A condenser has a capacity `2muF` and is charged to a voltage of 50 V. The energy stored is

Options

A. `25xx10^(5)J`
B. 25 J
C. `25xx10" erg"`
D. `25xx10^(3)" erg"`

Explanation

To find the energy stored in a capacitor, we can use the formula:

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

where:
- \(E\) is the energy stored in joules,
- \(C\) is the capacitance in farads,
- \(V\) is the voltage in volts.

### Step-by-Step Solution:

1. **Identify the given values**:
- Capacitance, \(C = 2 \, \mu F = 2 \times 10^{-6} \, F\)
- Voltage, \(V = 50 \, V\)

2. **Substitute the values into the energy formula**:
\[ E = \frac{1}{2} \times (2 \times 10^{-6}) \times (50)^2 \]

3. **Calculate \(V^2\)**:
\[ V^2 = 50^2 = 2500 \]

4. **Substitute \(V^2\) back into the energy equation**:
\[ E = \frac{1}{2} \times (2 \times 10^{-6}) \times 2500 \]

5. **Simplify the equation**:
\[ E = \frac{1}{2} \times 2 \times 2500 \times 10^{-6} \]
\[ E = 2500 \times 10^{-6} \, J \]

6. **Convert to a more standard form**:
\[ E = 2.5 \times 10^{-3} \, J = 0.0025 \, J \]

### Final Answer:
The energy stored in the capacitor is \(0.0025 \, J\) or \(2.5 \, mJ\).