A capacitor when charged by a potential difference of 200 Volts, stores a charge of 0.1C. By discharging energy liberated by the capacitor is-

Options

A. `-30 J`
B. `-15 J`
C. 10 J
D. 20 J

Answer

To solve the problem of finding the energy liberated by a capacitor when it is discharged, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Potential difference (V) = 200 Volts
- Charge (Q) = 0.1 Coulombs

2. **Use the formula for energy stored in a capacitor:**
The energy (U) stored in a capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]
However, we do not have the capacitance (C) directly.

3. **Relate capacitance to charge and voltage:**
We know that the capacitance (C) can be expressed as:
\[ C = \frac{Q}{V} \]
where Q is the charge and V is the voltage.

4. **Substitute C in the energy formula:**
Substitute \( C = \frac{Q}{V} \) into the energy formula:
\[ U = \frac{1}{2} \left(\frac{Q}{V}\right) V^2 \]

5. **Simplify the expression:**
Simplifying the expression gives:
\[ U = \frac{1}{2} Q V \]
Here, one V from \( V^2 \) cancels out with the V in the denominator.

6. **Substitute the known values:**
Now, substitute the values of Q and V into the equation:
\[ U = \frac{1}{2} \times 0.1 \, \text{C} \times 200 \, \text{V} \]

7. **Calculate the energy:**
- First, calculate \( \frac{1}{2} \times 200 = 100 \).
- Then, multiply by 0.1:
\[ U = 100 \times 0.1 = 10 \, \text{Joules} \]

8. **Conclusion:**
The energy liberated by the capacitor when discharged is **10 Joules**.