Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and for

Question

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 :

Accepted Answer

Initially, $F = \frac{kQ^2}{r^2}$. After transferring 25% of charge from A to B, the new charges are $q_A = Q - \frac{Q}{4} = \frac{3Q}{4}$ and $q_B = -Q + \frac{Q}{4} = -\frac{3Q}{4}$. The new force is $F_1 = \frac{k|q_A||q_B|}{r^2} = \frac{k(\frac{3Q}{4})(\frac{3Q}{4})}{r^2} = \frac{9}{16} \frac{kQ^2}{r^2} = \frac{9F}{16}$.

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