C}$, the capacitor stores one millicoulomb. The charge scales linearly with capacitance and voltage, so multiplying five microfarads by 200 volts gives $10^{-3}$ coulombs. 🔉⇢

Source: authored

Q79 A dielectric slab is inserted into a capacitor that stays connected to a battery. The charge on the plates hard
Step solution + source
With the battery connected, $V$ is fixed. Inserting a dielectric raises $C$ to $KC$, so $Q=CV$ increases to $KCV$: the battery pushes extra charge onto the plates. This differs from the isolated case, where $Q$ is fixed and $V$ instead falls; here voltage is held constant so charge rises. 🔉⇢

Source: NCERT XII Ch 2

Q80 A dielectric is inserted into an isolated (disconnected) charged capacitor. The voltage across it hard
Step solution + source
An isolated capacitor keeps its charge $Q$ fixed. Inserting a dielectric raises $C$ to $KC$, so $V=\dfrac{Q}{C}$ drops to $\dfrac{V}{K}$: the voltage falls. The dielectric's polarisation partly cancels the field, lowering the potential difference for the same stored charge, and the stored energy also decreases. 🔉⇢

Source: NCERT XII Ch 2

Q81 A capacitor is half-filled parallel to the plates with a dielectric $K$, forming two layers each of thickness $d/2$. The combination behaves as advanced
Step solution + source
Layers stacked along the field direction share the same charge but split the voltage, which is the series condition. Each half-gap forms a capacitor ($C_1$ vacuum, $C_2=KC_1$), and $\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$, giving an effective value between $C_0$ and $KC_0$, less than fully filling the gap. 🔉⇢

Source: authored

Q82 A dielectric slab of constant $K$ and thickness $t$ is inserted in a capacitor of gap $d$ (with $t\lt d$). The capacitance becomes advanced
Step solution + source
The slab and the remaining vacuum act as capacitors in series, so the effective gap becomes $d-t$ of vacuum plus $\dfrac{t}{K}$ from the slab, giving $C=\dfrac{\varepsilon_0 A}{d-t+t/K}$. For $K\to\infty$ it reduces to $\dfrac{\varepsilon_0 A}{d-t}$, and for $t\to0$ it returns to the empty value. 🔉⇢

Source: authored

Q83 The capacitance of an isolated spherical conductor of radius $R$ is medium
Step solution + source
For an isolated sphere $V=\dfrac{kQ}{R}$, so $C=\dfrac{Q}{V}=\dfrac{R}{k}=4\pi\varepsilon_0 R$. Capacitance grows linearly with radius and needs no second plate; the other plate is effectively at infinity, which is why larger spheres store more charge at the same potential. 🔉⇢

Source: NCERT XII Ch 2

Q84 The dielectric strength of a material is advanced
Step solution + source
Dielectric strength is the largest electric field a dielectric can tolerate before it ionises and conducts (breaks down), quoted in volts per metre. It limits the maximum voltage a capacitor can hold for a given gap and is distinct from the dielectric constant $K$, which measures how much the material boosts capacitance. 🔉⇢

Source: NCERT XII Ch 2

Q85 The capacitance of a capacitor depends on easy
Step solution + source
Capacitance is fixed by plate area, separation, shape and dielectric constant, as in $C=\dfrac{K\varepsilon_0 A}{d}$. It does not depend on the actual $Q$ or $V$, since these adjust together to keep the ratio $\dfrac{Q}{V}$ constant, so only changing geometry or medium changes $C$. 🔉⇢

Source: NCERT XII Ch 2

Q86 A capacitor blocks steady direct current because medium
Step solution + source
In steady DC the capacitor charges until its voltage matches the source, after which no further charge flows and the insulating gap prevents conduction. So it blocks steady DC while passing AC, where the continually changing voltage keeps charging and discharging the plates. 🔉⇢

Source: NCERT XII Ch 2

Q87 Capacitors in parallel have an equivalent capacitance equal to easy
Step solution + source
In parallel each capacitor sees the same voltage and the charges add, so $C_{eq}=C_1+C_2+\dots$. Adding capacitors in parallel is like enlarging the plate area, increasing total charge stored per volt, so the equivalent is always larger than the largest individual capacitor. 🔉⇢

Source: NCERT XII Ch 2

Q88 Capacitors in series have an equivalent capacitance given by easy
Step solution + source
In series each capacitor carries the same charge while the voltages add, so the reciprocals of capacitance add: $\dfrac{1}{C_{eq}}=\sum\dfrac{1}{C_i}$. The equivalent capacitance is therefore smaller than the smallest member, analogous to increasing the effective plate separation. 🔉⇢

Source: NCERT XII Ch 2

Q89 Two capacitors $2\,\mu F$ and $3\,\mu F$ connected in parallel give medium
Step solution + source
Parallel capacitances simply add, so $C_{eq}=2+3=5\,\mu F$. Each capacitor experiences the full applied voltage and their stored charges combine, giving a result that exceeds either individual value, as expected for a parallel combination that effectively increases plate area. 🔉⇢

Source: authored

Q90 Two capacitors $2\,\mu F$ and $3\,\mu F$ connected in series give medium
Step solution + source
For series, $\dfrac{1}{C_{eq}}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$, so $C_{eq}=\dfrac{6}{5}=1.2\,\mu F$. The equivalent value is less than the smaller capacitor, since the same charge must cross a larger effective separation; series stacking reduces overall capacitance. 🔉⇢

Source: authored

Q91 The energy stored in a charged capacitor is easy
Step solution + source
The energy is $U=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV=\dfrac{Q^2}{2C}$, the factor of one-half arising because the voltage builds up gradually from zero as charge accumulates. All three forms are equivalent through $Q=CV$, and this energy resides in the electric field between the plates. 🔉⇢

Source: NCERT XII Ch 2

Q92 The energy of a capacitor with charge $Q$ can also be written as medium
Step solution + source
Substituting $V=\dfrac{Q}{C}$ into $U=\dfrac{1}{2}CV^2$ gives $U=\dfrac{Q^2}{2C}$. This form is handy for an isolated capacitor where $Q$ is fixed. All equivalent forms $\dfrac{1}{2}CV^2$, $\dfrac{1}{2}QV$ and $\dfrac{Q^2}{2C}$ give the same energy, differing only in which variables are held. 🔉⇢

Source: NCERT XII Ch 2

Q93 A $4\,\mu F$ capacitor is charged to $100\,\text{V}$. The energy stored is hard
Step solution + source
Using $U=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times4\times10^{-6}\times(100)^2=\dfrac{1}{2}\times4\times10^{-6}\times10^4=0.02\,\text{J}$. The energy scales with capacitance and the square of voltage, so substituting the values gives 20 millijoules. 🔉⇢

Source: authored

Q94 The energy density (energy per unit volume) stored in the electric field of a capacitor is medium
Step solution + source
The electric field itself stores energy with density $u=\dfrac{1}{2}\varepsilon_0 E^2$. Multiplying by the volume between the plates ($Ad$) recovers $\dfrac{1}{2}CV^2$, showing energy resides in the field, not merely on the plates, and the density grows with the square of the field strength. 🔉⇢

Source: NCERT XII Ch 2

Q95 A capacitor charged to voltage $V$ is connected in parallel to an identical uncharged capacitor. The final common voltage is hard
Step solution + source
Charge is conserved and redistributes over the doubled capacitance. Initial charge $CV$ now sits on $2C$, giving $V_f=\dfrac{CV}{2C}=\dfrac{V}{2}$. Notably some energy is lost as heat in the connecting wires even though charge is conserved, a classic result of sharing charge between capacitors. 🔉⇢

Source: authored

Q96 In the sharing of charge between two identical capacitors above, the total energy of the system hard
Step solution + source
Initial energy is $\dfrac{1}{2}CV^2$; final is $\dfrac{1}{2}(2C)\left(\dfrac{V}{2}\right)^2=\dfrac{1}{4}CV^2$, so half the energy is lost, dissipated as heat and radiation in the wires. Charge is conserved but energy is not, because the sudden redistribution drives transient currents through the wire resistance. 🔉⇢

Source: authored

Q97 Three capacitors, each $C$, are all connected in series. The equivalent capacitance is advanced
Step solution + source
For identical series capacitors $\dfrac{1}{C_{eq}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{3}{C}$, so $C_{eq}=\dfrac{C}{3}$. Series combination lowers capacitance below any single member, here to one-third; each carries the same charge and the three equal voltage drops add to the source voltage. 🔉⇢

Source: authored

Q98 Two capacitors $C_1$ and $C_2$ are connected in series across a voltage $V$. The charge on each is advanced
Step solution + source
Series capacitors carry the same charge because the charge on the inner plates is induced and equal. That common charge is $Q=C_{eq}V$ with $C_{eq}=\dfrac{C_1 C_2}{C_1+C_2}$. The voltages then split inversely with capacitance, the smaller capacitor taking the larger share of the voltage. 🔉⇢

Source: authored

Q99 The force of attraction between the plates of a parallel-plate capacitor (charge $Q$, plate area $A$) is advanced
Step solution + source
Each plate sits in the field of the other, $\dfrac{\sigma}{2\varepsilon_0}=\dfrac{Q}{2\varepsilon_0 A}$, so the force is $F=Q\times\dfrac{Q}{2\varepsilon_0 A}=\dfrac{Q^2}{2\varepsilon_0 A}$. The factor of one-half appears because a plate exerts no force on itself; only the other plate's field acts, and the force is attractive. 🔉⇢

Source: authored

Q100 In a series combination of capacitors, the capacitor with the smallest capacitance has medium
Step solution + source
Series capacitors share equal charge $Q$, and $V=\dfrac{Q}{C}$, so the smallest $C$ has the largest voltage. This matters for breakdown, since the smallest capacitor is most likely to fail first. The voltages are inversely proportional to capacitance and sum to the applied voltage. 🔉⇢

Source: NCERT XII Ch 2

⏱️ Mock Test 30 Q · 60 min · +4 / −1 per question

Rules: Attempt all 30 questions in 45 minutes. Each question has exactly one correct option, and there is no negative marking in this practice mock. Where needed use $k=9\times10^{9}\,\text{N\,m}^2/\text{C}^2$ and $\varepsilon_0=8.85\times10^{-12}\,\text{F/m}$.

🎬 Video Lectures clip-indexed · NPTEL / IIT / MIT

MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.

Capacitors | Formulae and Concept REVISION in 22 min | JEE Physics by Mohit Sir (IITKGP) 🔉⇢
Eduniti - Physics by Mohit Goenka_IITKGP

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Eduniti - Physics by Mohit Goenka_IITKGP); found via yt-dlp search 'capacitors and capacitance class 12 physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Capacitors, dielectrics and capacitance — segment 1Hello hello kids welcome back to another form revision series a series where we discuss the concept along with the whole family in very less time in this video we are going to discuss the parameters which is your…capacitors, dielectrics and capacitance
  • 2:31–5:02Capacitors, dielectrics and capacitance — segment 2that the pass stand to my apps wala not a glutton cylinder upon lock the base will be here only and 281 is good now in this entire chapter we will discuss the thing which is a parallel plate capacitor read so this is…capacitors, dielectrics and capacitance
  • 5:02–7:32Capacitors, dielectrics and capacitance — segment 3zero, so whatever you had, you would have final only, fine, if you look carefully then what is the force between unbroken parallax, please charge parallax play store on this you will get plus to 4 [ __ ] - some charge,…capacitors, dielectrics and capacitance
  • 7:32–10:03Capacitors, dielectrics and capacitance — segment 4Now if you look carefully, there is a potential of one and one between one and one. There is a lining between two and one and there is a tip answer between two and air. So you have coded the circuit in this way. Can you…capacitors, dielectrics and capacitance
  • 10:03–12:34Capacitors, dielectrics and capacitance — segment 5to three and you for okay good one more thing if someone asks you how much is the charge charge then basically choice is basically equal to one plus why code it well and once you know why then you can find out what is…capacitors, dielectrics and capacitance
  • 12:34–15:04Capacitors, dielectrics and capacitance — segment 6and that is why the intensity found outside is more and inside it is a little less. Because there is a decrease in the intensity inside, so we can say here that the net inside it is nothing but title key - and these…capacitors, dielectrics and capacitance
Electric potential and potential difference || 3D animated explanation || class 12th & 10th Physics 🔉⇢
Visual Learning

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Visual Learning); found via yt-dlp search 'electrostatic potential and potential difference class 12 physics', oEmbed-verified live.

📑 Clips (1)
  • 0:00–2:24Electric potential and potential difference — segment 1In this video, we will understand electric potential and potential difference. As we know, electrons do not move on their own in a wire. To move electrons in a particular direction, an external force is required, and…electric potential and potential difference
The potential produced by a point charge is `V = kQ// r`. Use this information to determine the shap 🔉⇢
Doubtnut

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Doubtnut); found via yt-dlp search 'electric potential due to a point charge V=kq/r physics', oEmbed-verified live.

📑 Clips (1)
  • 0:00–1:38Electric potential and potential difference — segment 1Download it today and all your matches will be cleared by the doctor Chemistry Physics Biology Diet Just take a photo of the animals in the bus and get instant video solution Download now Jai Hind This question is the…electric potential and potential difference
Potential Due to an Electric Dipole Class 12 Physics chapter 2 🔉⇢
Mandeep Education Academy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Mandeep Education Academy); found via yt-dlp search 'electric potential due to a dipole class 12 physics', oEmbed-verified live.

