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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 is best given as:
A
$\frac{3F}{5}$
B
$\frac{2F}{3}$
C
$\frac{F}{2}$
D
$\frac{3F}{8}$
Explanation
After touching A, charge on A and C becomes $q/2$. Touching B gives $(q/2 + q)/2 = 3q/4$. Force becomes $F' \propto (q/2)(3q/4) = \frac{3}{8} F$.
Detailed Solution
Initially, force between spheres A and B is $F = \frac{k q^2}{r^2}$. When an uncharged identical sphere C touches A, charge is shared equally: $q_A' = q/2$, $q_C' = q/2$. Next, C touches B (which has charge $q$): the total charge $(q/2 + q) = 3q/2$ is shared equally, so $q_B' = \frac{3q/2}{2} = \frac{3q}{4}$. Finally, sphere C is removed. The new electrostatic repulsive force between A and B is $F' = \frac{k q_A' q_B'}{r^2} = \frac{k (q/2)(3q/4)}{r^2} = \frac{3}{8} \frac{k q^2}{r^2} = \frac{3F}{8}$.