A capacitor of capacity `2muF` is charged to `100 V`. What is the heat generated when this capacitor is connected in parallel to an another capacitor of same capacity?

Options

A. `2.5mJ`
B. `5.0mJ`
C. `10mJ`
D. `4mJ`

Explanation

To solve the problem of finding the heat generated when a charged capacitor is connected in parallel to another identical capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Initial Charge on the Capacitor:**
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
Where:
- \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- \( V = 100 \, V \)

Substituting the values:
\[ Q = 2 \times 10^{-6} \, F \times 100 \, V = 2 \times 10^{-4} \, C = 200 \, \mu C \]

2. **Determine the Final Charge on Each Capacitor:**
When the charged capacitor is connected in parallel to another identical capacitor, the total charge will redistribute equally between the two capacitors. Thus, each capacitor will have:
\[ Q_1 = Q_2 = \frac{Q}{2} = \frac{200 \, \mu C}{2} = 100 \, \mu C \]

3. **Calculate the Initial Potential Energy:**
The initial potential energy \( U_i \) stored in the first capacitor can be calculated using the formula:
\[ U_i = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U_i = \frac{1}{2} \times 2 \times 10^{-6} \, F \times (100 \, V)^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 = 10 \times 10^{-3} \, J = 0.01 \, J \]

4. **Calculate the Final Potential Energy:**
After connecting the two capacitors in parallel, the total capacitance becomes:
\[ C_{total} = C_1 + C_2 = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F \]
The final voltage \( V_f \) across the capacitors can be calculated using the total charge and total capacitance:
\[ V_f = \frac{Q}{C_{total}} = \frac{200 \, \mu C}{4 \, \mu F} = 50 \, V \]
Now, we can calculate the final potential energy \( U_f \):
\[ U_f = \frac{1}{2} C_{total} V_f^2 = \frac{1}{2} \times 4 \times 10^{-6} \, F \times (50 \, V)^2 = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 = 5 \times 10^{-3} \, J = 0.005 \, J \]

5. **Calculate the Heat Generated:**
The heat generated \( Q_{heat} \) when the capacitors are connected is the difference between the initial and final potential energies:
\[ Q_{heat} = U_i - U_f = 0.01 \, J - 0.005 \, J = 0.005 \, J \]
Converting to millijoules:
\[ Q_{heat} = 0.005 \, J \times 1000 = 5 \, mJ \]

### Final Answer:
The heat generated when the capacitor is connected in parallel to another capacitor of the same capacity is **5 mJ**.