Question

Two identical particles of mass $m$ carry charge $Q$ each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first from a large distance with an initial speed $V$. Find the closest distance of approach.

Answer 1

As there is no external force present hence the system's total energy will remain constant. So, the initial Kinetic Energy will be totally converted to the system's PE when they are closest to each other. Hence from there, we get,

$\left(\frac{1}{4\pi \epsilon }\right)\frac{Q^2}{R}=\left(\frac{1}{2}\right)m{v}^2⟹R=\frac{Q^2}{2\pi \epsilon v^2}$