📑 Clips (2)
  • 0:00–2:31Potential due to a dipole — segment 1Hello students, in this video we are going to do a very important derivation from the chapter Tu of class 12th Physics, which is to find electric potential, the electric dipole and the electric dipole. Let's start by…potential due to a dipole
  • 2:31–3:27Potential due to a dipole — segment 2find the potential. If it is complete then it is half a, so what is this distance from Pythagoras theorem and what is this distance. Obviously both the distances are equal, so both these distances are inside root a² -…potential due to a dipole
Electric Potential due to an Electric dipole | 12th Physics Handwritten Notes #cbse 🔉⇢
PHYSICS with Umesh Rajoria

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (PHYSICS with Umesh Rajoria); found via yt-dlp search 'electric potential due to a dipole class 12 physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:30Potential due to a dipole — segment 1That's a Venus diet plan, so gas is used in this system of equal and opposite words for electric cycle system in which both are equal but for each other, how much distance is there between the electric dipole and the…potential due to a dipole
  • 2:30–5:01Potential due to a dipole — segment 2and what is the name of this, position matter. So what did we see here that the electric dipole moment and the position which is restless, there is a funnel closed between them. Manglik had passed. Okay, so what did we…potential due to a dipole
  • 5:01–7:32Potential due to a dipole — segment 3both these lines and this length equal, play this curve has to be applied when the angle is small, now whether the angle here is small or not, it is small, how will you do this, so we said that your distance is very big…potential due to a dipole
  • 7:32–10:03Potential due to a dipole — segment 4that now you know how much is the MO, so what will you do, if you subtract the army from - then the MA distance will come which will be almost equal to 4, so what will you write it, hey, say art to equal to 4 - r total…potential due to a dipole
  • 10:03–12:46Potential due to a dipole — segment 5I will also tell you, so why this point of - and V - put that negative charge and the sum of this post is 2.2 here, it was - added, then what will be the net potential for OnePlus V2K belt? So how much is V1 V2? This is…potential due to a dipole
  • 12:46–15:19Potential due to a dipole — segment 6subscribe and I had subscribed for one like I invented this but its very small so see the volume it happens you can open it and remove it so instead of appointment only and this scientist who has the final answer this…potential due to a dipole
12. Equipotential Surfaces | Pledge 2023 | Electrostatics | CBSE | NCERT | Physics Baba 2.0 🔉⇢
Rankplus

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Rankplus); found via yt-dlp search 'equipotential surfaces class 12 physics explained', oEmbed-verified live.

📑 Clips (5)
  • 0:00–2:30Equipotential surfaces and field-potential relation — segment 1If you don't know numericals, no problem, if you do n't remember the formula, no problem, even after this this topic will be given marks and the chances of it coming are very high this time too, be it MCQ or theoretical…equipotential surfaces and field-potential relation
  • 2:30–5:03Equipotential surfaces and field-potential relation — segment 2perpendicular, that is why the potential surface, I just showed you here, I just made a potential surface here by joining all these points, if you look carefully, then all the electric field lines are always making an…equipotential surfaces and field-potential relation
  • 5:03–7:34Equipotential surfaces and field-potential relation — segment 3go from one point to another on an equipotential surface, then what work will I get? I will get zero work. If I move from one to the other, I will get something, but if I move from one to the other, I will get nothing.…equipotential surfaces and field-potential relation
  • 7:34–10:04Equipotential surfaces and field-potential relation — segment 4repair, it means here the electric field lines and electric field intensity are weak because these two are repelling each other, what happens outside, it will also repair it, if you place a charge, it is also repairing,…equipotential surfaces and field-potential relation
  • 10:04–11:35Equipotential surfaces and field-potential relation — segment 5distance and the charge is also the same, so the potential of point B will be the same as the potential of A, that is, if work done is equal to zero, this is the question that has come, keep it in your mind, anyway it…equipotential surfaces and field-potential relation
5.E = −dV/dr Explained | Relation Between Electric Field & Potential |Class 12 Physics| NEET +BOARDS 🔉⇢
Neeva Physics Classes

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Neeva Physics Classes); found via yt-dlp search 'relation between electric field and potential E=-dV/dr physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Equipotential surfaces and field-potential relation — segment 1Hello Bachcha Party Welcome Back to Neeva Physics Classes. So today we have brought an important concept for you. Relation between electric field and electric potential. Ok? In the last video we learned about potential.…equipotential surfaces and field-potential relation
  • 2:31–5:02Equipotential surfaces and field-potential relation — segment 2act on it and gravity will pull it down. but what are you doing? Against Gravity is getting the work done. So the work done to take this object 10 meters up gets stored in potential energy. Work done is potential…equipotential surfaces and field-potential relation
  • 5:02–7:32Equipotential surfaces and field-potential relation — segment 3what should you do by dividing it? We have divided the value of the charge you are bringing by the work done for that charge. So what will happen son? Potential V. Ok? Only work done is potential energy. and work done…equipotential surfaces and field-potential relation
  • 7:32–10:02Equipotential surfaces and field-potential relation — segment 4potential energy it will have will be more. Ok? Slowly, slowly, slowly the potential energy will decrease. So what happens to the electric field of positive charges ? It goes away. Meaning it goes away. So what happened…equipotential surfaces and field-potential relation
  • 10:02–12:33Equipotential surfaces and field-potential relation — segment 5giving an examination and your examination is of total 500 marks. So what is that 500, son? Total is your energy. But if you take out this 500 as a subject. Ok? But if the subject is 500, then 100 of one subject is…equipotential surfaces and field-potential relation
  • 12:33–15:05Equipotential surfaces and field-potential relation — segment 6energy. Ok? I can also write this as del u. So from here I can also write the work done - del u of the electric field. del u = q * e * d ok? And U which is potential energy. What is potential energy ? What is the…equipotential surfaces and field-potential relation
Potential Energy of a System of Charges | NCERT CBSE Class 12 Physics | Muruga MP 🔉⇢
Open Your Mind With Muruga MP

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Open Your Mind With Muruga MP); found via yt-dlp search 'electrostatic potential energy of a system of charges class 12 physics', oEmbed-verified live.

📑 Clips (6)
  • 0:01–2:57Electrostatic potential energy — segment 1hi everyone NC CB 12th standard physics MC soal energy of a system of charges so potential energy is equal W soti work done per un charge W which is equal to Q * of V General but so this is the work done it is nothing…electrostatic potential energy
  • 2:57–5:30Electrostatic potential energy — segment 2particular superp football or Cricket whatever it is for system of charges collection of charges so v k q by R so KQ by R charge potential one Charelectrostatic potential energy
  • 5:30–8:28Electrostatic potential energy — segment 3let me obviously potential V2 which is equal to Q * V Q q1 V1 understood right so this is nothing but in the second poal so potential can work done on Q2 k q1 r the distance and in the Q2 second let me say that is R1 so…electrostatic potential energy
  • 8:28–11:01Electrostatic potential energy — segment 4potential corre it may be positive positive or negative negative giving against this force electric field against electric field I'm applying this [Music] I'm doing against this negative q1 Q2 Q3 total potential which…electrostatic potential energy
  • 11:01–14:18Electrostatic potential energy — segment 5right q1 R1 2 plus q1 q1 Q3 by r13 plus q1 Q2 Q3 by even r r 23313 q1 Q3 plus Q2 Q3 R23 Q2 Q3 R 2.4 yes 2.4 2. likeelectrostatic potential energy
  • 14:18–15:03Electrostatic potential energy — segment 6maybe I think I don't remember this actually if it is a positive test charge Z basic understanding thank you help this with aelectrostatic potential energy
Potential Energy of a Dipole in a Uniform Electric Field || By: Mr. Bipin Kumar Singh 🔉⇢
Bipin Kumar Singh #physics

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Bipin Kumar Singh #physics); found via yt-dlp search 'potential energy of a dipole in a uniform electric field physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:35Electrostatic potential energy — segment 1A potential energy of and Udaipur in uniform electric considered a pole and taken place in uniform electric field in playlist a delightful movement Mick Jagger ok with electric field that dipole moment is observed from…electrostatic potential energy
  • 2:41–5:13Electrostatic potential energy — segment 2that cos theta 0 - the cost you that world color is stored in the form of potential energy which gives me equal to one plus cos theta one minus cos theta a to go okay okay now that this is 589 that is equal to 19th that…electrostatic potential energy
  • 5:15–6:29Electrostatic potential energy — segment 3talk about net question we have question 123 equal to 180 degrees you only at what time will you be plus the previous entries cord dispositions college unstable cloud computer we and stable that the previous two that I…electrostatic potential energy
Electrostatics of conductors | Electric Potential & Capacitance | 12 Physics #cbse #umeshrajoria 🔉⇢
PHYSICS with Umesh Rajoria

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (PHYSICS with Umesh Rajoria); found via yt-dlp search 'electrostatics of conductors class 12 physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:30Electrostatics of conductors — segment 1[Music] What is the next topic? Electrostatics of Conductors. Like the name of the hat is Electrostatics of Conductors. What do we have to study in this? We have to talk specifically about conductors. What are…electrostatics of conductors
  • 2:30–5:05Electrostatics of conductors — segment 2I resumed this conductor, okay, so what did I do, I charged it, so when you charge it, what will be the direction of the electric field produced around it, electric The field direction will always be perpendicular to…electrostatics of conductors
  • 5:05–7:35Electrostatics of conductors — segment 3potential is equal, that much potential will be there. How do we prove this on its surface? So to prove this, I am assuming that we have a charge, a positive charge, and I am taking it from A to B. So, for this, I had…electrostatics of conductors
  • 7:35–10:06Electrostatics of conductors — segment 4charged it, then if we take any point on its surface, for example, if we take any point, what is the survey chart above this point, then the surface charge at this point on the surface of this conductor is sigma, so the…electrostatics of conductors
  • 10:06–12:36Electrostatics of conductors — segment 5okay, so if I apply Koshish, then I can write by Koshish, we will apply integration, now see and tell me what will come out if you open this, the intensity for the cylinder will be different for different surfaces, for…electrostatics of conductors
  • 12:36–13:54Electrostatics of conductors — segment 6point just near it, I have zoomed it on the cylinder, so a will be a constant here because this point is very close, so I wrote a outside the constant, integration for the first surface went to ten a and it has become…electrostatics of conductors
Electrostatic Potential and Capacitance Class 12 Physics - Electrostatics of Conductors 🔉⇢
LearnFatafat

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (LearnFatafat); found via yt-dlp search 'electrostatics of conductors class 12 physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:33Electrostatics of conductors — segment 1Jhal that this traffic saunf conductors the substances battle of law of karma r conductor that a conductors characterized by president of movable barrier in case of metal gear electronic filing of electrolyte positive…electrostatics of conductors
  • 2:33–5:04Electrostatics of conductors — segment 2Subscribe Subscribe Button Electrostatic Vibrations 2018 201 Aap Busy Raho Conductor Can Have No Excess Charge In 10 Traffic Situations That Think About You Should Conduct Up The Receiver Mode Of Positive And Negative…electrostatics of conductors
  • 5:04–5:48Electrostatics of conductors — segment 3Now Electronics Fielding Male And Conductors And Electric Field Inside The Video then subscribe to the Page if you liked The Video then subscribe to theelectrostatics of conductors
Capacitors|Capacitance|NCERT|CBSE|Physics 12|Tamil|Muruga MP#murugamp#ncert#physics12 🔉⇢
Open Your Mind With Muruga MP

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Open Your Mind With Muruga MP); found via yt-dlp search 'capacitors and capacitance class 12 physics', oEmbed-verified live.

📑 Clips (4)
  • 0:30–3:02Capacitors, dielectrics and capacitance — segment 1capacitors and capacitance so capacitor and capacitance on our understanding down so capacitor updated parallel conductors are number capacitor everyday rendez parallel conductors two parallel conductors so wrinkles so…capacitors, dielectrics and capacitance
  • 3:02–5:32Capacitors, dielectrics and capacitance — segment 2the potential difference right when i increase the potential charge increases when i decrease the potential charge decreases upon input ram in the v and q and the directly proportional upper when i increase the voltage…capacitors, dielectrics and capacitance
  • 5:32–8:05Capacitors, dielectrics and capacitance — segment 3oppose c which is equal to q by v so c is nothing but capacitance and the capacitance will depend upon the material which is made setting up in the conductor numbers only in the parallel conductor insulator and…capacitors, dielectrics and capacitance
  • 8:05–9:21Capacitors, dielectrics and capacitance — segment 4potential increases to this range you will arrange on the potential increase it will not exceed than this will be in the limit it will be safe in the topic of and of course we have a representation for a capacitance…capacitors, dielectrics and capacitance
Capacitance of Parallel Plate Capacitor with Dielectric Slab Derivation | Class 12 Physics 🔉⇢
Mandeep Education Academy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Mandeep Education Academy); found via yt-dlp search 'parallel plate capacitor with dielectric physics derivation', oEmbed-verified live.

📑 Clips (2)
  • 0:00–2:33Capacitors, dielectrics and capacitance — segment 1Hello students, in this video we are going to do a very important derivation from Chapter 2 of Class 12th Physics to find the capacitance of a capacitor. The dielectric slab is inserted between its plates. Let's start.…capacitors, dielectrics and capacitance
  • 2:33–3:40Capacitors, dielectrics and capacitance — segment 2put e 0 0 ba instead of this e we put e 0 0 ba we took the common inside minus the remainder d - t p t ba the field between the plates of a capacitor is e 0 that is sigma by a no so we put sigma by a no sigma you know…capacitors, dielectrics and capacitance
Combination of Capacitors in Series & Parallel | Class 12 Physics Chapter 2 | CBSE Board 🔉⇢
Mandeep Education Academy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Mandeep Education Academy); found via yt-dlp search 'combination of capacitors series and parallel physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:30Combination of capacitors and energy stored — segment 1Hello student, in this video we are going to discuss a very important topic from class 12th Physics chapter 2 which is called series and parallel combination of capacitors. Let's start first of all, here we have three…combination of capacitors and energy stored
  • 2:32–5:05Combination of capacitors and energy stored — segment 2nothing is connected from here till here then there will be V here also and if we take the negative as zero then there will be zero here also, that is, the total potential difference across these three capacitors from…combination of capacitors and energy stored
  • 5:05–6:14Combination of capacitors and energy stored — segment 3different. But because this battery is connected and our total charge is q1 plus q2, then in place of q1 we write c1v and in place of q2 we write c2v. Now if we replace these two, if a single capacitor is installed…combination of capacitors and energy stored
Series and Parallel Grouping of Capacitors || animated hindi explanation || 12th class Physics || 🔉⇢
Visual Learning

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Visual Learning); found via yt-dlp search 'combination of capacitors series and parallel physics', oEmbed-verified live.

