A parallel plate capacitor with spacing $d$ and area of cross-section $A$ is connected to a source of voltage $V$. If the plates are pulled apart quasistatically to a spacing of $2d$, then which of the following statements are correct?
Options:
(A) The force between the plates at spacing $2d$ is $\frac{1}{8} \left(\frac{\varepsilon_0 A V^2}{d^2}\right)$
(B) The work done in moving the plates is $\frac{1}{8} \left(\frac{\varepsilon_0 A V^2}{d}\right)$
(C) The energy transferred to the voltage source is $\frac{1}{2} \left(\frac{\varepsilon_0 A V^2}{d}\right)$
(D) The energy of the capacitor reduces by $\frac{1}{4} \left(\frac{\varepsilon_0 A V^2}{d}\right)$
Official Answer Key:
A,C,D
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.53)**
**1. Electrostatics Formulation (Capacitor Connected to Constant Voltage Source)**:
- Initial state: spacing $d$, capacitance $C_1 = \frac{\epsilon_0 A}{d}$, energy $U_1 = \frac{1}{2} C_1 V^2 = \frac{\epsilon_0 A V^2}{2d}$.
- Final state: spacing $2d$, capacitance $C_2 = \frac{\epsilon_0 A}{2d}$, energy $U_2 = \frac{1}{2} C_2 V^2 = \frac{\epsilon_0 A V^2}{4d}$.
**2. Detailed Step-by-Step Derivation**:
- **Force at Spacing $2d$**:
$F = \frac{1}{2} \frac{\epsilon_0 A V^2}{(2d)^2} = \frac{1}{8} \left(\frac{\epsilon_0 A V^2}{d^2}\right)$
**Statement (A) is TRUE**.
- **Energy of Capacitor**:
$\Delta U_{\text{cap}} = U_2 - U_1 = \frac{\epsilon_0 A V^2}{4d} - \frac{\epsilon_0 A V^2}{2d} = -\frac{1}{4}\left(\frac{\epsilon_0 A V^2}{d}\right)$
The energy stored in the capacitor reduces by $\frac{1}{4}\left(\frac{\epsilon_0 A V^2}{d}\right)$. **Statement (D) is TRUE**.
- **Charge and Energy Transferred to Battery**:
$\Delta Q = (C_2 - C_1)V = -\frac{\epsilon_0 A V}{2d}$
Energy transferred into battery:
$W_{\text{battery}} = -\Delta Q \cdot V = \frac{1}{2}\left(\frac{\epsilon_0 A V^2}{d}\right)$
**Statement (C) is TRUE**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, C, D)**.
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