GATE 2022 PH – Question 53

Correct answer: (A) $F(2d)=\epsilon_0AV^2/(8d^2)$; (C) Energy returned to source is $\epsilon_0AV^2/(2d)$; (D) Stored energy falls by $\epsilon_0AV^2/(4d)$

Explanation

The capacitor stays connected to the battery, so the voltage $V$ is constant. Its capacitance is $C(x)=\epsilon_0A/x$ for plate separation $x$, and the stored energy is $U=\tfrac12CV^2$.

**Force between the plates at separation $x$:**
$$F=\frac12V^2\left|\frac{dC}{dx}\right|=\frac{\epsilon_0AV^2}{2x^2}.$$

• **A. $F(2d)=\epsilon_0AV^2/(8d^2)$.** Yes: $\dfrac{\epsilon_0AV^2}{2(2d)^2}=\dfrac{\epsilon_0AV^2}{8d^2}$. ✓
• **B. $W=\epsilon_0AV^2/(8d)$.** The work done by the external agent against the force is $\int_d^{2d}F\,dx=\dfrac{\epsilon_0AV^2}{2}\left(\dfrac1d-\dfrac1{2d}\right)=\dfrac{\epsilon_0AV^2}{4d}$, not $\tfrac1{8d}$. ✗
• **C. The energy returned to the source is $\epsilon_0AV^2/(2d)$.** The charge falls from $\epsilon_0AV/d$ to $\epsilon_0AV/(2d)$, so the charge $\dfrac{\epsilon_0AV}{2d}$ flows back to the battery at voltage $V$, returning energy $\dfrac{\epsilon_0AV^2}{2d}$. ✓
• **D. The stored energy falls by $\epsilon_0AV^2/(4d)$.** $U_i=\dfrac{\epsilon_0AV^2}{2d}$ and $U_f=\dfrac{\epsilon_0AV^2}{4d}$, so the fall is $\dfrac{\epsilon_0AV^2}{4d}$. ✓

(Energy balance: returned to source $\tfrac{\epsilon_0AV^2}{2d}$ = fall in stored energy $\tfrac{\epsilon_0AV^2}{4d}$ + work done by the agent $\tfrac{\epsilon_0AV^2}{4d}$ ✓.)

Answer **A, C and D**.