A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

Options

A. A) `(3 epsi_0 AV_(0)^(2))/( d) `
B. B) `(epsi_0 AV_(0)^(2))/( 2 d)`
C. C) `(epsi_0 AV_(0)^(2) )/(3d)`
D. D) `(epsi_0 AV_(0)^(2) )/( d)`

Answer

To solve the problem of finding the work required to separate the plates of a parallel plate capacitor, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The capacitor has a plate area \( A \) and an initial separation \( d \).
- It is charged to a potential difference \( V \).
- The charging battery is disconnected, meaning the charge \( Q \) on the plates remains constant.

2. **Calculate Initial Energy**:
- The initial energy \( U_i \) stored in the capacitor can be expressed as:
\[ U_i = \frac{Q^2}{2C} \]
- The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
- Therefore, substituting \( C \) into the energy equation:
\[ U_i = \frac{Q^2}{2 \left(\frac{\epsilon_0 A}{d}\right)} = \frac{Q^2 d}{2 \epsilon_0 A} \]

3. **Change the Separation**:
- The plates are pulled apart to three times the initial separation, so the new separation \( d' = 3d \).

4. **Calculate Final Capacitance**:
- The new capacitance \( C' \) when the plates are separated by \( 3d \) is:
\[ C' = \frac{\epsilon_0 A}{3d} \]

5. **Calculate Final Energy**:
- The final energy \( U_f \) stored in the capacitor after the plates are separated can be expressed as:
\[ U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2 \left(\frac{\epsilon_0 A}{3d}\right)} = \frac{3Q^2 d}{2 \epsilon_0 A} \]

6. **Calculate Work Done**:
- The work done \( W \) to separate the plates is the change in energy:
\[ W = U_f - U_i \]
- Substituting the expressions for \( U_f \) and \( U_i \):
\[ W = \left(\frac{3Q^2 d}{2 \epsilon_0 A}\right) - \left(\frac{Q^2 d}{2 \epsilon_0 A}\right) \]
- Simplifying this gives:
\[ W = \frac{2Q^2 d}{2 \epsilon_0 A} = \frac{Q^2 d}{\epsilon_0 A} \]

7. **Substituting for Charge \( Q \)**:
- Since \( Q = CV \) and \( C = \frac{\epsilon_0 A}{d} \):
\[ Q = \frac{\epsilon_0 A}{d} V \]
- Therefore, \( Q^2 = \left(\frac{\epsilon_0 A}{d} V\right)^2 = \frac{\epsilon_0^2 A^2 V^2}{d^2} \).

8. **Final Work Expression**:
- Substitute \( Q^2 \) back into the work equation:
\[ W = \frac{\frac{\epsilon_0^2 A^2 V^2}{d^2} d}{\epsilon_0 A} = \frac{\epsilon_0 A V^2}{d} \]

### Final Result:
The work required to separate the plates is:
\[ W = \frac{\epsilon_0 A V^2}{d} \]