A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$ . It approaches $Q$ upto the closest distance $r$ and then returns. If $q$ was given a speed $2v$ , the closest distance of approach would be:
(A) $r$
(B) $2r$
(C) $r/2$
(D) $r/4$

Hint - To calculate $q$ we're given a speed $2v$ , the distance of approach, the formula is used as mentioned below. We will solve it by taking different cases. The first case is when a charged particle is launched towards another charged particle and the second case is when it returns.
Formula used
Electrostatic potential energy,
$P.E. = \dfrac{{kQq}}{r}$
where, $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$
The kinetic energy of a particle,
$K.E. = \dfrac{1}{2}m{v^2}$

Complete Step-by-step solution
For the first case,
Change in potential energy at the closest distance $r$ is,
$P.E = \dfrac{{kQq}}{r}$
As we know that the kinetic energy of the charged particle $q$is,
$K.E. = \dfrac{{m{v^2}}}{2}$
where $m$is the mass of the charged particle and its velocity is $v$ .
By conservation of Energy,
Change in potential energy =change in kinetic energy
On putting the derived values in the above equation we get,
$\dfrac{{kQq}}{r} = \dfrac{{m{v^2}}}{2}$
$ \Rightarrow r = \dfrac{{2kQq}}{{m{v^2}}}$ ............. $\left( 1 \right)$
In the second case:
When charged particle returns with speed $2v$ , let the closest distance be ${r_1}$
Similar to the above case.
Change in potential energy at the closest distance ${r_1}$
$P.E = \dfrac{{kQq}}{{{r_1}}}$
And the kinetic energy of charged particle $q$when it returns is,
$K.E. = \dfrac{{m{{\left( {2v} \right)}^2}}}{2}$
According to the conservation of energy,
Change in potential energy = change in kinetic energy
On substituting the derived values we get,
$\dfrac{{kQq}}{{{r_1}}} = \dfrac{{m{{\left( {2v} \right)}^2}}}{2}$
$ \Rightarrow {r_1} = \dfrac{{2kQq}}{{4m{v^2}}}$ ..................$\left( 2 \right)$
From equation $\left( 1 \right)$,
${r_1} = \dfrac{1}{4}\left( {\dfrac{{2kQq}}{{m{v^2}}}} \right)$
$ \Rightarrow {r_1} = \dfrac{r}{4}$

Hence, the correct answer is option (D) $r/4$ .

Additional Information
Potential energy is the energy stored by an object because of its position compared to other objects, stresses within itself, its electric charge, or other traits. Electric potential energy, or electrostatic potential energy, is a potential energy (measured in joules) that outcomes from conservative Coulomb forces and are interrelated with the configuration of a particular set of point charges within a defined system.

Note When we say, the distance of closest approach we refer to the distance between their centers. The nature of the charge is not mentioned in the question. If the charge is positive then it will move in the same direction as the applied electric field. But, if the charge is negative, then it will be in the opposite direction to the applied electric field.