The plate separation in a parallel plate condenser and plate area is A. If it is charged to V volt battery is diconnected then the work done increasing the plate separation to 2d will be

Options

A. `(3)/(2)(epsilon_(0)AV^(2))/(d)`
B. `(epsilon_(0)AV^(2))/(d)`
C. `(2epsilon_(0)AV^(2))/(d)`
D. `(epsilon_(0)AV^(2))/(2d)`

Answer

To solve the problem of calculating the work done in increasing the plate separation of a parallel plate capacitor from \( d \) to \( 2d \) after disconnecting the battery, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The capacitor is charged to a voltage \( V \) when connected to a battery.
- The initial plate separation is \( d \) and the area of the plates is \( A \).

2. **Calculate the Initial Charge**:
- The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
- The charge \( Q \) on the capacitor when charged to voltage \( V \) is:
\[ Q = C \cdot V = \frac{\epsilon_0 A V}{d} \]

3. **Calculate Initial Potential Energy**:
- The initial potential energy \( U_i \) stored in the capacitor is given by:
\[ U_i = \frac{Q^2}{2C} = \frac{Q^2}{2 \left( \frac{\epsilon_0 A}{d} \right)} = \frac{Q^2 d}{2 \epsilon_0 A} \]
- Substitute \( Q = \frac{\epsilon_0 A V}{d} \):
\[ U_i = \frac{\left( \frac{\epsilon_0 A V}{d} \right)^2 d}{2 \epsilon_0 A} = \frac{\epsilon_0 A V^2}{2d} \]

4. **Calculate Final Capacitance**:
- When the plate separation is increased to \( 2d \), the new capacitance \( C_f \) is:
\[ C_f = \frac{\epsilon_0 A}{2d} \]

5. **Calculate Final Potential Energy**:
- The final potential energy \( U_f \) is:
\[ U_f = \frac{Q^2}{2C_f} = \frac{Q^2}{2 \left( \frac{\epsilon_0 A}{2d} \right)} = \frac{Q^2 \cdot 2d}{2 \epsilon_0 A} = \frac{Q^2 d}{\epsilon_0 A} \]
- Substitute \( Q = \frac{\epsilon_0 A V}{d} \):
\[ U_f = \frac{\left( \frac{\epsilon_0 A V}{d} \right)^2 d}{\epsilon_0 A} = \frac{\epsilon_0 A V^2}{d} \]

6. **Calculate Work Done**:
- The work done \( W \) in increasing the plate separation is the change in potential energy:
\[ W = U_f - U_i = \frac{\epsilon_0 A V^2}{d} - \frac{\epsilon_0 A V^2}{2d} \]
- Simplifying:
\[ W = \frac{\epsilon_0 A V^2}{d} \left( 1 - \frac{1}{2} \right) = \frac{\epsilon_0 A V^2}{d} \cdot \frac{1}{2} = \frac{\epsilon_0 A V^2}{2d} \]

### Final Answer:
The work done in increasing the plate separation from \( d \) to \( 2d \) is:
\[ W = \frac{\epsilon_0 A V^2}{2d} \]