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Question

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


A
F
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B
9F16
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C
16F9
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D
4F3
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Solution

The correct option is B $$\dfrac{9 F}{16}$$
$$F=\dfrac{KQ^2}{r^2}$$
If 25% of charge of A transferred to B then 
$$q_A = Q - \dfrac{Q}{4} = \dfrac{3 Q}{4} $$ and $$q_B = -Q + \dfrac{Q}{4} = \dfrac{-3Q}{4}$$
$$F_1 = \dfrac{kq_A q_B}{r^2}$$
$$F_1 = \dfrac{k \left(\dfrac{3Q}{4} \right)^2}{r^2}$$
$$F_1 = \dfrac{9}{16} \dfrac{kQ}{r^2}$$
$$\implies F_1 = \dfrac{9F}{16}$$
1259393_1618853_ans_03538f43d65c4dc59d3a6d52ca8a6ca3.png

Physics

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