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Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes:

1. F
2. 9F/16
3. 16F/9
4. 4F/3

Correct answer: 9F/16

Solution

Initially, the force is F = k(Q)(-Q)/r². After transferring 25% of charge from A to B, charges become (3Q/4) and (-3Q/4). The new force is F' = k(3Q/4)(-3Q/4)/r² = (9/16)F.

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