If the potential of a capacitor having capacity of `6 muF` is increased from 10 V to 20 V,then increase in its energy will be

Options

A. `4 xx 10^(-4)J`
B. `4 xx 10^(-14)J`
C. `9 xx 10^(-4)J`
D. `12 xx 10^(-6)J`

Answer

To find the increase in energy of the capacitor when the potential is increased from 10 V to 20 V, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Capacitance, \( C = 6 \, \mu F = 6 \times 10^{-6} \, F \)
- Initial Voltage, \( V_1 = 10 \, V \)
- Final Voltage, \( V_2 = 20 \, V \)

2. **Use the Formula for Energy Stored in a Capacitor:**
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]

3. **Calculate Initial Energy \( U_1 \):**
Substitute \( C \) and \( V_1 \) into the formula:
\[ U_1 = \frac{1}{2} \times 6 \times 10^{-6} \times (10)^2 \]
\[ U_1 = \frac{1}{2} \times 6 \times 10^{-6} \times 100 \]
\[ U_1 = 3 \times 10^{-6} \times 100 = 3 \times 10^{-4} \, J \]

4. **Calculate Final Energy \( U_2 \):**
Substitute \( C \) and \( V_2 \) into the formula:
\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times (20)^2 \]
\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times 400 \]
\[ U_2 = 3 \times 10^{-6} \times 400 = 1.2 \times 10^{-3} \, J \]

5. **Calculate the Increase in Energy \( \Delta U \):**
The increase in energy is given by:
\[ \Delta U = U_2 - U_1 \]
\[ \Delta U = (1.2 \times 10^{-3}) - (3 \times 10^{-4}) \]
\[ \Delta U = 1.2 \times 10^{-3} - 0.3 \times 10^{-3} = 0.9 \times 10^{-3} \, J \]
\[ \Delta U = 9 \times 10^{-4} \, J \]

### Final Answer:
The increase in energy of the capacitor is \( \Delta U = 9 \times 10^{-4} \, J \).