Capacitor Stored Energy Derivation

The energy density of an electric field (Joules per cubic meter) is:

$\eta_E = \frac{1}{2} \epsilon \, E^2$

Suppose we have a parallel plate capacitor with plate area $A$ and plate separation $d$ . If the capacitor carries a charge $Q$ , the following holds for an applied voltage $V$ :

$Q = C \, V$

Assuming the potential gradient is constant between the plates, the electric field is:

$E = \frac{V}{d}$

The energy density, volume, and stored energy are:

$\eta_E = \frac{1}{2} \epsilon \, E^2 = \frac{1}{2} \epsilon \, \frac{V^2}{d^2} \\ \text{volume} = A \, d \\ \text{stored energy} =\eta_E \times \text{volume} = \frac{1}{2} \frac{\epsilon \, A}{d} V^2 = \frac{1}{2} \, C \, V^2$

Note that there are several idealizations in this derivation (uniform potential gradient, no stored energy outside of the footprint of the plates, etc.). The stored energy in an actual capacitor may therefore deviate from this somewhat.

#ElectricityAndMagnetism

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