📑 Clips (2)
  • 0:00–2:30Combination of capacitors and energy stored — segment 1Grouping of capacitors Capacitors can be done in two ways - serial grouping and series grouping. So let us understand how to calculate the total capacitance of capacitors in series and series grouping. A. Serial…combination of capacitors and energy stored
  • 2:30–5:01Combination of capacitors and energy stored — segment 2we get some such value, this is the total capacitance of the individual capacitor. First grouping: When we connect the left terminal of all the capacitors to one end of the battery and the right terminal of all the…combination of capacitors and energy stored
वैद्युत विभव तथा विभवांतर | स्थिर वैद्युत विभव तथा धारिता | Class 12 Physics Chapter 2 🔉⇢
MK Sir Inspiration

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (MK Sir Inspiration); found via yt-dlp search 'विद्युत विभव और विभवांतर electric potential Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:01–2:33Electric potential and potential difference — segment 1हेलो मेरे प्यारे बच्चों क्या हाल चाल बढ़िया सब प्यारे बच्चों आज के इस वीडियो में हम फिजिक्स के सेकंड चैप्टर स्थिर वैद्युत विभव तथा विभवांतर को देखने वाले हैं और इसके अंतर्गत बच्चों आज हम देखेंगे वैद्युत विभव तथा…electric potential and potential difference
  • 2:33–5:04Electric potential and potential difference — segment 2दूर चलता है 300 मीटर दूर चलता है कार्य कितना किया जीरो किया रसगुल्ला कुछ भी नहीं क्यों क्योंकि विस्थापन नहीं हुआ ना भाई सर पे वो बोझा है टैची है पड़ी हुई है आप धक्का मार रहे हो दीवाल को दिन भर आप भले ही चाहे जितनी…electric potential and potential difference
  • 5:04–7:35Electric potential and potential difference — segment 3कार्य किया गया कार्य वैद्युत विभव कहलाता है वैद्युत विभव कहलाता है ठीक वैद्युत विभव कहलाता है अब यहां पर भाई क्या होता है इसे लिखेंगे कि इसे वि से दर्शाते हैं इसे क्या होता है बच्चों व से दर्शाते हैं ठीक है इसे व से…electric potential and potential difference
  • 7:35–10:05Electric potential and potential difference — segment 4आवेश तो आवेश का होता है ए ठीक इसका विमा क्या हो जाता है ए हो जाता है यही जब ऊपर जाता है भाई यही जब ऊपर जाएगा तो इसका डायमेंशन फार्मूला क्या हो जाएगा ये हो जाएगा m ए स्क्वायर t की पावर देखिए यहां पर माइ दो है ये प्लस है…electric potential and potential difference
  • 10:05–12:35Electric potential and potential difference — segment 5को धन परीक्षण आवेश को लिखेंगे या धन आवेश को कुछ भी कहले धन को एक बिंदु से एक बिंदु से दूसरे बिंदु तक एक बिंदु से दूसरे बिंदु तक द्युत क्षेत्र में एका धन परीक्षण आवेश को एक बिंदु से दूसरे बिंदु तक ले जाने में ले जाने में…electric potential and potential difference
  • 12:35–13:37Electric potential and potential difference — segment 6बीच दो बिंदुओं के बीच और बिंदु कहां रहेंगे वैद्युत क्षेत्र में दो बिंदुओं के बीच विभव के अंतर को विभव के अंतर को विभवांतर कहते हैं ये नाम में ही बहुत कुछ है भा विभवांतर यानी विभव का अंतर तो दो बिंदुओं क्लियर हो गया…electric potential and potential difference
विद्युत विभव | विद्युत विभवांतर | स्थिर वैद्युत विभव तथा धारिता | Class 12 Physics 🔉⇢
MK Sir Inspiration

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (MK Sir Inspiration); found via yt-dlp search 'विद्युत विभव और विभवांतर electric potential Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:05–2:35Electric potential and potential difference — segment 1हेलो बच्चों, कैसे हो आप सब? आशा करता हूं आप सभी अच्छे होंगे और आपकी तैयारी भी एकदम बढ़िया तरीके से चल रही होगी। बहुत सारे बच्चे बार-बार यही पूछते हैं कि सर आप रेगुलर नहीं बन रहे हैं। बीच-बीच में एक दो दिन गैप हो जा रहा…electric potential and potential difference
  • 2:35–5:05Electric potential and potential difference — segment 2कितना कार्य किया जीरो क्योंकि वो सर पे ही पड़ा है विस्थापन नहीं हो रहा है। लेकिन जैसे आप किसी चीज को दीवार को धक्का दे रहे हो दीवार गिर गई तो कार्य किया जाएगा। दीवार वैसे की वैसी पड़ी है तो आप चाहे जितना जूझ हो कार्य…electric potential and potential difference
  • 5:05–7:36Electric potential and potential difference — segment 3किया गया कार्य वैद्युत विभव कहलाता है। सिंपल सा है भैया याद रखना वैद्युत विभव कहलाता है। इसे P से दर्शाते हैं। विभव को भैया किससे दर्शाते हैं? V से दर्शाते हैं। अब फार्मूला देखिए क्या हो जाएगा? तो भैया यहां पर हमने…electric potential and potential difference
  • 7:36–10:07Electric potential and potential difference — segment 4रहेगा और A का क्या हो जाएगा? LT इनवर्स 2। मान लेते हैं A का नहीं पता। तो त्वरण बराबर वेग परिवर्तन बटे समय यहां से निकाल लीजिए। तो मतलब ऐसे अगर हम तोड़ते चले जाएं तो बीमा निकल आता है। अगर नहीं भी पता तो भैया एक बहुत…electric potential and potential difference
  • 10:07–12:38Electric potential and potential difference — segment 5मिलताजुलता है। बस थोड़ा सा अंतर है। मात्रक वही होगा, विमा वही होगा, राशि भी। ठीक है? अब अंतर क्या है समझते हैं। नाम ही है विभवांतर। ठीक? जैसे भैया मान लेते हैं कि वैद्युत क्षेत्र के भीतर हमने एक दो बिंदु ले लिया। a और b…electric potential and potential difference
  • 12:38–15:08Electric potential and potential difference — segment 6एक बिंदु से दूसरे बिंदु तक अगर हम ले जाएंगे धन परीक्षण आवेश को तो ले जाने में किया गया कार्य वैदुत विभवांतर कहलाता है। वैद्युत विभवांतर कहलाता है, ठीक है भईया, क्या कहलाता है? वैद्युत विभवांतर कहलाता है। ठीक, जैसे भी…electric potential and potential difference
वैद्युत द्विध्रुव के कारण किसी बिन्दु पर वैद्युत विभव का व्यंजक । vaidyut dvidhrav ke karan vibhav 🔉⇢
Balaji Study coaching

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Balaji Study coaching); found via yt-dlp search 'द्विध्रुव के कारण विभव potential due to dipole Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:01–2:32Potential due to a dipole — segment 1तो क्लास 12 की फिजिक्स विषय के अंतर्गत आप सभी लोगों को वैद्युत ध्रु के कारण उत्पन्न वैद्युत विभव का व्यंजक ज्ञात करना बताया जा रहा था जिसमें मैंने आप सभी लोगों को दो स्थितियां अभी तक बताई हैं पिछली क्लास में मैंने अक्षी…potential due to a dipole
  • 2:32–5:02Potential due to a dipole — segment 2हमें प्लस क आवेश के कारण p पर वैद्युत विभव निकालना है और माइनस क आवेश के कारण p पर वैद्युत विभव निकालना है लेकिन हमको ये डिस्टेंस सबसे पहले पता होनी चाहिए ये डिस्टेंस ना बराबर है और ना कोई निश्चित है ना कि हम समकोण…potential due to a dipole
  • 5:02–7:32Potential due to a dipole — segment 3भैया x नाम दे दीजिए या कुछ भी नाम दे दीजिए ए इसको नाम दीजिए c ये जो पॉइंट है ये c पॉइंट है ठीक है तो जो दूरी देखिए देखिए ये जो एक देखिए समकोण त्रिभुज बन रहा है ये देखिए समकोण त्रिभुज हम बना लेते हैं ये ऐसे बना हुआ है…potential due to a dipole
  • 7:32–10:04Potential due to a dipole — segment 4ओ की वैल्यू हमने निकाली कितनी निकाली l कॉस थीटा तो इस तरीके से हमें प्लस क आवेश से p तक की दूरी पता चल चुकी है अब हमको निकालना है इस माइनस क आवेश से p तक की दूरी तो हम यहां से इस वाली भुजा से सामने वाली भुजा पर लंब…potential due to a dipole
  • 10:04–12:35Potential due to a dipole — segment 5कारण बहुत आसान डेरिवेशन है माइनस क आवेश के कारण कहां पर भैया बिंदु प पर क्या निकालेंगे विभव निकालेंगे चलिए भाई माइनस क्य आवेश के कारण निकाल लेते हैं इसको v1 नाम दे देंगे इसको v2 नाम दे देंगे तो ये हो जाएगा 1 अप 4 पा एन…potential due to a dipole
  • 12:35–15:05Potential due to a dipole — segment 6के क्य l कस थीटा l कस थीटा कितना हो जाएगा 2l कस थीटा बटे में कितना हो जाएगा r स् - l स् कितना हो जाएगा क स्क्वा थीटा अब आप सभी लोगों ने पढ़ा है आवेश और 2l का जो गुणनफल होता है वो वैद्युत ध्रुव आगोर के बराबर होता है तो v…potential due to a dipole
12th Phy | L-3 : - Potential due to electric dipole | Potential and Capacitors hindi medium by As... 🔉⇢
ashish singh lectures

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (ashish singh lectures); found via yt-dlp search 'द्विध्रुव के कारण विभव potential due to dipole Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:11–2:43Potential due to a dipole — segment 1हेलो हेलो हेलो हेलो हां जी हेलो स्टूडेंट स्वागत है आप सभी का आशीष सिंह लेक्चर की इस लेटेस्ट लाइव स्ट्रीम में बच्चों नमस्कार कैसे हो आप सभी लोग उम्मीद करता हूं बहुत ही बेहतरीन होंगे और एसल के साथ आपकी पढ़ाई एकदम शानदार…potential due to a dipole
  • 2:49–5:19Potential due to a dipole — segment 2विभव मतलब हम एक विद्युत द्विध्रुव के कारण विभव निकालने जाएंगे आज की क्लास में विद्युत द्विध्रुव के बारे में बेसिक नॉलेज तो आपके पास होगी फर्स्ट चैप्टर में आपको याद है विद्युत द्रु क्या होता है हम लोगों ने पढ़ा था इसको…potential due to a dipole
  • 5:24–7:57Potential due to a dipole — segment 3चित्रास हमें एक विद्युत द्विध्रुव के विद्युत द्विध्रुव से स्मल आर दूरी पर स्थित उसके अक्ष पर बिंदु प पर विद्युत विभव की गणना करनी है माना चित्रा अनुसार हमें विद्युत द्विध्रुव के अक्ष पर विद्युत द्विध्रुव के अक्ष पर उससे…potential due to a dipole
  • 7:57–10:27Potential due to a dipole — segment 4है चलो देख ध्यान से बहुत इजी लगेगा आपको डेरिवेशन इनके बीच की दूरी हो गई 2a मतलब सेंटर पॉइंट से ये भी a होगा और ये भी a होगा जो पॉइंट p अपन ले रहे हैं ये अक्ष है ये आप लोग फर्स्ट चैप्टर में भी देख चुके हो कि विद्युत…potential due to a dipole
  • 10:27–13:01Potential due to a dipole — segment 5है जो मैंने आपके सामने अभी हा ट की है कैसे देखेंगे देखो यहां से यहां तक की दूरी a है तो यहां से यहां तक की दूरी r है तो कितना हो गया a प् आ कितनी दूरी हो गई बेटा कितनी दूरी हो गई बेटा ये दूरी हो गई r प् ए सेट इक्वेशन…potential due to a dipole
  • 13:01–15:32Potential due to a dipole — segment 6समीकरण एक व दो से जोड़ दो इसको v प किसके इक्वल हो जाएगा v प्स क प्लस v माइनस क बस सिंपल जोड़ना है भाई अदिश है ना यार सदिश में लफड़े होते हैं अदिश में तो बिल्कुल सिंपल जोड़ना है खत्म करना है अंकुश पढ़ते रहो बस यही तरीका…potential due to a dipole
12th Phy | L-6 : - Equipotential Surface | Potential and Capacitor hindi medium 🔉⇢
ashish singh lectures

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (ashish singh lectures); found via yt-dlp search 'समविभव पृष्ठ equipotential surfaces Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:06–2:52Equipotential surfaces and field-potential relation — segment 1हेलो हेलो हेलो हेलो हेलो हां जी हेलो स्टूडेंट स्वागत है आप सभी का आशीष सिंह लेक्चर के इस लेटेस्ट लाइव स्ट्रीम में प्यारे बच्चों नमस्कार कैसे हो आप सभी लोग उम्मीद करता हूं बहुत ही बेहतरीन होंगे और एसर के साथ आपकी पढ़ाई…equipotential surfaces and field-potential relation
  • 2:52–5:34Equipotential surfaces and field-potential relation — segment 2रहा देखो अब अगर आप आज का डेरिवेशन देखो तो आज का डे डेरिवेशन किसके लिए होने जा रहा है आज का डेरिवेशन होने जा रहा है हेडिंग लगाए ठोस अचा गोले के कारण ठोस अचा गोले के कारण ठोस अचानक गोले के कारण विद्युत विभव है ना ठोस…equipotential surfaces and field-potential relation
  • 5:34–8:05Equipotential surfaces and field-potential relation — segment 3अरे अंदर जीरो नहीं होता है गधों यह अचानक है अचानक में जीरो नहीं होता चालक नहीं है अ चालक है बताओ यार तुम इतना भी याद नहीं कर पा रहे हो [संगीत] के कल आ डिवाइड बाय कैपिटल आ क अब सही पकड़े हो k क आ डिवाइड बाय कपि आ क के कल…equipotential surfaces and field-potential relation
  • 8:05–10:41Equipotential surfaces and field-potential relation — segment 4सुनो स थ्योरी निकली एनसीआरटी यह थोड़ा एक्स्ट्रा चीज पढ़ रहे हैं अपन तो इसलिए इसको बस एस फार्मूला याद रखना है आपको चलो काम की बात पर आते हैं तो आपने क्या ऑब्जर्व किया चालक गोला हो फर्स्ट चैप्टर याद करो बेटा चालक गोला हो…equipotential surfaces and field-potential relation
  • 10:41–13:11Equipotential surfaces and field-potential relation — segment 5ठीक है देखें देखें इसको बहुत बढ़िया बहुत बढ़िया चलो अब थर्ड केस पे आते हैं जिस केस में क्या है भीतर स्थित बिंदु पर विद्युत विभव केस थ्री अलग है उसी को डायरेक्ट कर रहे हैं आप अपनी कॉपी में दो पेज छोड़ सकते हो साथ-साथ…equipotential surfaces and field-potential relation
  • 13:11–15:42Equipotential surfaces and field-potential relation — segment 6क्षेत्र का मान है यह मान लेते हैं यह बिंदु जो है व स्मल आर दूरी पर स्थित है कितनी दूरी पर स्थित हैल आर पर बाहर जो विद्युत क्षेत्र का मान है अचानक गोले के लिए वह है के क अपन आर स्क्वा और कैपिटल आर सेमल आर तक कैपिटल आर…equipotential surfaces and field-potential relation
Equipotential Surface Explained | Class 12 Physics | Full Concept in Hindi with Animation 🔉⇢
Physics and animation हिंदी में

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Physics and animation हिंदी में); found via yt-dlp search 'समविभव पृष्ठ equipotential surfaces Hindi physics class 12', oEmbed-verified live.

