Two equal and opposite charges are placed at a certain distance apart and force of attraction between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes:

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$\frac{F}{16} (attraction)$

$\frac{7 F}{16} (repulsion)$

$\frac{F}{16} (repulsion)$

$\frac{7 F}{16} (attraction)$

answer is B.

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Detailed Solution

$\begin{array}{rcl} F & \propto & q_{1} q_{2}; \\ F_{2} & \propto & \frac{q}{4} \times \frac{q}{4} (∵ q - \frac{3 q}{4} = + \frac{q}{4}) \\ \Rightarrow & F_{2} = \frac{F}{16} (Attraction) \end{array}$

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