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) $\dfrac{3F}{8}$
Concept. Identical conducting spheres in contact share their total charge equally; $F \propto q_Aq_B$.
Step 1. Third sphere touches A. Each gets $\dfrac{q}{2}$.
Step 2. Third sphere touches B. Each gets $\dfrac12\left(\dfrac{q}{2} + q\right) = \dfrac{3q}{4}$.
Step 3. New force.
$$F' = F \cdot \frac{(q/2)(3q/4)}{q^2} = \frac{3F}{8}$$
Answer. $\dfrac{3F}{8}$ (Option D).
Solution by Sreeraj P, M.Sc Physics