Two parallel metal plates carry equal and opposite charge, +Q on top and −Q below; between them the electric field is uniform (the evenly spaced brown arrows). The capacitance — how much charge the plates hold per volt — is C = ε₀A/d in vacuum, so bigger plates (area A) or a smaller gap d give more capacitance. Slide a dielectric slab of constant K between the plates and it multiplies the capacitance to C = Kε₀A/d. Slide A, d and K and read C.
Two parallel plates carry +Q (top) and −Q (bottom). Between them the electric field is uniform — the evenly spaced brown arrows.
C = K ε₀ A / d · ε₀ = 8.85×10⁻¹²
A = — m² · d = — mm · K = — · C = — pF