Question

A point charge $q$ having mass $m$ is projected from a long distance with speed $v$ towards another stationary particle of same mass and charge. The distance of closest approach of the particles is :

• A. $\frac{q^2}{2\pi {\epsilon }_{0}m{v}^2}$
• B. $\frac{2q^2}{\pi {\epsilon }_{0}m{v}^2}$
• C. $\frac{q^2}{\pi {\epsilon }_{0}m{v}^2}$
• D. Zero

Answer 1

The correct option is C $\frac{q^2}{\pi {\epsilon }_{0}m{v}^2}$

Given, for the both point charges,
mass$=m$
charge$=q$

For the first point charge initial velocity,$v_i=v$
For the second point charge initial velocity, $u_i=0$

During minimum separation (closest approach) both particles will move with same velocity. $($Let be $u)$


Applying the energy conservation principle,

$\text{Initial total energy=Final total energy}$

${P.E}_i+{K.E}_i={P.E}_f+{K.E}_f...\left(1\right)$

Here,
Since the particle are at infinite distance so, initial potential energy, $P.E_i=0$

initial kinetic energy, $K.E_i=\frac{1}{2}m{v}^2$

final potential energy when they are at $r_{min}$
$P.E_f=\frac{K{q}^2}{r_{min}}$

final kinetic energy, $K.E_f=\frac{1}{2}m{u}^2+\frac{1}{2}m{u}^2=m{u}^2$

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

$\Rightarrow 0+\frac{1}{2}m{v}^2=\frac{K{q}^2}{r_{min}}+m{u}^2$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{K{q}^2}{r_{min}}+m{u}^2....\left(2\right)$

Applying momentum conservation:

Initial momentum ${\left(P\right)}_i=$ Final momentum ${\left(P\right)}_f$

$\Rightarrow mv=mu+mu$

$\Rightarrow u=\frac{v}{2}....\left(3\right)$

Using $\left(2\right)$ in $\left(3\right)$ we get,

$\frac{1}{2}m{v}^2=\frac{K{q}^2}{r_{min}}+\frac{m{v}^2}{4}$

$\Rightarrow \frac{K{q}^2}{r_{min}}=\frac{m{v}^2}{4}$

$\Rightarrow \frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r_{min}}=\frac{m{v}^2}{4},[K=\frac{1}{4\pi {\epsilon }_0}]$

$\therefore r_{min}=\frac{q^2}{\pi {\epsilon }_{0}m{v}^2}$

Hence option (c) is the correct answer.