1. F 2. 9F/16 3. 16F/9 4. 4F/3

Answer: 2. 9F/16 25% of A's charge is 0.25Q . After transfer: q_A=0.75Q and q_B=-Q+0.25Q=-0.75Q . Coulomb force F∝ q_Aq_B at the same distance, so F'/F=0.75×0.75/1×1=dfrac916 . F'=9F/16 .

Topic: Electric charge, conservation, Coulomb's law, superposition· Electrostatics

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 :

1. F

2. \(\frac{9F}{16}\)

3. \(\frac{16F}{9}\)

4. \(\frac{4F}{3}\)

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 2. \(\frac{9F}{16}\)

25% of A's charge is \(0.25Q\). After transfer: \(q_A=0.75Q\) and \(q_B=-Q+0.25Q=-0.75Q\).

Coulomb force \(F\propto q_Aq_B\) at the same distance, so \(\dfrac{F'}{F}=\dfrac{0.75\times0.75}{1\times1}=\dfrac9{16}\).

\(F'=\dfrac{9F}{16}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 :

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