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NEET PHYSICSMedium

Two point charges A and B, having charges +Q and −Q respectively, are placed at a certain distance apart and the force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes:

A

4F/3

B

F

C

9F/16

D

16F/9

Step-by-Step Solution

According to Coulomb's Law, the initial force is F=k(Q)(Q)r2=kQ2r2F = \frac{k(Q)(Q)}{r^2} = \frac{kQ^2}{r^2}. When 25% of charge from A (+Q) is transferred to B (-Q):

  1. Charge transferred = 25100Q=Q4\frac{25}{100}Q = \frac{Q}{4}.
  2. New charge on A: qA′=Q−Q4=3Q4q_A' = Q - \frac{Q}{4} = \frac{3Q}{4}.
  3. New charge on B: qB′=−Q+Q4=−3Q4q_B' = -Q + \frac{Q}{4} = -\frac{3Q}{4}.
  4. New force F′=k(3Q/4)(3Q/4)r2=916kQ2r2F' = \frac{k(3Q/4)(3Q/4)}{r^2} = \frac{9}{16} \frac{kQ^2}{r^2}. Since F=kQ2r2F = \frac{kQ^2}{r^2}, the new force is F′=9F16F' = \frac{9F}{16}.
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