Question

A charge particle $q$ is shot from a large distance with speed $v$ towards a fixed charge particle $Q$. It approaches $Q$ upto a closest distance $r$ and then turns. 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 $\frac{r}{4}$

According to question,
When speed is $v$.
$\frac{1}{2}m{v}^2=\frac{KQq}{r}-------------------\left(1\right)$
When speed is $2v$.
$\frac{1}{2}m{4v}^2=\frac{KQq}{r^{\prime }}--------------\left(2\right)$
Divide equation $1$ with equation $2$-
$4r^{\prime }=r$
$r^{\prime }=\frac{r}{4}$