📑 Clips (4)
  • 0:00–2:30Equipotential surfaces and field-potential relation — segment 1हेलो स्टूडेंट्स। आज के इस वीडियो में हम एक्सप्लोर करने जा रहे हैं इक्विपोटेंशियल सरफेसेस का कांसेप्ट एनिमेटेड विज़ुअल्स की मदद से। इस वीडियो में हम कुछ इंपॉर्टेंट टॉपिक्स कवर करेंगे। सबसे पहले इक्विपोटेंशियल सरफेस क्या…equipotential surfaces and field-potential relation
  • 2:30–5:01Equipotential surfaces and field-potential relation — segment 2को इक्विपोटेंशियल सरफेस के एक पॉइंट से दूसरे पॉइंट तक मूव करते हैं तो नेट वर्क डन हमेशा जीरो होता है। इस बात को हम समझ सकते हैं पोटेंशियल डिफरेंस और वर्क डन के इक्वेशन से। वर्क डन WBA डिवाइडेड बाय चार्ज इक्व VM - VB…equipotential surfaces and field-potential relation
  • 5:01–7:32Equipotential surfaces and field-potential relation — segment 3इंपॉसिबल है। और उस पॉइंट पर दोनों सरफेससेस के अलग-अलग पोटेंशियल वैल्यूस होगी। V1 S1 के लिए और V2 S2 के लिए। लेकिन किसी भी सिंगल पॉइंट के लिए दो अलग-अलग पोटेंशियल वैल्यूस होना पॉसिबल नहीं है। इसलिए दो इक्विपोटेंशियल…equipotential surfaces and field-potential relation
  • 7:32–10:05Equipotential surfaces and field-potential relation — segment 4लाइंस क्लोजर होती है। फाइनली जब हम इस पैटर्न को 3D में विज़ुलाइज़ करते हैं तो हमें एक प्रॉपर स्ट्रक्चर मिलता है जो इलेक्ट्रिक फील्ड लाइंस और इक्विपोटेंशियल सरफेसेस के बीच के रिलेशन को क्लियरली एक्सप्लेन करता है। अब चलिए…equipotential surfaces and field-potential relation
Force Batch -12th Physics :- Ch.02 - L-08 स्थिर विद्युत - स्थितिज ऊर्जा by Ashish sir 🔉⇢
ashish singh lectures

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (ashish singh lectures); found via yt-dlp search 'स्थिरवैद्युत स्थितिज ऊर्जा potential energy of charges Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:33Electrostatic potential energy — segment 1झाल अजय को हेलो हेलो हां जी तो चलिए फिर शुरू करते हैं हेलो फ्रेंड स्वागत है आप सभी का सेटिंगस कि इस लैटर लव स्टोरी में दूसरा आइटम हम लोग जो है इससे विद्युत स्थितिज ऊर्जा है उसका नाम बिल्कुल ऐसे नहीं है इस फिर वृध्दि…electrostatic potential energy
  • 2:33–5:03Electrostatic potential energy — segment 2कि एकल आवेश की कमी भी विद्युत स्थितिज ऊर्जा नहीं होती आपको पता जब मैं स्टिचिंग करता हूं तो मैं स्टैंडर्ड आफ टीचिंग हैं हम लोग जो इस साल मेंटेन कर रहे हैं वह आईआईटी जेईई और नेट का लेवल का तो मैं आपको दो कि मैं नहीं रह…electrostatic potential energy
  • 5:03–7:33Electrostatic potential energy — segment 3प्रारंभिक स्थितिज ऊर्जा कि प्रारंभिक कि स्थितिज ऊर्जा अच्छा ठीक है तो आपने क्या किया इन दोनों आवेशों के बीच में जो दूरी है उस दूरी को बदल दिया आपने आवश्यक हो सकता और पास में लेकर कि आगे वह सकता है निवेशकों को दूर ले…electrostatic potential energy
  • 7:33–10:03Electrostatic potential energy — segment 4चाहे इस रास्ते से जाओ काली डमरु योगा चाहिए आपके इस रास्ते से होकर जब लोकायुक्त इससे ज्यादा गाढ़ी डब्लू होगा चाहे आप ऐसे जाओ काली डब्लू होगा और यह बताया था कि संरक्षित अव्वल होते हैं जिन लिए अगर आप एक से दो तक गए अपराधों…electrostatic potential energy
  • 10:03–12:35Electrostatic potential energy — segment 5परिवर्तन जब कर रहे हो कि निवेशकों को पास में लेकर आया हूं चाय दूर लेकर जा रहे हो आपको एक काम बहुत धीरे-धीरे करना है इस प्रकार से करना है कि आप की गतिज ऊर्जा आवेश की बदलने नहीं चाहिए तो यहां पर एक कंडीशन है कंडीशन है कि…electrostatic potential energy
  • 12:35–15:06Electrostatic potential energy — segment 612 आवेशों के निकाय की स्थितिज ऊर्जा हो मैं ऑफिस नोट कर लो बेटा तुरंत अभी के अभी दो आवेशों के निकालने की स्थितिज ऊर्जा दो आवेशों के निकाय की स्थितिज ऊर्जा शुरू करें देखो आप जानते हैं लिए सबसे पहले हम यह मान लेते हैं कि…electrostatic potential energy
दो आवेशों के निकाय की स्थितिज ऊर्जा | Class 12 Physics | By Monu Sir 🔉⇢
MK Sir Inspiration

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (MK Sir Inspiration); found via yt-dlp search 'स्थिरवैद्युत स्थितिज ऊर्जा potential energy of charges Hindi physics class 12', oEmbed-verified live.

📑 Clips (4)
  • 0:02–2:34Electrostatic potential energy — segment 1हेलो मेरे प्यारे बच्चों क्या हाल चाल बढ़िया सब भाई तबीयत एक दिन खराब थी इसलिए बच्चों कल वीडियो नहीं आ पाई उसके लिए दिल से सॉरी लेकिन गर्मी इतनी ज्यादा है कि भाई दिक्कत हो ही जाती है क्योंकि इतना ज्यादा गर्मी होती है इस…electrostatic potential energy
  • 2:35–5:06Electrostatic potential energy — segment 2है क्लियर इतना तो समझ में आ गया भाई कि हमने बताया कि एक दूसरे से आर दूरी पर है फिर आगे लिखेंगे कि ये क्रमश ए तथा बी पर स्थित है एक तथा बी पर स्थित है ठीक है तो भाई ये जो लिखने वा वाली चीजें होती है भैया स्टार्टिंग में…electrostatic potential energy
  • 5:06–7:38Electrostatic potential energy — segment 3तक लाने में किया गया कार्य लाने में किया गया कार्य अब कितना कार्य होगा भैया वही चीज देखना है तो कार्य बराबर डब् बराबर होता क्या है बच्चों विभव बराबर क्या बताया था भाई बराबर डब बा सॉरी व बराबर ी बराबर ड बा क ये लिखे थे ई…electrostatic potential energy
  • 7:38–9:02Electrostatic potential energy — segment 4में निहित होगा तो भाई स्थितिज ऊर्जा के रूप में निहित होगा तो स्थिति ऊर्जा के रूप में अब स्थितिज ऊर्जा को हम किससे प्रदर्शित करते हैं तो य से तो डब् के स्थान पर अपन क्या लिख देंगे बच्चों य लिख देंगे क्या लिखेंगे भैया य…electrostatic potential energy
BCS12Th-#24 NCERT | कक्षा-12 | अध्याय-2 | चालक स्थिरवैद्युतिकी | ELECTROSTATICS OF CONDUCTORS | 🔉⇢
BCS Academy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (BCS Academy); found via yt-dlp search 'चालकों का स्थिरवैद्युतिकी electrostatics of conductors Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Electrostatics of conductors — segment 1कि नमस्कार दोस्तों आप सभी का स्वागत है हमारे YouTube चैनल ऋषि स्पेसिफिक नहीं और आज के इस वीडियो में हम है कक्षा 12 जो एनसीईआरटी की पाठ्य पुस्तक उसका सेकंड चैप्टर है का एक महत्वपूर्ण टॉपिक लेकर चल रहे हैं चालक स्थिर…electrostatics of conductors
  • 2:31–5:02Electrostatics of conductors — segment 2कहलाते हैं तब चालक हैं उनको भी हम अलग भागों में डिवाइड कर सकते हैं देखिए जैसे हमें अब बात करें धात्विक चालक की धात्विक मेज़ चालक की बात करें तो धात्विक चालकों के अंदर क्या होता है धातु चालक का मतलब है धातु से बने हुए…electrostatics of conductors
  • 5:03–7:33Electrostatics of conductors — segment 3अनुपस्थितों नागरिक से दूर बल्कि द्वारा बना होता है थोड़ी सी भुजिया पापड़ सिद्धार्थ ने स्वतंत्र रूप से विचरण करता है शुद्ध रूप से गति करता है इसलिए उन्हें इलेक्ट्रॉनों को हम मैं मुक्त इलेक्ट्रॉन कहते हैं क्या कहा जाता है…electrostatics of conductors
  • 7:33–10:07Electrostatics of conductors — segment 4मुक्त इलेक्ट्रॉन होते हैं और वह विचित्र की विपरीत दिशा में गति करते हैं केप्लर इजरत में दिया गया अब देखिए दोस्तों अ कि चालक इस त्रिविध की के अंतर्गत यह मैं बहुत सारी महत्वपूर्ण चीजों का हमने अध्ययन करने वाले हैं 5 बिंदु…electrostatics of conductors
  • 10:07–12:39Electrostatics of conductors — segment 5क्षेत्र के की तीव्रता थे सुनने होती है इस स्थिति स्थिति में चालक के अंदर विद्युत क्षेत्र की तीव्रता सुनने भी होती है यह तीन दो एकदम क्लियर हो जाना चाहिए यह अंदर है विचित्र की तीव्रता सुनने होती है अब यहां पर हम है किसी…electrostatics of conductors
  • 12:39–15:10Electrostatics of conductors — segment 6आकर इस पृष्ठ के ऊपर इकट्ठे होते चले जाते हैं और तब तक होते रहेंगे जब तक पर इनके कारण से उत्पन्न विद्युत क्षेत्र की तीव्रता बाहे विद्युत क्षेत्र की तीव्रता के बराबर ना हो जाए देखी कि देबो एक तो यह हमने लगाई है यह चैनल जो…electrostatics of conductors
चालक स्थिरवैद्युतिकी || electrostatics of conductors class 12 physics in hindi 🔉⇢
nawendu classes hindi

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (nawendu classes hindi); found via yt-dlp search 'चालकों का स्थिरवैद्युतिकी electrostatics of conductors Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:32Electrostatics of conductors — segment 1अजय को झाल में आप सभी का स्वागत है हमारे चैनल दोस्तों क्लास हिंदी में आज हम बात करेंगे चालक स्थिर वैद्युत कि मैं आपको बताऊं जब किसी चालक पर हम जब आवेश देते हैं तो यह फेस सबसे पहले ही चालक पूरे सतह पर ब्रिटिश होता है और…electrostatics of conductors
  • 2:32–5:03Electrostatics of conductors — segment 2कराती है स्थिर विद्युत् यह साम्यावस्था इलेक्ट्रोस्टेटिक इक्विलिब्रियम जब भी किसी चालक को वित्तीय क्षेत्र में रख या उस पर आवेश देख लें यदि उस चालक पर आवेश का वितरण होने के बाद जब आवेश की स्थिर अवस्था में आ जाता है उस…electrostatics of conductors
  • 5:03–7:35Electrostatics of conductors — segment 3आप यहां पर भी देख सकते हैं हमने हर जगह चालक पर जो मिलती है सिर की दिशा दिखाने के लिए जो द्वितीय व रखा है हम तो यार विरोध किया है हर जगह विषय पर लिखा हूं कि हमने चालक के तौर पर कैसा रखा है लोगों रखा है तो तीसरा आपने…electrostatics of conductors
  • 7:35–10:11Electrostatics of conductors — segment 4मोदी साहब अलग-अलग यह सभी की और ब्लडी और कोई भी आप गैस पर बॉईल नहीं लगेगा क्योंकि हम लोग इस चीज की जाती है कानून व्यवस्था की स्थिति के लिए बात कर रहे हैं जब ऑफिस इसकी हो आवेश का हिस्सा रहता है सफर चालक के साथ आप कि उस पर…electrostatics of conductors
  • 10:11–12:41Electrostatics of conductors — segment 5हैं अब आप देख सकते हैं अगर चालक अनियमित आकार का हो तो इसके पृष्ठ के हर बिंदु पर प्रीस्ट वास गलत जो है सीखना जिसको लोग बोलते हैं सर पर चांद सिटी मुसलमान नहीं होगा यहां पर जहां भी इस तक कर्वेचर देखिए यहां पर यह जो यह इसका…electrostatics of conductors
  • 12:41–14:19Electrostatics of conductors — segment 6हो विभाग में ज्यादातर आएगा तो आप जानते हैं आवेश का प्रवाह धन आवेश का प्रवाह विद्यालय परिवार कुछ भी बहुत से मिल Vivo हाई वोल्टेज से 2 घंटे तक होता है तो आवेश की स्थिर अवस्था में एक चालक और वैसे प्रॉब्लम विचार कर रहे हैं…electrostatics of conductors
C2L19 संधारित्रों का संयोजन | कक्षा 12 | Combination of Capacitors in Hindi | NCERT Class 12 Physics 🔉⇢
CLS Study

