Two point charges A and B, having charges +Q and −Q respectively, are placed at a certain distance apart and t

Question

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

Accepted Answer

According to Coulomb's Law, the initial force is $F = \frac{k(Q)(Q)}{r^2} = \frac{kQ^2}{r^2}$. 
When 25% of charge from A (+Q) is transferred to B (-Q):
1. Charge transferred = $\frac{25}{100}Q = \frac{Q}{4}$.
2. New charge on A: $q_A' = Q - \frac{Q}{4} = \frac{3Q}{4}$.
3. New charge on B: $q_B' = -Q + \frac{Q}{4} = -\frac{3Q}{4}$.
4. New force $F' = \frac{k(3Q/4)(3Q/4)}{r^2} = \frac{9}{16} \frac{kQ^2}{r^2}$.
Since $F = \frac{kQ^2}{r^2}$, the new force is $F' = \frac{9F}{16}$.

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