Two identical charged conducting spheres A and B have their centres… | NEET 2025

Q21NEET 2025Medium

Electric Charge & Coulomb's Law

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:

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Answer: (b)

Identical spheres share charge equally on contact.
C touches A: $A = C = \dfrac{q}{2}$ . C touches B: $B = C = \dfrac{q/2 + q}{2} = \dfrac{3q}{4}$ .
$F' = \dfrac{k(q/2)(3q/4)}{r^2} = \dfrac{3}{8}\dfrac{kq^2}{r^2} = \dfrac{3F}{8}$
So the answer is (b) $\dfrac{3F}{8}$ .

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