Charging a capacitor requires work because additional charge must be moved onto conductors whose potential difference is already growing.

For a capacitor of capacitance $C$, the instantaneous voltage magnitude when charge $q$ has accumulated is

$$V(q)=\frac{q}{C}.$$

Moving an additional small charge $dq$ onto the capacitor requires work

$$dW=V(q),dq=\frac{q}{C},dq.$$

Integrating from the uncharged state to final charge magnitude $Q$ gives the stored electrostatic energy:

$$U=\int_0^Q\frac{q}{C},dq =\boxed{\frac{Q^2}{2C}}.$$

Using $Q=CV$, equivalent forms are

$$\boxed{U=\frac12QV=\frac12CV^2}.$$

The factor $1/2$ appears because the voltage rises from zero to its final value while the capacitor is being charged.

Worked example

A capacitor has

$$C=10,\mu\mathrm F$$

and is charged to

$$V=12,\mathrm V.$$

Its stored energy is

$$U=\frac12CV^2 =\frac12(10\times10^{-6})(12)^2 =7.2\times10^{-4},\mathrm J.$$

Thus

$$\boxed{U=0.72,\mathrm{mJ}}.$$

Energy resides in the electric field

For an ideal vacuum parallel-plate capacitor,

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

and

$$V=Ed.$$

Substituting into $U=\tfrac12CV^2$ gives

$$U =\frac12\left(\varepsilon_0\frac{A}{d}\right)(Ed)^2 =\frac12\varepsilon_0E^2(Ad).$$

The volume between the plates is

$$\mathcal V=Ad.$$

Therefore the energy per unit volume is

$$\boxed{u_E=\frac{U}{\mathcal V}=\frac12\varepsilon_0E^2}.$$

This is the electric-field energy density in vacuum.

The capacitor is not best thought of as storing energy in the metal plates themselves. The separated charges establish an electric field, and the electrostatic energy can be associated with that field throughout space.

General electrostatic fields

For an arbitrary electrostatic field in vacuum, the total field energy can be written as

$$\boxed{U=\int \frac12\varepsilon_0E^2,dV},$$

where the integral extends over the region containing the field.

This field viewpoint becomes especially important when electric and magnetic fields vary in time: electromagnetic energy can occupy space and flow from one region to another.

Fixed charge versus fixed voltage

The equivalent capacitor-energy formulas are mathematically identical for one state, but different experimental constraints matter when geometry changes.

• If a capacitor is isolated, its charge $Q$ remains fixed.
• If it stays connected to an ideal voltage source, the voltage $V$ remains fixed and charge can flow to or from the source.

When capacitance changes, the stored field energy can therefore change differently depending on which quantity is held fixed. Correct energy reasoning must include the complete system, including any voltage source that exchanges energy with the capacitor.