Two identical metal balls, A and B, with similar charges Q, are suspended with insulating threads. They repel each other with a force F. Another identical uncharged metal ball C is first touched with metal ball A and then touched with metal ball B and finally removed. The new force of repulsion between A and B will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \frac{3}{8}F \)
When two identical conducting balls touch, their total charge is shared equally. The force then follows from Coulomb's law, \( F \propto q_A q_B \) at a fixed separation.
\( \displaystyle \text{C touches A:}\ q_A = q_C = \frac{Q + 0}{2} = \frac{Q}{2} \)
\( \displaystyle \text{C touches B:}\ q_B = q_C = \frac{Q/2 + Q}{2} = \frac{3Q}{4} \)
\( \displaystyle \frac{F'}{F} = \frac{q_A\,q_B}{Q \cdot Q} = \frac{(Q/2)(3Q/4)}{Q^{2}} = \frac{3}{8} \)
\( \displaystyle F' = \frac{3}{8}F \)