Question

A parallel plate capacitor having plate area \( A' \) and separation \( d' \) is charged to a potential difference \( V \). The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is

• A. \( \frac{3}{4} \varepsilon_0 A' V^2 \)
• B. \( \frac{4}{4} \varepsilon_0 A' V^2 \)
• C. \( 2 \varepsilon_0 A' V^2 \)
• D. \( 3 \varepsilon_0 A' V^2 \)

Hint:

When increasing the distance between the plates of a capacitor, the capacitance decreases, and work is done to overcome the electrostatic forces. The energy stored in the capacitor depends on the capacitance and voltage.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Work done in changing the plate separation is equal to the change in electrostatic potential energy. Since the battery is disconnected, the charge \( Q \) remains constant.
Step 2: Key Formula or Approach:
1. Energy \( U = \frac{Q^2}{2C} \).
2. \( Q = CV = \frac{\epsilon_0 A V}{d} \).
3. Work \( W = U_{final} - U_{initial} \).
Step 3: Detailed Explanation:
Initial energy:
\[ U_i = \frac{1}{2} C_i V^2 = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) V^2 \]
Initial charge \( Q = C_i V = \frac{\epsilon_0 A V}{d} \).
Final separation \( d' = 4d \). Final capacitance \( C_f = \frac{\epsilon_0 A}{4d} = \frac{C_i}{4} \).
Final energy:
\[ U_f = \frac{Q^2}{2 C_f} = \frac{Q^2}{2 (C_i / 4)} = 4 \left( \frac{Q^2}{2 C_i} \right) = 4 U_i \]
Work done \( W \):
\[ W = U_f - U_i = 4 U_i - U_i = 3 U_i \]
\[ W = 3 \cdot \left[ \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) V^2 \right] = \frac{3 \epsilon_0 A V^2}{2d} \]
Step 4: Final Answer:
The work required is \( \frac{3 \epsilon_0 A V^2}{2d} \).