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:
- F2
- 3F8
- 3F5
- 2F3
Answer
(2) 3F8
Solution
Initially: F = kq2r2
Third sphere touches A: charge is shared equally, qA = q2, third sphere has q2.
Third sphere touches B: qB = 12 (q + q2 ) = 3q4
F' = k ( q2 ) ( 3q4 )r2 = 38 kq2r2 = 3F8
Answer: (2)