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 capacitance of a parallel plate capacitor.
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A'}{d'}, \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A' \) is the area of the plates,
- \( d' \) is the distance between the plates.

Step 2: Energy stored in the capacitor.
The energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2. \]
Substituting the formula for \( C \) into this equation:
\[ U = \frac{1}{2} \cdot \frac{\varepsilon_0 A' V^2}{d'}. \]

Step 3: Energy after the separation is increased.
When the plates are pulled apart to four times the initial separation, the new distance between the plates is \( 4d' \). The new capacitance \( C_{\text{new}} \) becomes:
\[ C_{\text{new}} = \frac{\varepsilon_0 A'}{4d'}. \]
The energy stored in the capacitor after the separation is increased is:
\[ U_{\text{new}} = \frac{1}{2} \cdot \frac{\varepsilon_0 A' V^2}{4d'}. \]

Step 4: Work done to increase the distance.
The work done to increase the separation is the difference in the stored energies:
\[ W = U_{\text{new}} - U = \frac{1}{2} \cdot \frac{\varepsilon_0 A' V^2}{4d'} - \frac{1}{2} \cdot \frac{\varepsilon_0 A' V^2}{d'}. \]
Simplifying:
\[ W = \frac{\varepsilon_0 A' V^2}{4d'} \left( \frac{1}{2} - 1 \right) = \frac{3}{4} \varepsilon_0 A' V^2. \]
Final Answer:
Thus, the work required to increase the distance between the plates is:
\[ \boxed{3 \varepsilon_0 A' V^2}. \]