A `2 muF` capacitor is charged to `100 V`, and then its plates are connected by a conducting Wire. The heat produced is .

Options

A. 1 J
B. 0.1 J
C. 0.01 J
D. 0.001 J

Answer

To solve the problem, we need to calculate the heat produced when a charged capacitor is short-circuited by connecting its plates with a conducting wire. The energy stored in the capacitor will be converted into heat.

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance (C) = 2 µF = \(2 \times 10^{-6} \, F\)
- Voltage (V) = 100 V

2. **Calculate the energy stored in the capacitor:**
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U = \frac{1}{2} \times (2 \times 10^{-6}) \times (100)^2 \]

3. **Calculate \(V^2\):**
\[ V^2 = 100^2 = 10000 \]

4. **Substituting \(V^2\) back into the energy formula:**
\[ U = \frac{1}{2} \times (2 \times 10^{-6}) \times 10000 \]

5. **Simplifying the expression:**
\[ U = \frac{1}{2} \times 2 \times 10^{-6} \times 10^4 \]
\[ U = 10^{-6} \times 10^4 = 10^{-2} \, J \]

6. **Convert to Joules:**
\[ U = 0.01 \, J \]

7. **Conclusion:**
The heat produced when the plates of the capacitor are connected by a conducting wire is \(0.01 \, J\).

### Final Answer:
The heat produced is \(0.01 \, J\).

---