Question

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

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

Answer 1

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

As initially $q$ is at infinite distance from $Q$, thus $P.E_i=0$

For case $1:$ the closest distance given is $r$ and velocity of $q$ is $v$ initially.

Applying conservation of energy,
$P.E_i+K.E_i=P.E_f+K.E_f...\left(1\right)$ ​

At the point of closest approach, the $K.E$ of the particle is momentarily zero.

Substituting the values in $\left(1\right)$, we get

$0+\frac{1}{2}m{v}^2=\frac{KQq}{r}+0$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{KQq}{r}...\left(2\right)$

For case $2:$ Let the closest distance be $r^{\prime }$ and velocity of $q$ is $2v$ initially.

Applying conservation of energy,
$P.E_i+K.E_i=P.E_f+K.E_f...\left(3\right)$ ​

At the point of closest approach, the $K.E$ of the particle is momentarily zero.

Substituting the values in $\left(3\right)$, we get

$0+\frac{1}{2}m(2v{)}^2=\frac{KQq}{r^{\prime }}+0$

$\Rightarrow \frac{1}{2}m(2v{)}^2=\frac{KQq}{r^{\prime }}...(4)$

Using eqaution $\left(2\right)$ and $\left(4\right)$ we get,

$r^{\prime }=\frac{r}{4}$

Hence, option (d) is the correct answer.