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Electrostatics question

2019 · Q120
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Electrostatics question

2019 · Q120

NEETPhysicsElectrostaticsMCQ+4 / −1
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. A
    F
  2. B
    16F9{{16F} \over 9}916F​
  3. C
    9F16{{9F} \over {16}}169F​
  4. D
    4F3{{4F} \over 3}34F​
View written solutionFree

Correct answer: C

Initially F = −KQ2r2 - {{K{Q^2}} \over {{r^2}}}−r2KQ2​

If 25% of charge of A transferred to B then after the transformation of charge.

qA = Q−Q4Q - {Q \over 4}Q−4Q​ = 3Q4{{3Q} \over 4}43Q​

and qb = -Q + Q4{Q \over 4}4Q​ = −3Q4 - {{3Q} \over 4}−43Q​

New force, F1 = K(3Q4)(−3Q4)r2{{K\left( {{{3Q} \over 4}} \right)\left( { - {{3Q} \over 4}} \right)} \over {{r^2}}} r2K(43Q​)(−43Q​)​

=−916KQ2r2= - {9 \over {16}}{{K{Q^2}} \over {{r^2}}}=−169​r2KQ2​

= 9F16{{9F} \over {16}}169F​

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