Two point charges A and B having charges +Q and -Q are placed at certain distance and force between them is F, if 25 % of charge A is transferred to B, then force between the charges becomes

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$\frac{9 F}{16}$

$\frac{16 F}{9}$

$\frac{4 F}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} F α q_{1} q_{2} \\ \frac{F_{1}}{F_{2}} = \frac{q_{1}}{q_{1}^{1}} \times \frac{q_{2}}{{q_{2}}^{1}} \\ \frac{F_{1}}{F_{2}} = \frac{100}{(75)} \times \frac{100}{(75)} \\ \frac{F}{F_{2}} = \frac{16}{9} \Rightarrow F_{2} = \frac{9 F}{16} \end{array}$

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