Source: High school physics (Chinese)
Problem
A parallel-plate capacitor has plate area $S$, separation $d$, and carries charge $Q$. The capacitor is isolated (disconnected from any source), and the plate separation is then doubled.
- By how much does the electrostatic energy change?
- How much work do external forces do against the electric force on the plates?
(1) Energy increases by $\Delta W = \dfrac{Q^2 d}{2 \varepsilon_0 S}$. (2) $W_{\text{ext}} = \dfrac{Q^2 d}{2 \varepsilon_0 S}$.
Charge is fixed at $Q$. The capacitance changes from $C = \dfrac{\varepsilon_0 S}{d}$ to $C' = \dfrac{\varepsilon_0 S}{2d} = \dfrac{C}{2}$.
Energies (using $W = Q^2/(2C)$):
$$W_1 = \dfrac{Q^2}{2C} = \dfrac{Q^2 d}{2 \varepsilon_0 S}, \qquad W_2 = \dfrac{Q^2}{2 C'} = \dfrac{Q^2 d}{\varepsilon_0 S}.$$(1) Change in electrostatic energy:
$$\Delta W = W_2 - W_1 = \dfrac{Q^2 d}{2 \varepsilon_0 S}.$$(2) Since the system is isolated, no source supplies energy; the increase in stored electrostatic energy comes entirely from external work done against the attractive force between the oppositely charged plates:
$$W_{\text{ext}} = \Delta W = \dfrac{Q^2 d}{2 \varepsilon_0 S}.$$