A capacitor with capacitance `5 mu F` is chrged to `5 mu C`. If the plates are pulled apart to reduce the capacitance to `2 mu F`, how much work is done ?

Options

A. `6.25xx10^(-6)J`
B. `3.75xx10^(-6)J`
C. `2.16xx10^(-6)J`
D. `2.55xx10^(-6)J`

Answer

To solve the problem, we need to determine the work done when the plates of a capacitor are pulled apart, changing its capacitance from \(5 \mu F\) to \(2 \mu F\) while keeping the charge constant at \(5 \mu C\).

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- Initial capacitance, \(C_i = 5 \mu F = 5 \times 10^{-6} F\)
- Charge on the capacitor, \(Q = 5 \mu C = 5 \times 10^{-6} C\)

2. **Calculate the Initial Potential Energy:**
The potential energy (\(U\)) stored in a capacitor is given by the formula:
\[ U = \frac{Q^2}{2C} \]
Substituting the initial values:
\[ U_i = \frac{(5 \times 10^{-6})^2}{2 \times (5 \times 10^{-6})} \]
\[ U_i = \frac{25 \times 10^{-12}}{10 \times 10^{-6}} = 2.5 \times 10^{-6} \text{ Joules} \]

3. **Understand the Final Conditions:**
- Final capacitance, \(C_f = 2 \mu F = 2 \times 10^{-6} F\)

4. **Calculate the Final Potential Energy:**
Using the same formula for potential energy:
\[ U_f = \frac{Q^2}{2C_f} \]
Substituting the final values:
\[ U_f = \frac{(5 \times 10^{-6})^2}{2 \times (2 \times 10^{-6})} \]
\[ U_f = \frac{25 \times 10^{-12}}{4 \times 10^{-6}} = 6.25 \times 10^{-6} \text{ Joules} \]

5. **Calculate the Work Done:**
The work done (\(W\)) in changing the configuration of the capacitor is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values:
\[ W = 6.25 \times 10^{-6} - 2.5 \times 10^{-6} = 3.75 \times 10^{-6} \text{ Joules} \]

### Final Answer:
The work done when the plates are pulled apart is \(3.75 \times 10^{-6} \text{ Joules}\) or \(3.75 \mu J\).