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 of the capacitor,
- \( V \) is the voltage (emf) across the capacitor.

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

### Step 2: Substitute the values into the formula
Now, substituting the values into the formula:

\[ U = \frac{1}{2} \times (6 \times 10^{-6}) \times (3)^2 \]

### Step 3: Calculate \( V^2 \)
First, calculate \( V^2 \):

\[ V^2 = 3^2 = 9 \, V^2 \]

### Step 4: Substitute \( V^2 \) back into the formula
Now substitute \( V^2 \) back into the energy formula:

\[ U = \frac{1}{2} \times (6 \times 10^{-6}) \times 9 \]

### Step 5: Perform the multiplication
Now perform the multiplication:

\[ U = \frac{1}{2} \times 54 \times 10^{-6} = 27 \times 10^{-6} \, J \]

### Step 6: Convert to microjoules
Since \( 1 \, J = 10^6 \, \mu J \), we can express the energy in microjoules:

\[ U = 27 \, \mu J \]

### Conclusion
The electrostatic energy stored in the capacitor is:

\[ \boxed{27 \, \mu J} \]