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (CLS Study); found via yt-dlp search 'संधारित्रों का संयोजन ऊर्जा combination of capacitors energy Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:30Combination of capacitors and energy stored — segment 1हेलो हेलो स्टूडेंट्स तो आज के इस वीडियो में हम आपसे बात करेंगे कोंबिनेशन ऑफ के फीचर्स के बारे में तो यहां पर हमको कोंबिनेशन देखना है वह दो तरीके का देखना है पहले हम देखेंगे इस रेड कम मैं फिर देखेंगे हम समांतर क्रम में…combination of capacitors and energy stored
  • 2:30–5:01Combination of capacitors and energy stored — segment 2आपने इसको प्लस टू आदेश दिया तो यहां पर प्लस कोई चार्ज हो गया कोई जितना चाहो यहां पर आता है उतना ही यहां विपरीत ऑफिस उत्पन्न हो तय यह पहले सबको पता है तो यहां भी - को आवेश आएगा अच्छे जितना - को है बनेगा उतने इस पेस्ट को…combination of capacitors and energy stored
  • 5:01–7:31Combination of capacitors and energy stored — segment 3होगा क्यों यह तो अभी और ले सकता है इसके बाद तो और खाली है अभी तो बिलासपुर खा लिए भी यह और पानी ले सकता है लेकिन आपने कम दिया है लेकिन उसके पास तो बहुत ज्यादा है ना भाषण में आ रही है अब देखिए यहां पर कि क्या यह तीनों…combination of capacitors and energy stored
  • 7:31–10:01Combination of capacitors and energy stored — segment 4तब हम फिर लिखेंगे बन्ना पॉर्न सीडी बराबर वन प्वाइंट्स ई वन प्लस वन अपॉन टीटू टीटू ठीक है सीटू प्लस वन अपॉन चीथड़े तो यह हुआ था यह त्यौहार पर ग्रामीण झाला समय यह क्या है उधर यह पॉइंट में नियुक्त करना तब हमें बराबर निकाल…combination of capacitors and energy stored
  • 10:01–12:32Combination of capacitors and energy stored — segment 5संयत्रों दो हो तीन हो 50 हूं जितने भी संधारित्रों उनकी पहली भुजाओं को एक अलग बिंदु पर जोड़ दें बाकी की दूसरी मुद्राओं को एक अलग बिंदु पर जोड़ दें तो जब इस तरह से जोड़ देंगे तो ऐसा संयोजन समांतर क्रम संयोजन कहलाता है ठीक…combination of capacitors and energy stored
  • 12:32–15:10Combination of capacitors and energy stored — segment 6एक बाल्टी पानी भर रख जो बॉडी में आपके पास तीन तरीके से गिला स्माल साइज के एक बहुत बड़ा ग्लास है छोटा हो यह बिल्कुल छोटा हो आप एडिट भरेंगे उसे बाल्टी पानी तो क्या होगा कि जितने साइज के ग्लास है वह उतना पानी इसमें भर…combination of capacitors and energy stored
संधारित्र का संयोजन | श्रेणी क्रम | समांतर क्रम | Sandharitra ka Sanyojan | Class 12 Physics 🔉⇢
MK Sir Inspiration

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (MK Sir Inspiration); found via yt-dlp search 'संधारित्रों का संयोजन ऊर्जा combination of capacitors energy Hindi physics class 12', oEmbed-verified live.

📑 Clips (6)
  • 0:01–2:40Combination of capacitors and energy stored — segment 1हेलो मेरे प्यारे बच्चों क्या हाल चाल बढ़िया सब उम्मीद है आप सभी अच्छे होंगे और तैयारी भी बेहतर चल रही होगी प्यारे बच्चों आज के इस वीडियो में हम स्थिर वैद्युत विभव तथा धारिता चैप्टर के एक और मजेदार टॉपिक का निपटारा करने…combination of capacitors and energy stored
  • 2:40–5:11Combination of capacitors and energy stored — segment 2यह संधारित की प्लेट हम बना रहे हैं ठीक है अब देखिए इसमें होता क्या है ध्यान से समझना बच्चों बहुत ही सिंपल तरीके से हमने क्या बताया कि श्रेणी क्रम में संधारित्र की प्लेटों को इस प्रकार जोड़ते हैं कि प्रत्येक संधारित्र का…combination of capacitors and energy stored
  • 5:11–7:41Combination of capacitors and energy stored — segment 3दूसरी सिरा पर भीतरी सिरा पर विपरीत आवेश उत्पन्न हो जाता है ठीक ये इतना समझ में आ गया और इसका जो आवेश है आवेश है वो क्या हो जाएगा भाई समान इस पर आवेश होगा यानी इसका भी क्यों होगा ठीक है इस पर भी क्यों होगा इस पर भी क्यों…combination of capacitors and energy stored
  • 7:41–10:12Combination of capacitors and energy stored — segment 4लिखेंगे इसको इक्वेशन नंबर वन दे देते हैं बच्चों तो अब इसको हम लिख सकते हैं कि v1 बराबर v1 बराबर q अपन c1 ठीक है इसी तरह से v2 बराबर q अपन सी इसी तरह v3 बराबर q अपन स3 ठीक v3 बराबर क्या हो जाएगा बच्चों q अपन स3 तो अब यही…combination of capacitors and energy stored
  • 10:12–12:44Combination of capacitors and energy stored — segment 5हम कॉमन कर सकते हैं क्य भाई हो क्या आ गया 1 अपन सीव 1 अपन सी2 1 अपन स3 क्लियर भाई क्यू से क्यों कैंसिल हो गया आया क्या 1 अपन सी बराबर 1 अपन सीव प्सव अप c2 प्व अपन स3 मतलब ये क्या आ गया बच्चों तो ये आ गया श्रेणी क्रम में…combination of capacitors and energy stored
  • 12:44–15:15Combination of capacitors and energy stored — segment 6गया चलिए अब देखते हैं समांतर क्रम ठीक अब हम क्या लिखेंगे समांतर क्रम ठीक सीरीज कॉमिनेशन तो हो गया भाई अब क्या है भ अब हम क्या करेंगे इसको समांतर क्रम में जोड़ते हैं श्रेणी क्रम में तो हो गया अब देखिए समांतर क्रम में…combination of capacitors and energy stored
Electric potential || potential difference || electrostatic potential || class 12 electrostatics 🔉⇢
Physics with Muhammad Arafat Khan

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Physics with Muhammad Arafat Khan); found via yt-dlp search 'electrostatic potential and potential difference class 12 physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Electric potential and potential difference — segment 1Jhal 's welcome student and all the few lines on my YouTube channel Physics Mohammad Ashraf Khan and Sector-12's electrostatics today's topic which we will study and learn and the name of that topic is electric…electric potential and potential difference
  • 2:33–5:05Electric potential and potential difference — segment 2understand the registers against button for quote 10 GK science so gravity download and your order against the graphics then you allocate this energy in this body and it will be stored away and that is called potential…electric potential and potential difference
  • 5:05–7:35Electric potential and potential difference — segment 3potential energy and here also the potential energy is potential energy and it is converted into a school of 1.2 I will come to this topic when class 11th was introduced so for now x square of 1.2 now see what else is…electric potential and potential difference
  • 7:35–10:06Electric potential and potential difference — segment 4your warden did the body, now the work of electric field, there should not be any ordinary body in the electric field, we need charge here, we are doing the warden, but we have to do it on some chart and you know like…electric potential and potential difference
  • 10:06–12:36Electric potential and potential difference — segment 5finally, due to my warding, due to against, such potential energy or you can say that which unit positive charge is against the field. If it is released then because of that warden energy gets stored in that charge and…electric potential and potential difference
  • 12:36–15:07Electric potential and potential difference — segment 6field so the warden is gone means you have to work against the border so this has become a simple definition of whose electric potential energy done simple you UNESCO electric potential now these two decoration…electric potential and potential difference
Potential due to dipole (logical derivation) | Electric potential & cap. | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'potential due to an electric dipole derivation physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Potential due to a dipole — segment 1imagine we have an electric dipole basically a negative charge and a positive charge of same value separated by some distance let's say we call it 2a traditionally we call that 2a our goal is to figure out what the…potential due to a dipole
  • 2:31–5:01Potential due to a dipole — segment 2sign comes here itself when i substitute all right so if i do that if i substitute due to plus q it's going to be k q divided by r2 so r2 and you will have due to the negative q you'll have minus q the minus sign comes…potential due to a dipole
  • 5:01–7:31Potential due to a dipole — segment 3also same when i subtract i will get 0 so let me just write that down if i were to directly substitute so direct substitution what does that give me we get potential at point p to be zero and that is wrong why is that…potential due to a dipole
  • 7:31–10:01Potential due to a dipole — segment 4imagine r1 was say 11 and let's say r2 was a little smaller so let's call it nine and let's say r somewhere in between the two numbers is 10. okay let's write that down over here as to um what happens when you…potential due to a dipole
  • 10:01–12:32Potential due to a dipole — segment 5parallel to each other and that's the secret we're going to use look at them they look so parallel to each other so i'm going to go back i'm going to redraw this and this time i'm going to draw r1 and r2 parallel to…potential due to a dipole
  • 12:32–15:02Potential due to a dipole — segment 6hopefully you can see that r2 minus r1 minus r2 delta r automatically becomes 2a maximum all right notice what happens as this now comes off this particular axis and comes back what happens this will become smaller and…potential due to a dipole
Equipotential surfaces (& why they are perpendicular to field) | Electric potential | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'equipotential surfaces class 12 physics explained', oEmbed-verified live.

📑 Clips (5)
  • 0:00–2:30Equipotential surfaces and field-potential relation — segment 1we've learned how to visualize electric field by drawing field lines in this video let's explore how to visualize electric potentials and the way to do that or at least one way of doing that is by drawing something…equipotential surfaces and field-potential relation
  • 2:30–5:00Equipotential surfaces and field-potential relation — segment 2and farther away why is that well it's got something to do with the strength of the electric field close to the charge the field is very strong and that's where the potentials are equipotential surfaces will be closer…equipotential surfaces and field-potential relation
  • 5:00–7:31Equipotential surfaces and field-potential relation — segment 3big sheet of charge which has let's say negative charge then we know we've seen before it produces a uniform electric field can you think of what the equipotential surfaces here would look like can you draw try drawing…equipotential surfaces and field-potential relation
  • 7:31–10:01Equipotential surfaces and field-potential relation — segment 4so here we are seeing that the two are perpendicular to each other hmm let's look it over here hey here also we are seeing that the field lines are perpendicular to the equipotential surfaces interesting so can we say…equipotential surfaces and field-potential relation
  • 10:01–12:08Equipotential surfaces and field-potential relation — segment 5when you drop a ball gravitational field does positive work what happens to the potential energy it loses it what happens when you throw a ball up gravity does negative work what happens to the potential energy it gains…equipotential surfaces and field-potential relation
Potential energy of a system of 3 charges | Electrostatic potential & capacitance | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'electrostatic potential energy of a system of charges class 12 physics', oEmbed-verified live.

📑 Clips (5)
  • 0:00–2:30Electrostatic potential energy — segment 1suppose we have three charges kept like this our goal in this video is to figure out what the potential energy of this system is going to be so what do we even mean by potential energy over here well imagine at the…electrostatic potential energy
  • 2:30–5:00Electrostatic potential energy — segment 2and so that this total work done would now represent the total potential energy of this system so now we have to figure out what is the total work done so let's do that let's focus on the first one so let me dim the…electrostatic potential energy
  • 5:00–7:30Electrostatic potential energy — segment 3that we have a faster way of doing this because we've already done all the hard work in in the previous videos so if you remember we can bring back the concept of potential we know how to calculate potential at any…electrostatic potential energy
  • 7:30–10:00Electrostatic potential energy — segment 4two charges at this point and we can that's that'll be the potential due to this charge at this point plus the potential due to this charge at this point so the potential due to this charge at this point is going to be…electrostatic potential energy
  • 10:01–10:51Electrostatic potential energy — segment 5and if you look at this one q q2 q3 that is the potential energy that this expression is the potential energy of the system of these two charges alone potential energy of system of these two so what's interesting is…electrostatic potential energy
Electrostatic shielding & Faraday cage | Electrostatic potential & capacitance | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'electrostatic shielding Faraday cage physics explained', oEmbed-verified live.

📑 Clips (4)
  • 0:00–2:30Electrostatics of conductors — segment 1[Music] this picture shows a girl a couple of girls standing inside a metallic cage and somebody is trying to electrocute them using lightning and nothing seems to be happening to them she seems happy why now if you're…electrostatics of conductors
  • 2:30–5:01Electrostatics of conductors — segment 2nothing special about a spherical conductor this has to be true for any conductor which means we can now go ahead and write a general conclusion in electrostatic conditions electric field inside any conductor must be…electrostatics of conductors
  • 5:01–7:33Electrostatics of conductors — segment 3field is zero it should be an equipotential surface can you make try and make that link yourself all right here's how i like to think about it consider any two points on this conductor maybe say one point inside over…electrostatics of conductors
  • 7:33–9:55Electrostatics of conductors — segment 4at this this means that this conductor is sort of like sucking the electric field and that's why even these field lines are going to sort of sort of like bend towards their conductor because they can sort of like…electrostatics of conductors
Energy stored in capacitor derivation (why it's not QV) | Electrostatic potential | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'energy stored in a capacitor derivation physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:32Combination of capacitors and energy stored — segment 1suppose a capacitor is charged to a potential difference we having a total charge of q on each plates the goal of this video is to figure out what is the total potential energy stored in this capacitor so let's start by…combination of capacitors and energy stored
  • 2:32–5:03Combination of capacitors and energy stored — segment 2charge transferred from here to here is q the question now is what is the total work i did in transferring that charge can you pause the video and think a little bit about it all right so what i would do what my common…combination of capacitors and energy stored
  • 5:03–7:33Combination of capacitors and energy stored — segment 3going to call almost 0. so we're going to do it step by step okay now let's move the second charge now when i move the second charge do i have to do some work yes i do because now i feel repulsion from this charge in…combination of capacitors and energy stored
  • 7:33–10:06Combination of capacitors and energy stored — segment 4and come up with an explanation by looking at this summation it's it's mathematical now why is it not q times v all right hopefully they tried the main reason is because not all the charges went through the potential…combination of capacitors and energy stored
  • 10:06–12:38Combination of capacitors and energy stored — segment 5me write that down capacitance is charged by voltage so at any point voltage should equals charge divided by capacitance so i know that my voltage at this point should be the charge right now divided by capacitance so…combination of capacitors and energy stored
  • 12:38–13:23Combination of capacitors and energy stored — segment 6and if i substitute for c over here c is equal to q over v then i'll get and you can do that yourself you'll get it as half q v so half times v and you can check that all i'm doing is substituting these values over here…combination of capacitors and energy stored

💬 Doubt Solving Ask-any-doubt RAG + FAQ

🤖 Ask any doubt RAG · NCERT-grounded, cited

Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.

