A parallel-plate air capacitor has an initial plate separation $x_0$. An external agent slowly pulls

Question

A parallel-plate air capacitor has an initial plate separation $x_0$. An external agent slowly pulls the plates apart until the separation becomes $2x_0$. Consider two distinct processes. 
Process I: The capacitor is kept connected to a constant voltage source $V_0$ during the pulling.
Process II: The capacitor is initially charged to voltage $V_0$ and then disconnected from the source before pulling the plates. \Let $W_I$ and $W_{II}$ be the work done by the external agent in Process I and Process II, respectively. What is the ratio $W_{II} / W_I$?

Professor PC Thomas Classes · Answer

Let the initial capacitance be $C_0$. In Process I (constant voltage $V_0$), the final capacitance is $C_f = C_0 / 2$. The change in stored energy is $\Delta U = \frac{1}{2} C_f V_0^2 - \frac{1}{2} C_0 V_0^2 = -\frac{1}{4} C_0 V_0^2$. The work done by the battery is $W_{bat} = V_0 \Delta Q = V_0 (C_f V_0 - C_0 V_0) = -\frac{1}{2} C_0 V_0^2$. From energy conservation, $W_I + W_{bat} = \Delta U \implies W_I - \frac{1}{2} C_0 V_0^2 = -\frac{1}{4} C_0 V_0^2 \implies W_I = \frac{1}{4} C_0 V_0^2$.

In Process II (constant charge $Q_0 = C_0 V_0$), the energy change is solely due to external work: $W_{II} = \Delta U = \frac{Q_0^2}{2C_f} - \frac{Q_0^2}{2C_0} = \frac{Q_0^2}{2(C_0/2)} - \frac{Q_0^2}{2C_0} = \frac{Q_0^2}{C_0} - \frac{Q_0^2}{2C_0} = \frac{Q_0^2}{2C_0}$. Substituting $Q_0 = C_0 V_0$, $W_{II} = \frac{1}{2} C_0 V_0^2$. Therefore, the ratio $W_{II} / W_I = (\frac{1}{2} C_0 V_0^2) / (\frac{1}{4} C_0 V_0^2) = 2$.