Solution:

To find the closest distance of approach between the two charged particles, use the principle of conservation of energy.

Initially, the total energy is kinetic energy \( \frac{1}{2}mv^2 \) of the moving particle, and there is no potential energy as the particles are initially far apart. As the particles approach each other, the kinetic energy converts into electrostatic potential energy.

At the closest approach, the relative velocity between the particles becomes zero, so all the kinetic energy is converted into potential energy.

The potential energy between two charges \( Q \) separated by a distance \( r \) is:

\[
U = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r}
\]

At the closest approach, the total energy is:

\[
\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r}
\]

Solving for \( r \):

\[
r = \frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{mv^2}
\]

Thus, the closest distance of approach is:

\[
{\frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{mv^2}}
\]