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NEET PHYSICSMedium

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

\frac{9F}{16}

C

\frac{16F}{9}

D

\frac{4F}{3}

Step-by-Step Solution

Initially, F=kQ2r2F = \frac{kQ^2}{r^2}. After transferring 25% of charge from A to B, the new charges are qA=Q−Q4=3Q4q_A = Q - \frac{Q}{4} = \frac{3Q}{4} and qB=−Q+Q4=−3Q4q_B = -Q + \frac{Q}{4} = -\frac{3Q}{4}. The new force is F1=k∣qA∣∣qB∣r2=k(3Q4)(3Q4)r2=916kQ2r2=9F16F_1 = \frac{k|q_A||q_B|}{r^2} = \frac{k(\frac{3Q}{4})(\frac{3Q}{4})}{r^2} = \frac{9}{16} \frac{kQ^2}{r^2} = \frac{9F}{16}.

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