Frequently-asked doubts

What is the difference between electric potential and electric potential energy?
They are related but distinct. Electric potential $V$ is a property of a point in space set up by some source charges — it is the work done per unit charge in bringing a small positive test charge from infinity to that point, measured in volts ($\mathrm{J\,C^{-1}}$). Potential energy $U$ is a property of a specific charge $q$ placed at that point, namely the work done to bring that particular charge there, and it equals $U=qV$, measured in joules. So potential exists whether or not any charge is sitting there, while potential energy needs an actual charge to 'own' it. A useful analogy: potential is like the height of a hill (a property of the location), while potential energy is like $mgh$ for a specific mass placed on the hill. Doubling the test charge doubles its potential energy but leaves the potential unchanged.
How can the potential be zero at a point where the field is not, and vice versa?
Because $V$ is a scalar sum and $\vec{E}$ is a vector sum, they carry different information. At the midpoint between two equal and opposite charges (a dipole's centre), the two potentials $+\dfrac{kq}{r}$ and $-\dfrac{kq}{r}$ cancel to give $V=0$, yet both field vectors point the same way and add, so $E\neq 0$. Conversely, at the midpoint between two equal positive charges the two field vectors are equal and opposite and cancel to give $E=0$, but the two potentials are both positive and add to give $V\neq 0$. The clean way to remember it: $V=0$ does not force $E=0$, and $E=0$ does not force $V=0$. What is always true is the local relation $E=-\dfrac{dV}{dr}$ — the field measures how fast the potential changes with position, not its value. A flat but non-zero potential region has zero field; a zero-potential point on a steep slope has a large field.
Does inserting a dielectric increase or decrease the capacitance?
It always increases it, by the factor $K\gt 1$ (the dielectric constant): $C=KC_0$. The reason is that the dielectric polarises in the field, and its bound surface charges partly cancel the field of the free charge on the plates. For a fixed charge this means a smaller net field and hence a smaller voltage $V=Ed$, and since $C=Q/V$, a smaller $V$ for the same $Q$ means a larger $C$. Equivalently, the dielectric lets the capacitor hold more charge at the same voltage. What happens to the other quantities depends on what is held fixed — but the capacitance itself unambiguously goes up. This is exactly why real capacitors are filled with a dielectric rather than left with an air gap: you get far more capacitance in the same volume, and the dielectric also raises the breakdown voltage.
When a dielectric is inserted, what stays constant — the charge or the voltage?
It depends entirely on whether the battery is still connected. If the battery remains connected, the voltage $V$ is clamped at the battery's value; the capacitance rises to $KC_0$, so the charge $Q=CV$ increases $K$-fold (the extra charge flows in from the battery) and the field between the plates is unchanged. If the battery is first disconnected, the charge $Q$ is trapped and cannot change; then inserting the dielectric raises $C$, so the voltage $V=Q/C$ falls to $V_0/K$ and the field falls to $E_0/K$. The single most important habit for these problems is to ask first: is the source still connected (then $V$ is fixed) or disconnected (then $Q$ is fixed)? Every other quantity — field, energy, force — follows once you have pinned down which one is held constant. Mixing up the two cases is the most common source of wrong answers in capacitor questions.
Why does the stored energy go up when the battery is connected but down when it is disconnected?
With the battery connected, $V$ is fixed and $U=\tfrac12 CV^{2}$, so raising $C$ by inserting the dielectric raises $U$ by the factor $K$. The extra energy — and more — is supplied by the battery, which pushes additional charge onto the plates; in fact the battery does work $QV$ while the stored energy rises by only $\tfrac12(\Delta Q)V$, the rest being accounted for by the mechanical work of pulling the slab in. With the battery disconnected, $Q$ is fixed and $U=\dfrac{Q^{2}}{2C}$, so raising $C$ lowers $U$ by the factor $K$. The lost stored energy is converted into mechanical work: the capacitor actually pulls the dielectric into the gap, and that inward force does positive work on the slab. So both results are consistent with energy conservation — you just have to track who is doing work on whom.
What decides the sign of the work done in moving a charge in an electric field?
Compare the direction of motion with the electric force on the charge. The work done BY the electric field is positive when the charge moves in the same general direction as the force on it (a positive charge moving from high to low potential, 'downhill'), and negative when it moves against the force (a positive charge pushed from low to high potential, 'uphill'). The work done by the EXTERNAL agent is the opposite sign. In terms of potential, the work done by the field on a charge $q$ moving from A to B is $W_{field}=q(V_A-V_B)$; the work done by an external force moving it without acceleration is $W_{ext}=q(V_B-V_A)=\Delta U$. A positive charge naturally 'falls' toward lower potential, gaining kinetic energy; a negative charge does the reverse, moving spontaneously toward higher potential. Always fix the sign by asking whether the field is helping or opposing the motion of that particular sign of charge.
If the field inside a conductor is zero, is the potential inside also zero?
No — zero field means constant potential, not zero potential. Inside a conductor in electrostatic equilibrium, $\vec{E}=0$ everywhere, and since $E=-\dfrac{dV}{dr}$, a zero field means $V$ does not change: the entire conductor, surface and interior, is one single equipotential at some constant value. That constant is generally not zero; a charged isolated sphere, for instance, sits at potential $V=\dfrac{Q}{4\pi\varepsilon_0 R}$ throughout. This is why the whole body of a car struck by lightning can be at a huge potential while there is no field, and hence no force, inside it. The takeaway: 'field zero' tells you the potential is uniform, and you need a separate boundary condition (like grounding, which sets $V=0$, or the charge and geometry) to find the actual constant value.
Why is the electric field always perpendicular to an equipotential surface?
Because if it were not, there would be a component of $\vec{E}$ lying along the surface, and moving a test charge along that direction would require work $W=q\vec{E}\cdot d\vec{l}\neq 0$, changing its potential energy and hence the potential — but that contradicts the surface being 'equipotential' (constant $V$). So the tangential component must vanish, leaving only a component perpendicular to the surface. Equivalently, no work is done moving a charge along an equipotential (since $\Delta V=0$), which is exactly the condition that the force, and therefore the field, is at right angles to the path. This is why field lines and equipotential surfaces always form an orthogonal grid, just like the streamlines and contour lines on a topographic map, and why the surface of any conductor (an equipotential) has its field pointing straight out of it.
How can two capacitors 'lose' energy when they share charge if charge is conserved?
Charge conservation and energy conservation are two separate accounting rules, and both hold — but the stored electrostatic energy is not the whole energy budget. When a charged capacitor is connected to an uncharged one, charge flows through the connecting wires until the two reach a common potential; during that transient flow, a current runs through the small but non-zero resistance of the wires and dissipates energy as heat ($I^{2}R$ losses), and a little may be radiated as electromagnetic waves. Add up the final electrostatic energies and you find they total less than the initial stored energy — the difference is exactly the heat produced. The charge is fully accounted for (none is lost), but energy has simply changed form from electrostatic to thermal. Remarkably, in the idealised limit of very small resistance the fraction lost is still one half for two equal capacitors, because a smaller resistance carries a proportionally larger current.
Why does a capacitor have a maximum voltage rating?
Because the insulating material between the plates can only withstand a limited electric field before it breaks down. As you raise the voltage, the field $E=V/d$ in the gap grows; once it exceeds the dielectric strength of the insulator (about $3\times10^{6}\ \mathrm{V\,m^{-1}}$ for air, higher for good solid dielectrics), the medium ionises, becomes conducting, and the charge suddenly arcs across, usually destroying the capacitor. The rated voltage is set safely below this breakdown value. It is a geometry-and-material limit, not a charge limit as such: a capacitor with a thicker or tougher dielectric can be rated for a higher voltage. This is also why you should never connect a capacitor rated for, say, $16\ \mathrm{V}$ across a $50\ \mathrm{V}$ supply — the dielectric will puncture. Designers trade off a high dielectric constant (for large capacitance) against a high dielectric strength (for a high voltage rating).
In a series or parallel combination, which capacitors carry the same charge and which the same voltage?
In a series combination the capacitors carry the SAME charge, because the charge on the connected inner plates must be equal and opposite (they were neutral and only redistribute), and the voltages add: $\dfrac{1}{C_{series}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dots$, giving an equivalent capacitance smaller than the smallest member. In a parallel combination the capacitors share the SAME voltage (they are connected across the same two nodes), and their charges add: $C_{parallel}=C_1+C_2+\dots$, giving an equivalent capacitance larger than the largest member. A handy consequence for series: the smaller capacitor takes the larger share of the voltage (since $V=Q/C$ with $Q$ common). Getting this backwards — assuming equal voltage in series or equal charge in parallel — is a frequent mistake; anchor yourself on 'series shares charge, parallel shares voltage'.
Is the potential due to a dipole the same as that due to a single point charge?
No, they fall off differently and the dipole potential also depends on direction. A point charge gives $V=\dfrac{kq}{r}$, which falls as $1/r$ and is the same in all directions (spherical symmetry). A dipole gives $V=\dfrac{k\,p\cos\theta}{r^{2}}$, which falls off faster, as $1/r^{2}$, and depends on the angle $\theta$ between the position vector and the dipole moment $\vec{p}$. Along the axis ($\theta=0$) the potential is maximal and positive on the positive-charge side; on the perpendicular bisector ($\theta=90^{\circ}$) it is zero everywhere because the two equal-and-opposite contributions cancel. The faster $1/r^{2}$ decay reflects the near-cancellation of the two opposite charges as seen from far away, and the angular dependence is why a dipole's equipotential surfaces are not spheres but the characteristic figure-of-eight-derived shapes.
Where exactly is the energy of a charged capacitor stored?
It is stored in the electric field in the region between the plates, not 'on' the plates themselves. Although you can compute the energy from the charge and voltage as $U=\tfrac12 QV=\tfrac12 CV^{2}=\dfrac{Q^{2}}{2C}$, the physically illuminating form is the field picture: the energy density (energy per unit volume) at any point is $u=\tfrac12\varepsilon_0 E^{2}$, and integrating this over the volume of field gives the total stored energy. For a parallel-plate capacitor the field is confined to the gap of volume $Ad$, so $U=\tfrac12\varepsilon_0 E^{2}\cdot Ad$, which is exactly $\tfrac12 CV^{2}$. This field view is more than bookkeeping: the same $u=\tfrac12\varepsilon_0E^{2}$ holds for the field of any charge configuration, and it is the electrostatic ancestor of the idea that electromagnetic waves — including light — carry energy in their fields through empty space.

Trap-answer taxonomy

Trap: Confusing potential with potential energy

Students use $V$ and $U$ interchangeably, quoting a 'potential' in joules or a 'potential energy' in volts, and forget that $U=qV$.

Fix: Keep them distinct: potential $V$ (volts, $\mathrm{J\,C^{-1}}$) is a property of a location set by the source charges and exists even with no charge present; potential energy $U=qV$ (joules) belongs to a specific charge placed there. Height versus $mgh$ is the analogy.

Trap: Assuming $V=0$ implies $E=0$ (or the reverse)

Students conclude the field must vanish wherever the potential is zero, or that a point of zero field must be at zero potential.

Fix: They are independent because $V$ is a scalar sum and $\vec{E}$ a vector sum. At a dipole's centre $V=0$ but $E\neq0$; midway between two equal positive charges $E=0$ but $V\neq0$. Only $E=-dV/dr$ always holds — the field is the slope of $V$, not its value.

Trap: Forgetting to decide what is held constant during dielectric insertion

Students apply $U=\tfrac12CV^2$ blindly when the charge is actually fixed, or hold $V$ constant when the battery has been disconnected.

Fix: Ask first: is the battery connected ($V$ fixed) or disconnected ($Q$ fixed)? With $V$ fixed, $Q$ and $U$ rise by $K$; with $Q$ fixed, $V$, $E$ and $U$ fall by $K$. Choose the energy form ($\tfrac12CV^2$ vs $Q^2/2C$) that keeps the constant quantity explicit.

Trap: Using the full gap field for the force on a capacitor plate

Students compute the force on a plate as $F=QE$ with $E=\sigma/\varepsilon_0$, the full field between the plates, and get twice the right answer.

Fix: A plate feels only the field of the OTHER plate, $E_{other}=\sigma/2\varepsilon_0$, since a charge sheet exerts no net force on itself. The correct force per unit area is the electrostatic pressure $P=\tfrac12\varepsilon_0E^2$, giving $F=\dfrac{Q^2}{2\varepsilon_0 A}$ — note the factor of one half.

Trap: Swapping the series and parallel rules for capacitors

Students add capacitances in series and take reciprocals in parallel, copying the resistor rules with the roles reversed.

Fix: For capacitors it is the mirror image of resistors: parallel ADD directly, $C_{par}=C_1+C_2+\dots$ (same voltage, charges add); series add as RECIPROCALS, $1/C_{ser}=1/C_1+1/C_2+\dots$ (same charge, voltages add). Series gives less than the smallest; parallel gives more than the largest.

Trap: Believing charge is lost when capacitors share charge and 'lose' energy

Seeing the final stored energy come out less than the initial, students conclude charge has disappeared or the theory is wrong.

Fix: Charge is exactly conserved; it is stored ENERGY that is not. The deficit is dissipated as heat in the connecting-wire resistance (and slight radiation) during the transient current. Track energy separately from charge — the two obey different conservation statements here.

Trap: Thinking the field inside a conductor being zero means the potential is zero

Students set $V=0$ inside a conductor because $E=0$ there, and then get wrong surface-potential or capacitance answers.

Fix: Zero field means CONSTANT potential, not zero. A whole conductor is one equipotential at some value fixed by its charge and geometry (or by grounding, which does set $V=0$). An isolated charged sphere sits at $V=Q/4\pi\varepsilon_0R$ throughout, field-free but far from zero potential.

