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

- The work done in changing the separation of two charges is equal to the change in electrostatic potential energy: \[ W = k q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \]
- Given:
- \( k = 9 \times 10^9 \, N\,m^2/C^2 \)
- \( q_1 = 10 \times 10^{-6} \, C \)
- \( q_2 = 12 \times 10^{-6} \, C \)
- \( r_1 = 0.12 \, m \)
- \( r_2 = 0.04 \, m \)
- Substitute values: \[ W = 9 \times 10^9 \times (10 \times 10^{-6})(12 \times 10^{-6}) \left( \frac{1}{0.04} - \frac{1}{0.12} \right) \]
- Simplify charge product: \[ (10 \times 10^{-6})(12 \times 10^{-6}) = 120 \times 10^{-12} \]
- Now compute reciprocals: \[ \frac{1}{0.04} = 25,\quad \frac{1}{0.12} \approx 8.33 \]
- Difference: \[ 25 - 8.33 = 16.67 \]
- Now substitute: \[ W = 9 \times 10^9 \times 120 \times 10^{-12} \times 16.67 \]
- Simplify powers: \[ 9 \times 120 \times 10^{-3} \times 16.67 \]
- Final result: \[ W \approx 18 \, J \]
- Hence, the work done is: 18 J