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Question:

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?

3F/4

F/2

F

3F/8

Solution:

Initially charges on both sphere, q
F = Kq²/d² (1)
When sphere C will get touched with sphere A, then final charges on both will become q/2.
Now, when this sphere C will get touched to sphere B, then final charges on both of them will be
qc = qd = (q/2 + q/2)/2 = 3q/4
Now force between A and B will be
F' = k × (q/2) × (3q/4)/d²
F' = 3F/8