Trap: Getting the sign of work in moving a charge backwards

Students assume any charge 'falls' to lower potential and assign the sign of work without checking the charge's sign.

Fix: Only positive charges move spontaneously from high to low potential; negative charges do the opposite. The work done by the field is $W_{field}=q(V_A-V_B)$ and by an external agent $W_{ext}=q(V_B-V_A)=\Delta U$. Always insert the charge WITH its sign and check whether the field helps or opposes the motion.

🚪 Dive Deeper Mystery room · 40 discoveries

🎯 Put a charge outside a hollow conductor and the inside stays at exactly zero field — not nearly zero. Watch the shell's own charges rearrange until they have cancelled the intruder.
🔉⇢
field inside the cavity = 0.00 — for every Q, every distance, always
induced charge on the shell: near face —, far face — · net 0
What you are looking at
  • The red charge outside and the brown arrows — its field, arriving at the shell.
  • The dots on the shell — its own free charges, which slide round until their field cancels the intruder’s everywhere inside the metal.
  • The faint purple specks in the cavity — test charges. They never move, because there is nothing there to move them.
What to do
  1. Bring the outside charge closer. The induced pattern intensifies; the interior field stays at exactly 0.00.
  2. Reverse Q. The induced charges swap sides; the interior is still zero.
  3. Earth the shell and watch the far-face charge drain away — now the shielding works in BOTH directions.
What it means — this is a Faraday cage, and the zero is exact rather than approximate. It has to be: if any field remained inside the metal it would push the free charges, and they would keep moving until it did not. That same exactness is what lets physicists test the inverse-square exponent to one part in 10¹⁶ — looking for a residual interior field that should not be there.

🗝️ Mystery room · the Faraday cage — why the inside of a conductor is shielded, exactly and not approximately

Discovered 0 / 40

JEE Advanced archive — DISCOVER → ATTEMPT → REVEAL

A parallel-plate capacitor of plate area $A$ and separation $d$ has a dielectric slab of constant $K$ partially inserted a distance $x$ from one edge (the plates have width $L$, so total area $A=L\cdot\ell$ where $\ell$ is the plate length). With the battery of emf $V$ still connected, find the capacitance as a function of $x$ and the force pulling the slab further in.

Attempt, then reveal full solution
The filled and empty portions act as two capacitors in parallel. If the plates have width $b$ (into the page) so that inserted area is $bx$ and empty area is $b(\ell-x)$, then $C(x)=\dfrac{\varepsilon_0 b}{d}\big[Kx+(\ell-x)\big]=\dfrac{\varepsilon_0 b}{d}\big[\ell+(K-1)x\big]$. With the battery connected $V$ is fixed, so the energy expression to use for the force is the co-energy: $F=+\dfrac{1}{2}V^{2}\dfrac{dC}{dx}$. Here $\dfrac{dC}{dx}=\dfrac{\varepsilon_0 b (K-1)}{d}$, so $F=\dfrac{\varepsilon_0 b (K-1)V^{2}}{2d}$, a constant, directed so as to pull the slab further into the gap (since $K\gt 1$ makes $F\gt 0$). Physically the fringing field at the slab's edge has a component that drags the polarised dielectric inward — the system lowers its energy at fixed $V$ by increasing $C$. Note the sign: at fixed voltage the plate energy $\tfrac12CV^2$ increases as the slab enters, but the battery supplies twice that increase, half going to the field and half returned as mechanical work.

JEE-Advanced style (authored); grounded in NCERT XII §2.13 (effect of dielectric) and §2.15 (energy $U=\tfrac12 CV^2$)

A parallel-plate capacitor of capacitance $C_0$ (vacuum) is charged to voltage $V_0$ by a battery. Compare the charge $Q$, voltage $V$, field $E$, capacitance $C$ and stored energy $U$ after a dielectric slab of constant $K$ fully fills the gap, in two cases: (a) the battery stays connected, (b) the battery is disconnected before insertion.

Attempt, then reveal full solution
In both cases the capacitance rises to $C=KC_0$. Case (a), battery connected — $V$ is held at $V_0$. Then $Q=CV=KC_0V_0=KQ_0$ (charge increases $K$-fold, drawn from the battery); the field $E=V_0/d$ is unchanged; energy $U=\tfrac12 CV^{2}=\tfrac12 KC_0V_0^{2}=KU_0$ increases $K$-fold. Case (b), battery disconnected — $Q=Q_0$ is now fixed. Then $V=\dfrac{Q_0}{C}=\dfrac{V_0}{K}$ (voltage drops $K$-fold); the field $E=\dfrac{V}{d}=\dfrac{E_0}{K}$ drops $K$-fold (the bound polarisation charge partly cancels the free charge); energy $U=\dfrac{Q_0^{2}}{2C}=\dfrac{U_0}{K}$ decreases $K$-fold. The universal exam trap: decide first what is held constant — $V$ if the battery is connected, $Q$ if it is disconnected — and everything else follows. Energy rises with the dielectric when the battery pays for it, and falls when the isolated capacitor pulls the slab in and does mechanical work.

JEE-Advanced style (authored); NCERT XII §2.13 ($C=KC_0$, $E=E_0/K$) and §2.15 (energy relations)

Show that the outward electrostatic force per unit area (the electrostatic pressure) on the plate of a parallel-plate capacitor is $P=\dfrac{1}{2}\varepsilon_0 E^{2}$, and hence find the attractive force between plates of area $A$ carrying charge $Q$.

Attempt, then reveal full solution
A plate carrying surface charge density $\sigma$ sits in the field produced by the OTHER plate only, which is $E_{other}=\dfrac{\sigma}{2\varepsilon_0}$ — not the full gap field $E=\dfrac{\sigma}{\varepsilon_0}$, because a sheet cannot exert a force on itself. The force per unit area is therefore $P=\sigma E_{other}=\sigma\cdot\dfrac{\sigma}{2\varepsilon_0}=\dfrac{\sigma^{2}}{2\varepsilon_0}=\dfrac{1}{2}\varepsilon_0 E^{2}$, using $\sigma=\varepsilon_0 E$. This equals the energy density $u=\tfrac12\varepsilon_0E^{2}$ of the field — no coincidence, since pulling a plate out by $dx$ opens up a slab of new field-filled volume $A\,dx$, and the mechanical work $P\,A\,dx$ must equal the field energy $u\,A\,dx$ created. The total attractive force is $F=PA=\dfrac{1}{2}\varepsilon_0 E^{2}A=\dfrac{\sigma^{2}A}{2\varepsilon_0}=\dfrac{Q^{2}}{2\varepsilon_0 A}$. The factor of one half — the field acting on a plate is half the gap field — is the single most common error in this classic problem.

JEE-Advanced style (authored); energy density $u=\tfrac12\varepsilon_0E^2$ from NCERT XII §2.15 (Eq. 2.73)

Find the electrostatic potential energy (self-energy) of a uniformly charged conducting sphere of radius $R$ carrying total charge $Q$, by two methods, and comment.

Attempt, then reveal full solution
Method 1 — assembling the charge. The sphere's potential when it holds charge $q$ is $V=\dfrac{q}{4\pi\varepsilon_0 R}$. Bringing up a further $dq$ from infinity costs $dW=V\,dq=\dfrac{q\,dq}{4\pi\varepsilon_0 R}$. Integrating from $0$ to $Q$: $W=\dfrac{1}{4\pi\varepsilon_0 R}\cdot\dfrac{Q^{2}}{2}=\dfrac{Q^{2}}{8\pi\varepsilon_0 R}$. Method 2 — integrating the field energy. All charge on a conductor is on the surface, so the field is zero inside and $E=\dfrac{Q}{4\pi\varepsilon_0 r^{2}}$ outside. The energy is $U=\displaystyle\int_R^{\infty}\tfrac12\varepsilon_0E^{2}\,4\pi r^{2}\,dr=\dfrac{Q^{2}}{8\pi\varepsilon_0}\int_R^{\infty}\dfrac{dr}{r^{2}}=\dfrac{Q^{2}}{8\pi\varepsilon_0 R}$. The two agree exactly, confirming that the energy $\tfrac12\sum q_iV_i$ stored in assembling the charges is physically the same as the energy $\int u\,dV$ residing in the field. This self-energy can also be written $U=\dfrac{Q^{2}}{2C}$ with $C=4\pi\varepsilon_0 R$ the capacitance of an isolated sphere.

JEE-Advanced style (authored); energy density $u=\tfrac12\varepsilon_0E^2$ and $U=Q^2/2C$ from NCERT XII §2.15

A dielectric slab of constant $K$ and thickness $t$ ($t\lt d$) is inserted parallel to and between the plates of a vacuum parallel-plate capacitor of plate area $A$ and separation $d$. Find the new capacitance.

Attempt, then reveal full solution
The gap is now two regions in series: a dielectric layer of thickness $t$ and a vacuum layer of thickness $(d-t)$, since the same field lines pass through both. Treat them as two capacitors in series: $C_1=\dfrac{\varepsilon_0 K A}{t}$ (dielectric) and $C_2=\dfrac{\varepsilon_0 A}{d-t}$ (vacuum). For series, $\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}=\dfrac{t}{\varepsilon_0 KA}+\dfrac{d-t}{\varepsilon_0 A}=\dfrac{1}{\varepsilon_0 A}\left[(d-t)+\dfrac{t}{K}\right]$. Hence $C=\dfrac{\varepsilon_0 A}{(d-t)+\dfrac{t}{K}}=\dfrac{\varepsilon_0 A}{d-t\left(1-\dfrac{1}{K}\right)}$. Two useful checks: if $t=0$ (no slab) this gives $C=\varepsilon_0 A/d=C_0$; if $t=d$ (slab fills the gap) it gives $C=\varepsilon_0 KA/d=KC_0$, matching the full-dielectric result. A conductor slab is the limit $K\to\infty$, giving $C=\dfrac{\varepsilon_0 A}{d-t}$ — inserting a metal sheet is equivalent to reducing the effective gap by its thickness, regardless of where it sits.

JEE-Advanced style (authored); NCERT XII §2.13 (Example 2.8, partial dielectric) and §2.14 (series combination)

Two capacitors $C_1=2\ \mu\mathrm{F}$ and $C_2=3\ \mu\mathrm{F}$ are separately charged to $V_1=200\ \mathrm{V}$ and $V_2=100\ \mathrm{V}$. They are disconnected from their batteries and connected in parallel, positive plate to positive plate. Find the common voltage, the final energy, and the energy lost.

Attempt, then reveal full solution
Total charge is conserved: $Q=C_1V_1+C_2V_2=(2)(200)+(3)(100)=400+300=700\ \mu\mathrm{C}$. In parallel the equivalent capacitance is $C=C_1+C_2=5\ \mu\mathrm{F}$, so the common voltage is $V=\dfrac{Q}{C}=\dfrac{700}{5}=140\ \mathrm{V}$. Initial energy $U_i=\tfrac12 C_1V_1^{2}+\tfrac12 C_2V_2^{2}=\tfrac12(2)(200)^{2}+\tfrac12(3)(100)^{2}=40000+15000=55000\ \mu\mathrm{J}=55\ \mathrm{mJ}$ (working in $\mu\mathrm{F}$ and volts gives $\mu\mathrm{J}$). Final energy $U_f=\tfrac12 CV^{2}=\tfrac12(5)(140)^{2}=49000\ \mu\mathrm{J}=49\ \mathrm{mJ}$. Energy lost $=U_i-U_f=6\ \mathrm{mJ}$, dissipated as heat in the connecting wires. This generalises the equal-capacitor paradox: whenever charged capacitors at different potentials are joined, charge redistributes until the potentials equalise and some electrostatic energy is always converted to heat — connecting positive-to-negative instead would give an even larger loss.

JEE-Advanced style (authored); NCERT XII §2.14 (parallel combination) and §2.15 (energy, Example 2.10)

In the network, three capacitors $C_1=10\ \mu\mathrm{F}$, $C_2=5\ \mu\mathrm{F}$ and $C_3=4\ \mu\mathrm{F}$ are arranged so that $C_1$ and $C_2$ are in series, and that series combination is in parallel with $C_3$. A battery of $100\ \mathrm{V}$ is applied across the parallel network. Find the equivalent capacitance, the charge on each capacitor, and the voltage across $C_1$.

Attempt, then reveal full solution
Series of $C_1,C_2$: $\dfrac{1}{C_{12}}=\dfrac{1}{10}+\dfrac{1}{5}=\dfrac{1+2}{10}=\dfrac{3}{10}$, so $C_{12}=\dfrac{10}{3}=3.33\ \mu\mathrm{F}$. This is in parallel with $C_3$: $C_{eq}=C_{12}+C_3=3.33+4=7.33\ \mu\mathrm{F}$. Across the parallel network sits the full $100\ \mathrm{V}$, so $C_3$ carries $Q_3=C_3V=4\times100=400\ \mu\mathrm{C}$. The series branch also has $100\ \mathrm{V}$ across it, and series capacitors carry equal charge: $Q_1=Q_2=C_{12}\times100=\dfrac{10}{3}\times100=333\ \mu\mathrm{C}$. The voltage across $C_1$ is $V_1=\dfrac{Q_1}{C_1}=\dfrac{333}{10}=33.3\ \mathrm{V}$, and across $C_2$ it is $V_2=\dfrac{333}{5}=66.7\ \mathrm{V}$ (note $V_1+V_2=100\ \mathrm{V}$, as it must, and the smaller capacitor takes the larger share of voltage in series).

JEE-Advanced style (authored); NCERT XII §2.14 (series and parallel combination of capacitors)

An isolated (battery disconnected) charged parallel-plate capacitor holds charge $Q$ with a dielectric slab of constant $K$ filling the gap. Find the work an external agent must do to pull the slab completely out, and state where that energy comes from and goes.

