Single Choice

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$$F$$
B$$\dfrac{9 F}{16}$$
Correct Answer
C$$\dfrac{16 F}{9}$$
D$$\dfrac{4 F}{3}$$

Solution

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

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