Formulas/physics/Electrostatic Potential/Parallel Plate Capacitor

Parallel Plate Capacitor

Capacitance of a parallel plate capacitor with plate area A and separation d in vacuum. Increases with area, decreases with separation.

Class 11Class 12

Derivation

Setup

Two parallel conducting plates, each of area $A$ , separated by distance $d$ (with $d \ll \sqrt{A}$ so fringing is negligible). Plates carry charges $+Q$ and $-Q$ , giving surface charge density $\sigma = Q/A$ .

Field between the plates

From the field of two oppositely charged infinite sheets (formula ef17-field-two-planes):

E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}

Potential difference

The field is uniform, so:

V = Ed = \frac{Qd}{\varepsilon_0 A}

Capacitance

C = \frac{Q}{V} = \frac{Q}{\dfrac{Qd}{\varepsilon_0 A}}

\boxed{C = \frac{\varepsilon_0 A}{d}}

Key dependencies

$C \propto A$ : larger plates → more charge at same $V$
$C \propto 1/d$ : closer plates → stronger field, lower $V$ for same $Q$ , higher $C$
Independent of $Q$ and $V$ individually — confirms $C$ is purely geometric

Remember

Fringing fields at the edges are ignored in this derivation. The approximation $d \ll \sqrt{A}$ ensures the uniform-field result is accurate in the bulk of the capacitor.

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