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) → $\frac{3}{8}F$
Initially:
Charge on A = Q, Charge on B = Q
Force of repulsion, $F = \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{r^2}$
When uncharged ball C touches A:
Charges redistribute equally (since identical spheres)
Charge on A = $\frac{Q}{2}$, Charge on C = $\frac{Q}{2}$
Now C (with $\frac{Q}{2}$) touches B (with Q):
Total charge = $Q + \frac{Q}{2} = \frac{3Q}{2}$
Each gets $\frac{3Q}{4}$
After removing C:
Charge on A = $\frac{Q}{2}$, Charge on B = $\frac{3Q}{4}$
New force between A and B:
$F' = \frac{1}{4\pi\varepsilon_0}\frac{(\frac{Q}{2})(\frac{3Q}{4})}{r^2}$
$F' = \frac{3}{8}\frac{1}{4\pi\varepsilon_0}\frac{Q^2}{r^2}$
$F' = \frac{3F}{8}$
∴ New force of repulsion = $\frac{3F}{8}$