A condenser of capacitance `12muF` was initially charged to 20V. Now potential difference is made 40V. The increase in potential energy is

Options

A. `7.2 xx 10^(-3)J`
B. `4 xx 10^(-3)J`
C. `3 xx 10^(-4)J`
D. `5 xx 10^(-6)J`

Explanation

To solve the problem of finding the increase in potential energy of a capacitor when the potential difference is changed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance, \( C = 12 \, \mu F = 12 \times 10^{-6} \, F \)
- Initial voltage, \( V_1 = 20 \, V \)
- Final voltage, \( V_2 = 40 \, V \)

2. **Recall the formula for potential energy (U) stored in a capacitor:**
\[ U = \frac{1}{2} C V^2 \]
where \( U \) is the potential energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.

3. **Calculate the initial potential energy \( U_1 \) when the voltage is 20V:**
\[ U_1 = \frac{1}{2} C V_1^2 = \frac{1}{2} \times (12 \times 10^{-6}) \times (20)^2 \]
\[ U_1 = \frac{1}{2} \times (12 \times 10^{-6}) \times 400 = \frac{1}{2} \times 12 \times 10^{-6} \times 400 \]
\[ U_1 = 6 \times 10^{-6} \times 400 = 2400 \times 10^{-6} = 2.4 \times 10^{-3} \, J \]

4. **Calculate the final potential energy \( U_2 \) when the voltage is 40V:**
\[ U_2 = \frac{1}{2} C V_2^2 = \frac{1}{2} \times (12 \times 10^{-6}) \times (40)^2 \]
\[ U_2 = \frac{1}{2} \times (12 \times 10^{-6}) \times 1600 = \frac{1}{2} \times 12 \times 10^{-6} \times 1600 \]
\[ U_2 = 6 \times 10^{-6} \times 1600 = 9600 \times 10^{-6} = 9.6 \times 10^{-3} \, J \]

5. **Calculate the increase in potential energy \( \Delta U \):**
\[ \Delta U = U_2 - U_1 = (9.6 \times 10^{-3}) - (2.4 \times 10^{-3}) \]
\[ \Delta U = 7.2 \times 10^{-3} \, J \]

### Final Answer:
The increase in potential energy is \( 7.2 \times 10^{-3} \, J \).