Two small identical conducting balls A and B of charges `+10 mu C` and `+30 mu C` respectively, are kept at a separation of 50 cm. these balls have been connected by a wire for a short time. The final charge on each of the balls A and B will be

Options

A. `10 mu C ` and `30 mu C`, respectively `
B. `20 muC` on each ball
C. `30 mu C` and `10 mu C`, respectively
D. `-40 mu C` and `80 mu C`, respectively

Answer

To solve the problem, we need to determine the final charge on each of the two identical conducting balls A and B after they are connected by a wire.

### Step-by-Step Solution:

1. **Identify the Initial Charges**:
- Charge on ball A, \( Q_A = +10 \, \mu C = 10 \times 10^{-6} \, C \)
- Charge on ball B, \( Q_B = +30 \, \mu C = 30 \times 10^{-6} \, C \)

2. **Calculate the Total Charge**:
- The total charge when both balls are connected by a wire is the sum of their individual charges.
\[ Q_{\text{total}} = Q_A + Q_B = 10 \times 10^{-6} + 30 \times 10^{-6} = 40 \times 10^{-6} \, C \]

3. **Determine the Charge Distribution**:
- Since the balls are identical and conducting, the total charge will distribute equally between them when connected by the wire.
\[ Q_{\text{final}} = \frac{Q_{\text{total}}}{2} = \frac{40 \times 10^{-6}}{2} = 20 \times 10^{-6} \, C \]

4. **Final Charges on Each Ball**:
- Therefore, the final charge on each ball A and B will be:
\[ Q_A' = Q_B' = 20 \, \mu C \]

### Final Answer:
- The final charge on each of the balls A and B will be \( +20 \, \mu C \).