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

$\frac{3}{8}F$

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 = $\frac{3F}{8}$