Electrostatic Energy and Forces

Energy Stored in the Field

Assembling a set of charges takes work, and that work is recoverable, so it is stored energy. For point charges,

$$U = \frac{1}{2}\sum_i q_i V_i$$

where $V_i$ is the potential at charge $i$ due to all the others. The factor of $\tfrac{1}{2}$ prevents counting each pair twice.

The same energy can be attributed to the field itself, with an energy density

$$u = \tfrac{1}{2}\varepsilon_0 E^2$$

or $\tfrac{1}{2}\mathbf{D}\cdot\mathbf{E}$ in a dielectric. Integrating $u$ over all space gives the same total as the charge-based sum, the two are alternative bookkeeping for one quantity, and the field picture becomes essential once radiation is involved, since then the energy is demonstrably out there travelling.

For a capacitor the stored energy takes three equivalent forms:

$$U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}$$

Choosing among them matters, because which is constant depends on the experiment. Pull the plates apart at fixed charge and $Q^2/2C$ rises as $C$ falls, you do work against the attraction. Do it at fixed voltage, with a battery connected, and $\tfrac{1}{2}CV^2$ falls, because charge flows back into the battery.

Forces follow from differentiating the energy. At fixed charge, $F = -\partial U/\partial x$: the system moves so as to lower its energy. The attraction between capacitor plates comes straight out of this, and so does the fact that a dielectric slab is always pulled into the gap, doing so raises $C$ and lowers $Q^2/2C$.

Common pitfall: using $\tfrac{1}{2}CV^2$ when the charge is what is held fixed. All three expressions are correct at any instant, but only one has a constant in front during a given process. Pick the form whose variable your experiment actually holds still.