A capacitor of capacitance 6 `mu`F is connected with an emf sourface 3 V. Electrostaci energy stored in the capacitor is

Options

A. 3 `mu`j
B. 18 ` mu` J
C. 27 `mu`J
D. 36 `mu` J

Answer

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

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

where:
- \( U \) is the energy stored in the capacitor,
- \( C \) is the capacitance,
- \( V \) is the voltage (emf) across the capacitor.

### Step-by-Step Solution:

1. **Identify the given values**:
- Capacitance \( C = 6 \, \mu F = 6 \times 10^{-6} \, F \)
- Voltage \( V = 3 \, V \)

2. **Substitute the values into the formula**:
\[ U = \frac{1}{2} C V^2 \]
\[ U = \frac{1}{2} \times (6 \times 10^{-6}) \times (3)^2 \]

3. **Calculate \( V^2 \)**:
\[ V^2 = 3^2 = 9 \]

4. **Substitute \( V^2 \) back into the equation**:
\[ U = \frac{1}{2} \times (6 \times 10^{-6}) \times 9 \]

5. **Multiply the values**:
\[ U = \frac{1}{2} \times 54 \times 10^{-6} \]
\[ U = 27 \times 10^{-6} \, J \]

6. **Convert to microjoules**:
\[ U = 27 \, \mu J \]

### Final Answer:
The electrostatic energy stored in the capacitor is \( 27 \, \mu J \).

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