Q. 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:
A) $\frac{3F}{5}$
B) $\frac{2F}{3}$
C) $\frac{F}{2}$
D) $\frac{3F}{8}$
Explanation
[Solution by Claude - not from source] After touching A: both have $q/2$. Then touching B: $\frac{q/2+q}{2}=\frac{3q}{4}$. $F'=\frac{k(q/2)(3q/4)}{r^2}=\frac38F$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.