Two small identical conducting balls A and B of charges `+10 mu C` and `+30 mu C` respectively, are kept at a sepration of 50 cm. these balls have beeen 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

Explanation

To solve the problem of finding the final charge on each of the conducting balls A and B after they are connected by a wire, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Charges**:
- The charge on ball A is \( +10 \, \mu C \).
- The charge on ball B is \( +30 \, \mu C \).

2. **Calculate the Total Charge**:
- The total charge \( Q_{total} \) is the sum of the charges on both balls:
\[ Q_{total} = Q_A + Q_B = 10 \, \mu C + 30 \, \mu C = 40 \, \mu C \]

3. **Understand Charge Distribution**:
- When two identical conducting balls are connected by a wire, the charges redistribute equally because they are identical and the system will reach electrostatic equilibrium.

4. **Calculate the Final Charge on Each Ball**:
- Since the total charge is equally distributed between the two balls, the final charge \( Q_{final} \) on each ball is:
\[ Q_{final} = \frac{Q_{total}}{2} = \frac{40 \, \mu C}{2} = 20 \, \mu C \]

5. **Conclusion**:
- The final charge on each ball A and B after they are connected by the wire is \( +20 \, \mu C \).

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

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