Question

A parallel plate capacitor has two plates of area 'A' separated by distance 'd'. The capacitor is charged to a potential difference 'V' and the battery is disconnected. A metal plate with area 'A' and thickness \( \frac{d}{2} \) is inserted between the plates, so that it is always parallel to the plates. The work done on the metal slab while it was inserted is:

• A. \( -\frac{\epsilon_0 A V^2}{4d} \)
• B. \( -\frac{\epsilon_0 A V^2}{2d} \)
• C. \( -\frac{\epsilon_0 A V^2}{d} \)
• D. \( -\frac{2 \epsilon_0 A V^2}{d} \)

Hint:

Work done in a capacitor can be calculated from the change in stored energy when the distance between plates is altered.

Answer 1

The Correct Option is A

When the metal plate is inserted, the capacitance of the capacitor changes. Let $C_i$ be the initial capacitance and $C_f$ be the final capacitance.
The initial capacitance is \[C_i = \frac{\epsilon_0 A}{d}.\]After the metal plate is inserted, the capacitor is effectively divided into two capacitors in series, each with plate separation $\frac{d}{4}$. The capacitance of each of these capacitors is \[\frac{\epsilon_0 A}{d/4} = \frac{4 \epsilon_0 A}{d},\]so the final capacitance is \[\frac{1}{C_f} = \frac{1}{\frac{4 \epsilon_0 A}{d}} + \frac{1}{\frac{4 \epsilon_0 A}{d}} = \frac{d}{2 \epsilon_0 A},\]which means $C_f = \frac{2 \epsilon_0 A}{d}$. The initial energy stored in the capacitor is \[U_i = \frac{1}{2} C_i V^2 = \frac{1}{2} \cdot \frac{\epsilon_0 A}{d} \cdot V^2 = \frac{\epsilon_0 A V^2}{2d}.\]Since the battery is disconnected, the charge $Q$ on the capacitor remains constant. Then $Q = C_i V = C_f V_f$, where $V_f$ is the final potential difference across the capacitor. Hence, \[V_f = \frac{C_i}{C_f} V = \frac{\frac{\epsilon_0 A}{d}}{\frac{2 \epsilon_0 A}{d}} V = \frac{V}{2}.\]The final energy stored in the capacitor is \[U_f = \frac{1}{2} C_f V_f^2 = \frac{1}{2} \cdot \frac{2 \epsilon_0 A}{d} \cdot \left( \frac{V}{2} \right)^2 = \frac{\epsilon_0 A V^2}{4d}.\]The work done on the metal slab is equal to the change in energy stored in the capacitor: \[W = U_f - U_i = \frac{\epsilon_0 A V^2}{4d} - \frac{\epsilon_0 A V^2}{2d} = \boxed{-\frac{\epsilon_0 A V^2}{4d}}.\] Final Answer: \( -\frac{\epsilon_0 A V^2}{4d} \)