Electrostatics > Electrical pressure. The energy of an electric field > Problem 6.5.14

Statement

$6.5.14.$ In a field of strength $E_0$, two non‑conducting plane oppositely charged plates are placed perpendicular to the direction of the field. The field strength between the plates is $E$. What work must be done to orient these plates parallel to the external field? The area of each plate is $S$, the distance between the plates is $h$, which is much smaller than the dimensions of the plates.

Solution

Assume that the capacitor's own field (our plates indeed form a capacitor) is initially directed opposite to the external field,
$$
E_c = E_0 - E.
$$
We change the orientation of the capacitor; now $\vec E_c \perp \vec E_0$, so the field between the plates has magnitude:
$$
E' = \sqrt{E_c^2 + E_0^2}.
$$
The difference in field energy between the plates in these two states is
$$
\Delta W = \frac{\varepsilon_0}{2} (E'^2 - E^2) V = \frac{\varepsilon_0}{2} (E_c^2 + E_0^2 - E^2) V,
$$
$$
\Delta W = \frac{\varepsilon_0}{2} \bigl((E_0 - E)^2 + E_0^2 - E^2\bigr) S h,
$$
$$
\Delta W = \varepsilon_0 S h E_0 (E_0 - E).
$$
This energy difference equals the work that must be done against the force exerted by the external field, which tends to keep the capacitor in its original orientation. Such forces also act on a dipole in a field (and two oppositely charged plates resemble a dipole).

Answer

$$
\boxed{A = \varepsilon_0 S h E_0 (E_0 - E)}
$$