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Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to -

1. 3F/4
2. F/2
3. F
4. 3F/8

Correct answer: 3F/8

Solution

Initially A = B = q, F = kq^2/d^2. C (0) touches A -> both q/2. Then C (q/2) touches B (q) -> each (q + q/2)/2 = 3q/4. New force = k(q/2)(3q/4)/d^2 = (3/8)kq^2/d^2 = 3F/8.

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