Source: High school physics (Chinese)
Problem
A parallel-plate capacitor with plate area $S$ and separation $d$ is connected to a constant-voltage source of voltage $U$. With the source remaining connected, the plate separation is doubled.
- By how much does the electrostatic energy stored in the capacitor change?
- How much energy is returned to the source during this process?
- How much work do external forces do against the electric force on the plates?
(1) $\Delta W = -\dfrac{\varepsilon_0 S U^2}{4d}$ (capacitor energy decreases). (2) $W_{\text{returned}} = \dfrac{\varepsilon_0 S U^2}{2d}$. (3) $W_{\text{ext}} = \dfrac{\varepsilon_0 S U^2}{4d}$.
The voltage is fixed at $U$. The capacitance halves ($C \to C/2$), so the charge halves and the energy halves.
Initial: $C = \dfrac{\varepsilon_0 S}{d}$, $W_1 = \tfrac{1}{2} C U^2 = \dfrac{\varepsilon_0 S U^2}{2d}$, $Q_1 = CU = \dfrac{\varepsilon_0 S U}{d}$.
Final: $C' = \dfrac{C}{2}$, $W_2 = \tfrac{1}{2} C' U^2 = \dfrac{\varepsilon_0 S U^2}{4d}$, $Q_2 = C' U = \dfrac{Q_1}{2}$.
(1) Change in capacitor energy:
$$\Delta W = W_2 - W_1 = -\dfrac{\varepsilon_0 S U^2}{4d} \quad (\text{decreases}).$$(2) Charge $|\Delta Q| = Q_1 - Q_2 = \dfrac{\varepsilon_0 S U}{2d}$ flows back to the source. The energy returned is:
$$W_{\text{returned}} = U |\Delta Q| = \dfrac{\varepsilon_0 S U^2}{2d}.$$(3) Energy balance (no dissipation): $W_{\text{ext}} + W_{\text{source}\to\text{cap}} = \Delta W$, with $W_{\text{source}\to\text{cap}} = -W_{\text{returned}} = -\dfrac{\varepsilon_0 S U^2}{2d}$. Hence:
$$W_{\text{ext}} = \Delta W - W_{\text{source}\to\text{cap}} = -\dfrac{\varepsilon_0 S U^2}{4d} + \dfrac{\varepsilon_0 S U^2}{2d} = \dfrac{\varepsilon_0 S U^2}{4d}.$$