Q. Two point charges+$3\mu$C and +$8\mu$C repel each other with a force of 40 N. If a charge of +$5\mu$C is added to each of them, then the force between them will become
A) $- 10$ N
B) $+ 10$ N
C) $+ 20$ N
D) $- 20$ N
Explanation
(a) Here, $\ F = 40 = \frac{k(3)(8)}{r^{2}} = \frac{24k}{r^{2}}$
$F^{'} = \frac{k(3 - 5)(8 - 5)}{r^{2}} = \frac{- 6k}{r^{2}}$
$\frac{F^{'}}{F} = {\frac{- 6k}{r^{2}} \times {\frac{r^{2}}{24k} = - \frac{1}{4}}}$
$F^{'} = - \frac{F}{4} = - \frac{40}{4} = - 10N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.