If the energy of a capacitor of capacitance `2muF` is 0.16 joule. Then its potential difference will be

Options

A. 800V
B. 400V
C. `16 xx 10^(4)V`
D. `16 xx 10^(-4)V`

Explanation

To find the potential difference across a capacitor given its capacitance and energy, we can use the formula for the energy stored in a capacitor:

\[ U = \frac{1}{2} C V^2 \]

Where:
- \( U \) is the energy (in joules),
- \( C \) is the capacitance (in farads),
- \( V \) is the potential difference (in volts).

### Step-by-step Solution:

1. **Identify the given values:**
- Capacitance, \( C = 2 \mu F = 2 \times 10^{-6} F \)
- Energy, \( U = 0.16 J \)

2. **Substitute the known values into the energy formula:**
\[ U = \frac{1}{2} C V^2 \]
\[ 0.16 = \frac{1}{2} (2 \times 10^{-6}) V^2 \]

3. **Simplify the equation:**
Multiply both sides by 2 to eliminate the fraction:
\[ 0.32 = (2 \times 10^{-6}) V^2 \]

4. **Solve for \( V^2 \):**
Divide both sides by \( 2 \times 10^{-6} \):
\[ V^2 = \frac{0.32}{2 \times 10^{-6}} = \frac{0.32}{2} \times 10^{6} = 0.16 \times 10^{6} = 1.6 \times 10^{5} \]

5. **Calculate \( V \):**
Take the square root of both sides:
\[ V = \sqrt{1.6 \times 10^{5}} = \sqrt{16 \times 10^{4}} = 4 \times 10^{2} = 400 \text{ volts} \]

### Conclusion:
The potential difference across the capacitor is \( V = 400 \text{ volts} \).