Attempt, then reveal full solution
With the battery disconnected the charge $Q$ is fixed. The capacitance with the slab in is $C=KC_0$ and the stored energy is $U_{in}=\dfrac{Q^{2}}{2C}=\dfrac{Q^{2}}{2KC_0}$. With the slab fully removed the capacitance falls to $C_0$ and the energy rises to $U_{out}=\dfrac{Q^{2}}{2C_0}$. The change in stored energy is $\Delta U=U_{out}-U_{in}=\dfrac{Q^{2}}{2C_0}\left(1-\dfrac{1}{K}\right)\gt 0$. Since no battery is connected, no external electrical source supplies this; it must come entirely from the mechanical work done by the agent. The capacitor attracts the dielectric inward (the polarised slab is pulled toward the region of stronger fringing field), so pulling it out is working against that attraction: $W_{ext}=\Delta U=\dfrac{Q^{2}}{2C_0}\left(1-\dfrac{1}{K}\right)$, all of which is stored as increased field energy. The sign is the crucial check: because $K\gt 1$ the energy of an isolated capacitor is lower with the slab in, so the slab is sucked in and must be dragged out — the exact opposite intuition to the battery-connected case, where the battery pays.

JEE-Advanced style (authored); NCERT XII §2.13 (dielectric, $C=KC_0$) and §2.15 (energy $U=Q^2/2C$)

Three point charges $+q$, $+q$ and $+q$ are brought from infinity and fixed at the three corners of an equilateral triangle of side $a$. Find the total electrostatic potential energy of the configuration, and the work needed to assemble it.

Attempt, then reveal full solution
Potential energy of a system of charges is the total work done in assembling them from infinity, summing over each distinct pair once. Bringing the first charge to its corner costs no work (no field yet): $W_1=0$. Bringing the second to a corner distance $a$ away costs $W_2=\dfrac{kq^{2}}{a}$, where $k=\dfrac{1}{4\pi\varepsilon_0}$. Bringing the third, which now sits distance $a$ from each of the other two, costs $W_3=\dfrac{kq^{2}}{a}+\dfrac{kq^{2}}{a}=\dfrac{2kq^{2}}{a}$. The total is $U=W_1+W_2+W_3=\dfrac{3kq^{2}}{a}=\dfrac{3q^{2}}{4\pi\varepsilon_0 a}$. Equivalently, count the three distinct pairs directly: each pair contributes $\dfrac{kq^{2}}{a}$, and there are $\binom{3}{2}=3$ pairs, giving the same $U=\dfrac{3kq^{2}}{a}$. The energy is positive because all charges repel — external work had to be done to force them together, and that work would be released as kinetic energy if the charges were let go. The common error is to double-count pairs; each unordered pair must be included exactly once.

JEE-Advanced style (authored); NCERT XII §2.7 (potential energy of a system of charges, sum over distinct pairs)

A cylindrical (coaxial) capacitor has an inner conductor of radius $a=1.0\ \mathrm{mm}$, an outer shell of radius $b=3.0\ \mathrm{mm}$ and length $L=0.10\ \mathrm{m}$, with vacuum between. Derive its capacitance from Gauss's law and evaluate it numerically.

Attempt, then reveal full solution
Put charge $+Q$ on the inner conductor and $-Q$ on the outer shell, giving linear charge density $\lambda=Q/L$. A coaxial Gaussian cylinder of radius $r$ (with $a\lt r\lt b$) and length $L$ encloses charge $\lambda L$, and by symmetry $E$ is radial and uniform over its curved surface: $E\,(2\pi r L)=\dfrac{\lambda L}{\varepsilon_0}$, so $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$. The potential difference is $V=\displaystyle\int_a^b E\,dr=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{b}{a}$. Hence $C=\dfrac{Q}{V}=\dfrac{\lambda L}{V}=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$. Numerically, $\ln(b/a)=\ln 3=1.0986$, and $2\pi\varepsilon_0 L=2\pi(8.85\times10^{-12})(0.10)=5.56\times10^{-12}$, so $C=\dfrac{5.56\times10^{-12}}{1.0986}=5.06\times10^{-12}\ \mathrm{F}\approx5.1\ \mathrm{pF}$. Note the capacitance depends only on the geometry and grows only logarithmically with the radius ratio, so doubling $b$ changes $C$ only modestly.

JEE-Advanced style (authored); Gauss's law and cylindrical capacitor, NCERT XII §2.12 (geometry-only capacitance)

A spherical capacitor consists of an inner sphere of radius $a$ and a concentric outer shell of radius $b$, carrying $+Q$ and $-Q$ with vacuum between. Find its capacitance and the energy stored, and show the energy equals $Q^2/2C$.

Attempt, then reveal full solution
Between the conductors ($a\lt r\lt b$) Gauss's law gives the point-charge field $E=\dfrac{Q}{4\pi\varepsilon_0 r^{2}}$. The potential difference is $V=\displaystyle\int_a^b E\,dr=\dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=\dfrac{Q}{4\pi\varepsilon_0}\dfrac{b-a}{ab}$. Therefore $C=\dfrac{Q}{V}=4\pi\varepsilon_0\dfrac{ab}{b-a}$. The stored energy from the field is $U=\displaystyle\int_a^b \tfrac12\varepsilon_0 E^{2}\,4\pi r^{2}\,dr=\dfrac{Q^{2}}{8\pi\varepsilon_0}\int_a^b\dfrac{dr}{r^{2}}=\dfrac{Q^{2}}{8\pi\varepsilon_0}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)$. Compare with $\dfrac{Q^{2}}{2C}=\dfrac{Q^{2}}{2}\cdot\dfrac{1}{4\pi\varepsilon_0}\dfrac{b-a}{ab}=\dfrac{Q^{2}}{8\pi\varepsilon_0}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)$ — identical, confirming the field energy is exactly $Q^{2}/2C$. In the limit $b\to\infty$, $C\to4\pi\varepsilon_0 a$ (isolated sphere) and $U\to\dfrac{Q^{2}}{8\pi\varepsilon_0 a}$, the self-energy of a charged sphere.

JEE-Advanced style (authored); spherical capacitor and energy density, NCERT XII §2.12 and §2.15

A parallel-plate capacitor of plate area $A$ and separation $d$ is half-filled by two dielectric slabs placed side by side, each covering area $A/2$ and the full gap $d$, with constants $K_1$ and $K_2$. Find the equivalent capacitance. Then compare with the case where the two slabs, each of thickness $d/2$, are stacked to fill the full area.

Attempt, then reveal full solution
Side-by-side arrangement: the two halves share the same voltage $V$ (both plates are equipotentials), so they act as two capacitors in PARALLEL. Each has area $A/2$ and gap $d$: $C_1=\dfrac{\varepsilon_0 K_1 (A/2)}{d}$, $C_2=\dfrac{\varepsilon_0 K_2 (A/2)}{d}$. Adding, $C_{\parallel}=C_1+C_2=\dfrac{\varepsilon_0 A}{2d}(K_1+K_2)$. Stacked arrangement: the same field lines pass through both layers in turn, so they carry equal charge and act as two capacitors in SERIES. Each has area $A$ and thickness $d/2$: $C_1'=\dfrac{\varepsilon_0 K_1 A}{d/2}=\dfrac{2\varepsilon_0 K_1 A}{d}$, $C_2'=\dfrac{2\varepsilon_0 K_2 A}{d}$. In series $\dfrac{1}{C_{series}}=\dfrac{d}{2\varepsilon_0 A}\left(\dfrac{1}{K_1}+\dfrac{1}{K_2}\right)$, giving $C_{series}=\dfrac{2\varepsilon_0 A}{d}\dfrac{K_1 K_2}{K_1+K_2}$. The parallel (side-by-side) result is an arithmetic mean of the $K$'s and always exceeds the series (stacked) result, which is a harmonic mean dominated by the smaller constant.

JEE-Advanced style (authored); composite dielectrics as parallel/series capacitors, NCERT XII §2.13–2.14

An infinite ladder network is built from identical capacitors each of value $C$: at every stage one capacitor is in series along the line and one is in parallel (a shunt) to the return rail. Find the equivalent capacitance $C_{eq}$ looking into the input terminals.

Attempt, then reveal full solution
Use the self-similarity of an infinite ladder: adding one more stage in front leaves the equivalent capacitance unchanged. Let $C_{eq}=x$. The network is one series capacitor $C$ followed by (a shunt capacitor $C$ in parallel with the rest of the ladder, which is again $x$). The shunt $C$ and the remaining ladder $x$ are in parallel: $C+x$. This is in series with the leading series capacitor $C$: $x=\dfrac{C\,(C+x)}{C+(C+x)}=\dfrac{C(C+x)}{2C+x}$. Cross-multiplying: $x(2C+x)=C(C+x)\Rightarrow 2Cx+x^{2}=C^{2}+Cx\Rightarrow x^{2}+Cx-C^{2}=0$. Solving the quadratic and keeping the positive root, $x=\dfrac{-C+\sqrt{C^{2}+4C^{2}}}{2}=\dfrac{C(\sqrt5-1)}{2}\approx0.618\,C$. The equivalent capacitance is the golden-ratio fraction $0.618\,C$ of a single capacitor — a classic result showing how an infinite series/parallel combination settles to a finite self-consistent value.

JEE-Advanced style (authored); infinite ladder self-similarity with series/parallel rules, NCERT XII §2.14

A parallel-plate capacitor $C_0=\varepsilon_0 A/d$ is charged by a battery of emf $V_0$ and remains connected. The plate separation is now slowly increased from $d$ to $2d$. Find the new capacitance, charge, field and stored energy, the work done by the external agent pulling the plates apart, and the energy exchanged with the battery.

Attempt, then reveal full solution
With the battery connected, $V$ stays at $V_0$. New capacitance $C=\dfrac{\varepsilon_0 A}{2d}=\dfrac{C_0}{2}$. New charge $Q=CV_0=\dfrac{C_0 V_0}{2}$, half the original $Q_0=C_0V_0$ — so charge $\dfrac{C_0V_0}{2}$ flows BACK into the battery. New field $E=\dfrac{V_0}{2d}=\dfrac{E_0}{2}$. Stored energy falls from $U_i=\tfrac12 C_0V_0^{2}$ to $U_f=\tfrac12\cdot\dfrac{C_0}{2}V_0^{2}=\dfrac{1}{4}C_0V_0^{2}$, a drop of $\Delta U=-\dfrac{1}{4}C_0V_0^{2}$. Energy bookkeeping: the battery receives back charge $\Delta Q=\dfrac{C_0V_0}{2}$ at voltage $V_0$, so the battery GAINS $W_{batt}=V_0\,\Delta Q=\dfrac{1}{2}C_0V_0^{2}$. Energy conservation: $W_{ext}+W_{by\,battery\,to\,cap}=\Delta U$, with $W_{by\,battery\,to\,cap}=-\dfrac12 C_0V_0^2$ (battery absorbs energy). Thus $W_{ext}=\Delta U-W_{by\,battery}= -\dfrac14C_0V_0^2-\left(-\dfrac12C_0V_0^2\right)=+\dfrac{1}{4}C_0V_0^{2}$. The agent does positive work $\dfrac14C_0V_0^2$ pulling the attracting plates apart, the capacitor's stored energy drops by the same $\dfrac14C_0V_0^2$, and the battery pockets $\dfrac12C_0V_0^2$ — a balanced ledger.

JEE-Advanced style (authored); constant-voltage energy accounting, NCERT XII §2.12 and §2.15

$n$ identical small spherical mercury drops, each of radius $r$ and each charged to a potential $V$, coalesce into one large spherical drop. Assuming charge is conserved and no charge leaks, find the potential of the big drop in terms of $V$ and $n$.

Attempt, then reveal full solution
Each small drop has charge $q=4\pi\varepsilon_0 r V$ (since its potential is $V=\dfrac{q}{4\pi\varepsilon_0 r}$). Charge conservation: the big drop carries $Q=nq$. Volume conservation (mercury is incompressible): $\dfrac{4}{3}\pi R^{3}=n\cdot\dfrac{4}{3}\pi r^{3}$, so $R=n^{1/3}r$. The potential of the big drop is $V'=\dfrac{Q}{4\pi\varepsilon_0 R}=\dfrac{nq}{4\pi\varepsilon_0\,n^{1/3}r}=n^{2/3}\cdot\dfrac{q}{4\pi\varepsilon_0 r}=n^{2/3}V$. So coalescing raises the potential by the factor $n^{2/3}$: eight small drops ($n=8$) at potential $V$ merge into one drop at $8^{2/3}=4$ times the potential, $4V$. The charge grew by $n$ but the radius grew only by $n^{1/3}$, and since $V'\propto Q/R$ the net effect is the $n^{2/3}$ boost — a favourite JEE result.

JEE-Advanced style (authored); charge and volume conservation with $V=Q/4\pi\varepsilon_0 R$, NCERT XII §2.3

📊 Rank Predictor JoSAA/MCC-calibrated

Disclaimer: These bands are approximate and illustrative, built from publicly reported JoSAA 2023-24 closing-rank trends. Actual ranks depend on the number of candidates, paper difficulty and normalisation in a given year, and vary by category and shift. Use them for orientation, not as a guarantee.
What this does: Electrostatic Potential and Capacitance is a compact, high-yield chapter for JEE Main, typically contributing one to two questions each year, and it forms part of the substantial electrostatics-and-current-electricity block. In JEE Advanced it appears mostly inside multi-concept problems that chain capacitor networks with circuits or with the insertion of a dielectric. The bands below map an approximate overall JEE Main percentile to a JoSAA closing-rank range, to help you gauge where a given performance sits. They are indicative only.
How to read it: enter your score on a full chapter mock below. The tool maps it — via historical JEE marks→percentile→JoSAA closing-rank data — to the percentile and All-India-Rank band a student at that level typically lands in. It is a calibration signal for THIS chapter's mastery, not a full-exam rank.
Chapter-mock scorePercentile bandProjected AIR band
99.5+ percentile99.5+$\lt 1500$
99.0-99.5 percentile99.0-99.5$1500-4000$
98.0-99.0 percentile98.0-99.0$4000-9000$
95.0-98.0 percentile95.0-98.0$9000-25000$
90.0-95.0 percentile90.0-95.0$25000-55000$
80.0-90.0 percentile80.0-90.0$55000-120000$
$\lt 80$ percentile$\lt 80$$\gt 120000$

JoSAA 2023-24 closing-rank trends (indicative)

🔖 Bookmarks & Notes Saved to this browser

Bookmark any question or concept card (click the ☆ that appears on hover), and jot notes below. Everything is saved locally in your browser.

Bookmarked items

No bookmarks yet.

Authoritative & comprehensive JEE Main + Advanced resource · sources traced Tier 1–3 · SME-review state (append ?review=1)