Q. The distance between two point charges is increased by 10%. The force of interaction between them
A) Increased by 10%
B) decreased by 10%
C) decreased by 17%
D) decreased by 21%
Explanation
(c) The force of interaction $\ \ F \propto \frac{1}{r^{2}}$
When length is increased by 10%
$F = \frac{1}{\ {(110/100)}^{2}\ }$ times =$\ \frac{100}{121}\ $times
$= 0.03\ times$
$\therefore$ Decrease in force $= (1 - 0.83)100\% = 17\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.