A capacitor of capacitance `15 muF` has a charge `30 muC` and stored energy is W. If charge increased to `60 muC` the energy stored will be.

Options

A. W
B. 4W
C. 6W
D. 1/2 W`

Answer

To solve the problem, we need to calculate the energy stored in the capacitor when the charge is increased from 30 μC to 60 μC. We will use the formula for the energy stored in a capacitor.

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance (C) = 15 μF = \(15 \times 10^{-6} \, F\)
- Initial charge (Q1) = 30 μC = \(30 \times 10^{-6} \, C\)
- Final charge (Q2) = 60 μC = \(60 \times 10^{-6} \, C\)

2. **Use the formula for energy stored in a capacitor:**
The energy (W) stored in a capacitor is given by the formula:
\[ W = \frac{1}{2} \frac{Q^2}{C} \]
where \(Q\) is the charge and \(C\) is the capacitance.

3. **Calculate the initial energy (W1) with Q1:**
\[ W_1 = \frac{1}{2} \frac{(30 \times 10^{-6})^2}{15 \times 10^{-6}} \]
\[ W_1 = \frac{1}{2} \frac{900 \times 10^{-12}}{15 \times 10^{-6}} \]
\[ W_1 = \frac{1}{2} \times 60 \times 10^{-6} = 30 \times 10^{-6} \, J \]

4. **Calculate the final energy (W2) with Q2:**
\[ W_2 = \frac{1}{2} \frac{(60 \times 10^{-6})^2}{15 \times 10^{-6}} \]
\[ W_2 = \frac{1}{2} \frac{3600 \times 10^{-12}}{15 \times 10^{-6}} \]
\[ W_2 = \frac{1}{2} \times 240 \times 10^{-6} = 120 \times 10^{-6} \, J \]

5. **Final Result:**
The energy stored when the charge is increased to 60 μC is:
\[ W_2 = 120 \, \mu J \]

### Summary:
When the charge in the capacitor is increased from 30 μC to 60 μC, the energy stored in the capacitor increases to 120 μJ.