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 step by step, we will use the formula for the energy stored in a capacitor, which is given by:

\[ W = \frac{1}{2} \frac{Q^2}{C} \]

where:
- \( W \) is the energy stored,
- \( Q \) is the charge,
- \( C \) is the capacitance.

### Step 1: Calculate the initial energy stored (W1)

Given:
- Capacitance \( C = 15 \, \mu F = 15 \times 10^{-6} \, F \)
- Initial charge \( Q_1 = 30 \, \mu C = 30 \times 10^{-6} \, C \)

Using the formula:

\[ W_1 = \frac{1}{2} \frac{Q_1^2}{C} = \frac{1}{2} \frac{(30 \times 10^{-6})^2}{15 \times 10^{-6}} \]

Calculating \( W_1 \):

\[ W_1 = \frac{1}{2} \frac{(900 \times 10^{-12})}{15 \times 10^{-6}} = \frac{1}{2} \frac{900}{15} \times 10^{-6} = \frac{1}{2} \times 60 \times 10^{-6} = 30 \times 10^{-6} \, J \]

### Step 2: Calculate the new energy stored (W2)

Now, we increase the charge to \( Q_2 = 60 \, \mu C = 60 \times 10^{-6} \, C \).

Using the same formula:

\[ W_2 = \frac{1}{2} \frac{Q_2^2}{C} = \frac{1}{2} \frac{(60 \times 10^{-6})^2}{15 \times 10^{-6}} \]

Calculating \( W_2 \):

\[ W_2 = \frac{1}{2} \frac{(3600 \times 10^{-12})}{15 \times 10^{-6}} = \frac{1}{2} \frac{3600}{15} \times 10^{-6} = \frac{1}{2} \times 240 \times 10^{-6} = 120 \times 10^{-6} \, J \]

### Step 3: Relate W2 to W1

Now, we can find the ratio of \( W_2 \) to \( W_1 \):

\[ \frac{W_2}{W_1} = \frac{120 \times 10^{-6}}{30 \times 10^{-6}} = 4 \]

Thus, we can express \( W_2 \) in terms of \( W_1 \):

\[ W_2 = 4 W_1 \]

### Conclusion

If the initial energy stored is \( W \), then the new energy stored when the charge is increased to \( 60 \, \mu C \) will be:

\[ W_2 = 4W \]

### Final Answer

The energy stored when the charge is increased to \( 60 \, \mu C \) will be \( 4W \).

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