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 [2019]

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Solution

Q.
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 [2019]

1 $\frac{4 F}{3}$

2 $F$

3 $\frac{9 F}{16}$

4 $\frac{16 F}{9}$

Ans.
(3)

In Case I :

$F = - \frac{1}{4 π ε_{0}} \frac{Q^{2}}{r^{2}}$ ...(i)

In Case II : $Q_{A} = Q - \frac{Q}{4}, Q_{B} = - Q + \frac{Q}{4}$

$∴$ $F' = \frac{1}{4 π ε_{0}} \frac{(Q - \frac{Q}{4}) (- Q + \frac{Q}{4})}{r^{2}}$

$= \frac{1}{4 π ε_{0}} \frac{(\frac{3}{4} Q) (\frac{- 3}{4} Q)}{r^{2}} = - \frac{1}{4 π ε_{0}} \frac{9}{16} \frac{Q^{2}}{r^{2}}$ ...(ii)

From equations (i) and (ii), $F' = \frac{9}{16} F$

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