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\% c - Physics MCQ NEET 2019 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Initial force F = k \frac{∣ ( Q ) ( - Q ) ∣}{r ^{2}} = k \frac{Q ^{2}}{r ^{2}} .After transferring 25% (Q /4) from A to B :New charge of A : Q^{'} = Q - Q /4 = \frac{3 Q}{4} .New charge of B : - Q^{'} = - Q + Q /4 = - \frac{3 Q}{4} .New force F^{'} : F^{'} = k \frac{∣ ( \frac{3 Q}{4} ) ( - \frac{3 Q}{4} ) ∣}{r ^{2}} = k \frac{9 Q ^{2}}{16 r ^{2}} = \frac{9}{16} F

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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:

F

\frac{9 F}{16}

\frac{16 F}{9}

\frac{4 F}{3}

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