Question

Two positive point charges of \(10 \, \mu C\) and \(12 \, \mu C\) are kept in air with a separation of 12 cm. - To make the distance between the charges 4 cm, the work done is

• A. \(24 \, J\)
• B. \(18 \, J\)
• C. \(9 \, J\)
• D. \(12 \, J\)

Hint:

When moving point charges, use the formula: \[ W = k \cdot q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \] where \( k = 9 \times 10^9 \). Ensure units are consistent (meters, Coulombs).

Answer 1

The Correct Option is B

Step 1: Work Done in Moving Charges The work done in moving two point charges from an initial separation \(r_1\) to a final separation \(r_2\) is given by the formula: \[ W = k \cdot \frac{q_1 q_2}{r_2} - k \cdot \frac{q_1 q_2}{r_1} \] where: 
- \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) (Coulomb’s constant), 
- \( q_1 = 10 \times 10^{-6} C \), 
- \( q_2 = 12 \times 10^{-6} C \), 
- \( r_1 = 12 \, cm = 0.12 \, m \), 
- \( r_2 = 4 \, cm = 0.04 \, m \). 

Step 2: Substituting Values \[ W = 9 \times 10^9 \times \left( \frac{10 \times 10^{-6} \times 12 \times 10^{-6}}{0.04} - \frac{10 \times 10^{-6} \times 12 \times 10^{-6}}{0.12} \right) \] \[ = 9 \times 10^9 \times 120 \times 10^{-12} \times \left( \frac{1}{0.04} - \frac{1}{0.12} \right) \] \[ = 9 \times 10^9 \times 120 \times 10^{-12} \times \left( 25 - 8.33 \right) \] \[ = 9 \times 10^9 \times 120 \times 10^{-12} \times 16.67 \] \[ = 18 J \] Thus, the correct answer is \( \mathbf{(2)} \ 18 J \).