Question

A charged particle $q$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the closest distance of approach would be:

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

Answer 1

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

Given: initial velocity =v, final velocity=2v

Solution: We know the formula of distance of closest approach :

$\frac{1}{2}m{v}^2=k\times \frac{qQ}{r}$

$r\propto \frac{1}{v}$

$\frac{r_1}{r_2}:\frac{(2v{)}^2}{v^2}$

$r_2=\frac{r_1}{4}$

Hence the correct option is D