Q. 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) $\frac{4F}{3}$
Explanation
[Solution] Charges become +0.75Q and -0.75Q, so $F' = (0.75)^2F = \frac{9F}{16}$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.