Question

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. 2 r
• C. $\frac{r}{2}$
• D. $\frac{r}{4}$

Answer 1

The correct option is D $\frac{r}{4}$

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. $\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_0}.\frac{qQ}{r}$
Therefore the distance of closest approach is given by
$r=\frac{qQ}{4\pi {ϵ}_0}.\frac{2}{m{v}^2}\Rightarrow r\infty \frac{1}{v^2}$
Hence if v is doubled, r becomes one fourth.