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
B
9F16
C
16F9
D
4F3

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 $$q_B = -Q + \dfrac{Q}{4} = \dfrac{-3Q}{4}$$$$F_1 = \dfrac{kq_A q_B}{r^2}$$$$F_1 = \dfrac{k \left(\dfrac{3Q}{4} \right)^2}{r^2}$$$$F_1 = \dfrac{9}{16} \dfrac{kQ}{r^2}$$$$\implies F_1 = \dfrac{9F}{16}$$Physics

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