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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This question was previously asked in

NEET 2025 Official Paper (Held On: 04 May, 2025)

Answer 1

Calculation:
Let the charge on each sphere A and B be q and the separation be d.

Therefore, the force between spheres A and B is:

F = (1 / (4πɛ₀)) × (q² / d²) ... (1)

When spheres A and C are touched and then separated, charge on each will be:

(q + 0) / 2 = q / 2

Now sphere B is touched with sphere C. Charge on each will be:

(q + q/2) / 2 = (3q) / 4

Now the force between sphere A and sphere B will be:

F' = (1 / (4πɛ₀)) × (q/2 × 3q/4) / d²

= (3/8) × (1 / (4πɛ₀)) × (q² / d²)

⇒ F' = (3/8) × F