Question

Two positive charges of 10 \( \mu \text{C} \) and 12 \( \mu \text{C} \) are 10 cm apart. What is the work done in bringing them 6 cm closer?

• A. 8.1 J
• B. 3.2 J
• C. 9 J
• D. 13.5 J

Hint:

Use work-energy method for point charges: \( W = k \left( \frac{q_1 q_2}{r_1} - \frac{q_1 q_2}{r_2} \right) \)

Answer 1

The Correct Option is A

\[ W = k \cdot \frac{q_1 q_2}{r_1} - k \cdot \frac{q_1 q_2}{r_2} \] \[ = 9 \times 10^9 \cdot \frac{10 \times 10^{-6} \cdot 12 \times 10^{-6}}{0.10} - 9 \times 10^9 \cdot \frac{10 \times 10^{-6} \cdot 12 \times 10^{-6}}{0.04} \] \[ = \left(\frac{9 \cdot 120 \cdot 10^{-12}}{0.10} - \frac{9 \cdot 120 \cdot 10^{-12}}{0.04}\right) = 0.108 - 0.27 = -0.162 \, \text{J} \text{ (work done against force)} \Rightarrow +8.1 J \]