Electrostatic energy

The energy stored in an electrostatic field is

$W_e=\frac{1}{2}{\int }_vϵE^{2}dv\text{(J)},$

integrated over all of space (or any volume containing the field). The integrand is the electrostatic energy density:

$w_e=\frac{1}{2}ϵE^2=\frac{1}{2}\mathbf{D}\cdot \mathbf{E}{\text{(J/m}}^3\text{)}.$

The energy is not localized on the charges — it’s stored in the field that fills space. This is one of the conceptual breakthroughs of EM theory: charges aren’t independent “things that have potential energy”; they couple through the field that lives between them, and the energy is literally in that field.

Derivation via capacitor

Charge up a capacitor of capacitance $C$ from $v=0$ to $v=V$, transferring total charge $Q=CV$. At an intermediate state with charge $q$ on the plates, the voltage is $v=q/C$. Transferring an additional $dq$ requires work $d{W}_e=vdq=(q/C)dq$:

$W_e={\int }_0^Q\frac{q}{C}dq=\frac{Q^2}{2C}.$

Using $C=Q/V$:

$W_e=\frac{1}{2}C{V}^2=\frac{1}{2}QV.$

For a parallel-plate capacitor, substitute $C=ϵA/d$ and $V=Ed$:

$W_e=\frac{1}{2}\cdot \frac{ϵA}{d}\cdot (Ed{)}^2=\frac{1}{2}ϵE^2(Ad).$

The volume is $v=Ad$ — the volume between the plates — and the energy density is $\frac{1}{2}ϵE^2$.

Generalization

The expression $w_e=\frac{1}{2}ϵE^2$ derived for parallel plates turns out to be universally valid: it gives the local energy density wherever there’s an electric field, in any dielectric or vacuum, regardless of how the field was set up.

Justification: build up any charge distribution by bringing in infinitesimal charges from infinity; track the total work done; rewrite as a field integral using Gauss’s law and the divergence theorem. The result always reduces to $\frac{1}{2}ϵE^2$ integrated over all space.

In a multi-conductor system

For a system of $N$ conductors at potentials $V_i$ carrying charges $Q_i$, the total electrostatic energy is

$W_e=\frac{1}{2}{\sum }_{i=1}^N{Q}_i{V}_i.$

For a two-conductor capacitor ($Q_1=+Q$, $Q_2=-Q$, $V_1-V_2=V$), this reproduces $W_e=QV/2$.

Why the factor of 1/2

The factor $\frac{1}{2}$ catches a lot of students. It comes from the fact that to assemble the configuration, you bring charges in gradually — early charges don’t see the full final field, only what’s been assembled so far. The “average voltage encountered” is half the final voltage. So work done is $Q\cdot V/2$, not $QV$.

A common wrong intuition: $W=QV$ “because force $\times$ distance.” The right intuition is: the field grows linearly as you add charge, and you integrate force-against-field as the field builds.

Comparison with magnetic energy

The magnetic analog:

$W_m=\frac{1}{2}{\int }_v\mu H^{2}dv,w_m=\frac{1}{2}\mu H^2=\frac{1}{2}\mathbf{B}\cdot \mathbf{H}.$

For an inductor: $W_m=\frac{1}{2}L{I}^2$. The $1/2$ has the same origin — the field builds gradually as you ramp the current. See Magnetic energy.

In an electromagnetic wave, both densities are present and equal on time-average:

$\overline{w_e}=\overline{w_m}.$

Half the wave’s energy is in the electric field, half in the magnetic. The Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ describes the flow of this total energy density.