Q. A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2v,$ the closest distances of approach would be
A) $r$
B) $2r$
C) $r/2$
D) $r/4$
Explanation
(d) Charge $q$ will momentarily come to rest at a distance $r$ from charge $Q$ when all it’s kinetic energy converted to potential energy $i.e.\ \ \frac{1}{2}mv^{2} = \frac{1}{4\pi\varepsilon_{0}}.\frac{qQ}{r}$
Therefore the distance of closest approach is given by
$r = \frac{qQ}{4\pi\varepsilon_{0}}.\frac{2}{mv^{2}} \Rightarrow r \propto \frac{1}{v^{2}}$
Hence if $v$ is doubled, $r$ becomes one fourth
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.