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 and force of interaction will be

Options

A. `28.8N`
B. `32.6N`
C. `14.4N`
D. `72N`

Explanation

To solve the problem step by step, we will follow these actions:

### Step 1: Identify the initial charges on the balls
- Charge on ball A, \( Q_A = +10 \, \mu C = 10 \times 10^{-6} \, C \)
- Charge on ball B, \( Q_B = +30 \, \mu C = 30 \times 10^{-6} \, C \)

### Step 2: Calculate the total charge
The total charge \( Q_{total} \) when the balls are connected by a wire is given by:
\[ Q_{total} = Q_A + Q_B = 10 \times 10^{-6} + 30 \times 10^{-6} = 40 \times 10^{-6} \, C \]

### Step 3: Determine the final charge on each ball
Since the balls are identical and connected by a wire, they will share the total charge equally. Therefore, the final charge \( Q_f \) on each ball is:
\[ Q_f = \frac{Q_{total}}{2} = \frac{40 \times 10^{-6}}{2} = 20 \times 10^{-6} \, C = 20 \, \mu C \]

### Step 4: Calculate the force of interaction between the balls
The force of interaction \( F \) between two charges can be calculated using Coulomb's Law:
\[ F = k \frac{Q_1 Q_2}{R^2} \]
Where:
- \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) (Coulomb's constant)
- \( Q_1 = Q_f = 20 \times 10^{-6} \, C \)
- \( Q_2 = Q_f = 20 \times 10^{-6} \, C \)
- \( R = 50 \, cm = 0.5 \, m \)

Substituting the values into the formula:
\[ F = 9 \times 10^9 \frac{(20 \times 10^{-6})^2}{(0.5)^2} \]
Calculating:
\[ F = 9 \times 10^9 \frac{400 \times 10^{-12}}{0.25} \]
\[ F = 9 \times 10^9 \times 1600 \times 10^{-12} \]
\[ F = 14.4 \, N \]

### Final Result
- The final charge on each ball \( Q_f = 20 \, \mu C \)
- The force of interaction \( F = 14.4 \, N \)

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