Energy stored in a capacitor

Charging a capacitor: progressive work

A capacitor of capacitance C, charged under a voltage U, carries a charge Q = C*U on its plates. Progressively charging the capacitor from 0 to Q requires work, because at each instant the voltage across it grows with the charge already deposited.

By integrating the elementary work dW = v * dq (with v = q/C) from 0 to Q, we obtain the stored energy:

E = (1/2) * Q * U = (1/2) * C * U^2 = Q^2 / (2*C)

These three forms are equivalent; the factor 1/2 comes from the integration (the average voltage during charging is U/2, not U).

Numerical example

Consider a capacitor C = 100 microF charged under U = 12 V.

Quantity	Value
Charge Q = C*U	1.2.10^-3 C
Energy E = (1/2)CU^2	7.2.10^-3 J

Volumetric energy density

For a parallel-plate capacitor (plates of area S, distance d, uniform field E between them), it can be shown that the energy can be rewritten as an energy distributed within the field itself:

u = (1/2) * epsilon0 * E^2 (energy per unit volume, in J/m^3)

This formula is general: it shows that the electric field itself carries energy, not only the charges that create it.

Classic pitfall

The most common mistake is to write E = QU instead of E = (1/2)QU. The factor 1/2 is essential: even if a constant-voltage source U supplies the energy QU during charging, only half of it, (1/2)QU, is stored in the capacitor; the other half is dissipated by the Joule effect in the circuit.