Q111 / 180 · NEET 2019 (All India) · Physics

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 the force between the charges becomes

A
$F$ .
Correct answer
B
$\frac{9 F}{16}$ .
C
$\frac{16 F}{9}$ .
D
$\frac{4 F}{3}$ .

Solution

F = \frac{K Q^{2}}{r^{2}}

After transferring

25 %

of charge,

q_{1} = \frac{3 Q}{4}

and

q_{2} = - \frac{3 Q}{4}

F_{1} = \frac{9}{16} \frac{K Q^{2}}{r^{2}} = \frac{9}{16} F

Hence, the correct answer is option: 'B'.

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