Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q a…

Question

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. The new force of repulsion between spheres A and B (radii negligible compared to the separation, so they act as point charges) is best given as:

1. A. 3F/5
2. B. 2F/3
3. C. F/2
4. D. 3F/8 (Correct answer)

Correct Answer

Option D — 3F/8

Detailed Solution & Explanation

The correct answer is $\dfrac{3F}{8}$.

Key Points

• Initially $F=\dfrac{kq^2}{r^2}$.
• Third sphere touches $A$: charge shares equally, so $A$ and the third sphere each become $\dfrac{q}{2}$.
• The third sphere ($\dfrac{q}{2}$) then touches $B$ ($q$): they share equally to $\dfrac{q/2+q}{2}=\dfrac{3q}{4}$ each; so $B$ becomes $\dfrac{3q}{4}$.
• New force: $F'=\dfrac{k\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^2}=\dfrac{3}{8}\cdot\dfrac{kq^2}{r^2}=\dfrac{3F}{8}$.