Q. Two point charges repel each other with a force of 100 N. One of the charges is increased by 10% and other is reduced by 10%. The new force of repulsion at the same distance would be
A) 100 N
B) 121 N
C) 99 N
D) None of these
Explanation
(c) $F = \frac{1}{4\pi\varepsilon_{0}\ }\frac{q_{1}q_{2}}{r^{2}}\ (110/100)\ (90/100)$ times
$ie,\ \frac{99}{100}\ \ $times
$\therefore$ net force $= \frac{99}{100} \times 100 = 99N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.