Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
Answer: (D) $\dfrac{3F}{8}$
Touching identical spheres share their total charge equally.
Third sphere touches A: each gets $\dfrac{q}{2}$.
Third sphere (with $\dfrac{q}{2}$) touches B (with $q$): each gets $\dfrac{1}{2}\left(\dfrac{q}{2} + q\right) = \dfrac{3q}{4}$.
$$F' = F\cdot\frac{(q/2)(3q/4)}{q^2} = \frac{3F}{8}$$
Solution by Sreeraj P, M.Sc Physics