Question

A particle of mass $M$ and charge $q$, initially at rest, is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be :

• A. $\frac{qQ}{4\pi {ϵ}_{0}D}$
• B. $\frac{Q}{4\pi {ϵ}_{0}ED}$
• C. $\frac{qQ}{2\pi {ϵ}_0{D}^2}$
• D. $\frac{Q}{4\pi {ϵ}_{0}E}$

Answer 1

The correct option is B $\frac{Q}{4\pi {ϵ}_{0}ED}$

Acceleration of the particle $a=\frac{F_e}{M}=\frac{qE}{M}$

Initial speed of the particle is zero i.e. $u=0$

Let the speed of the particle after traveling a distance $D$ be $v$.

Using $v^2-u^2=2aD$

$\therefore$ $v^2-0=\frac{2qED}{M}$ $⟹v^2=\frac{2qED}{M}$

Kinetic energy of the particle $K=\frac{1}{2}M{v}^2=qED$

At closest approach, kinetic energy of the particle is equal to its potential energy i.e. $K=U$

Thus $K=\frac{qQ}{4\pi {ϵ}_{o}r}$

Distance of closest approach $r=\frac{qQ}{4\pi {ϵ}_{o}K}$

$\therefore$ $r=\frac{qQ}{4\pi {ϵ}_o\left(qED\right)}=\frac{Q}{4\pi {ϵ}_{o}ED}$