11.2 Potential, capacitance, and electric energy

The electric field stores energy. The cleanest place to see this is a capacitor — two conductors that hold equal and opposite charge and sustain a voltage between them. This lesson builds capacitance from Gauss’s law and shows that the energy resides in the field itself.

The electric potential

Because the electrostatic field is conservative, the work to move a charge between two points is path-independent and defines a potential difference,

V(\mathbf{b}) - V(\mathbf{a}) \;=\; -\int_\mathbf{a}^\mathbf{b} \mathbf{E}\cdot d\boldsymbol{\ell},

with $\mathbf{E} = -\nabla V$ . Potential is potential energy per unit charge; a charge $q$ at potential $V$ has energy $qV$ . Conductors in equilibrium are equipotentials, since any field along the surface would drive the free charges until it vanished.

Capacitance

Two conductors carrying $+Q$ and $-Q$ hold a voltage $V$ between them proportional to the charge. The constant of proportionality is the capacitance,

Q \;=\; CV.

▶ The parallel-plate capacitor Derivation

For two plates of area $A$ separated by a gap $d$ , the charge $\pm Q$ spreads with surface density $\sigma = Q/A$ . Gauss’s law for the field between the plates (a planar symmetry) gives a uniform field

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

The voltage is the field times the gap, $V = Ed = Qd/(\varepsilon_0 A)$ , so

C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}.

where

$C$: capacitance F
$A$: plate area m²
$d$: gap between plates m
$\varepsilon_0$: permittivity of free space F/m

Filling the gap with a dielectric of relative permittivity $\varepsilon_r$ multiplies the capacitance by $\varepsilon_r$ , because the polarised medium partly cancels the field. Capacitance grows with plate area and shrinks with gap: a large capacitance is a large area held a small distance apart.

Energy stored in the field

Charging a capacitor takes work, because each additional charge must be pushed against the voltage already present.

▶ Energy of a charged capacitor Derivation

Moving a charge $dq$ across the current voltage $V = q/C$ costs $dW = V\,dq = (q/C)\,dq$ . Integrating from $0$ to $Q$ ,

U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C} = \tfrac12 CV^2 = \tfrac12 QV.

Plate area A = 1000 cm² Plate separation d = 1.00 mm Voltage V = 10.0 V

The parallel-plate capacitor stores charge Q = CV on its plates and energy U = ½CV² in the field between them. C = ε₀A/d scales linearly with area and inversely with gap; E = V/d gives the operating field. Cell membranes are essentially nanoscale capacitors: with d ≈ 5 nm and dielectric ~5, the specific capacitance is ~1 μF/cm² — enough to support 100 mV potentials with manageable charge per cell.

This energy is not located on the plates but in the field between them. Writing $U = \tfrac12 CV^2$ with $C = \varepsilon_0 A/d$ and $V = Ed$ gives $U = \tfrac12\varepsilon_0 E^2 \times (Ad)$ , so the energy per unit volume is

u = \tfrac12\varepsilon_0 E^2.

The electric field carries an energy density $\tfrac12\varepsilon_0 E^2$ everywhere it exists — a result that will reappear, alongside its magnetic counterpart, when electromagnetic waves are shown to transport energy through empty space.