Q. The force between two charges $0.06m$ apart is$\ 5N$. If each charge is moved towards the other by $0.01m$, then the force between them will become
A) $7.20N$
B) $11.25N$
C) $22.50N$
D) $45.00N$
Explanation
(b) $F \propto \frac{1}{r^{2}\ } \Rightarrow \frac{F_{1}}{F_{2}\ } = \left( \frac{r_{2}}{r_{1}} \right)^{2} \Rightarrow \frac{5}{F_{2}} = \left( \frac{0.04}{0.06} \right)^{2\ } = F_{2} = 11.25N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.