Question

A particle of mass $m$ and charge $q$ is released from rest in an electric field $E$. Then the K.E. after time $t$ will be

• A. $\frac{2E^2{t}^2}{mq}$
• B. $\frac{E^2{q}^2{t}^2}{2m}$
• C. $\frac{E{q}^{2}m}{2t^2}$
• D. $\frac{Eqm}{2t}$

Answer 1

The correct option is B $\frac{E^2{q}^2{t}^2}{2m}$

Assuming $F$ force is acting on charge $q$ due to presence of electric field $E.$

$\therefore F=qE$

$\Rightarrow ma=qE$

$\Rightarrow a=\frac{qE}{m}$

Let $v$ be the velocity of charge after $t\text{seconds.}$

Using 1st equation of motion

$v=u+at=at(\text{given},u=0)$

$\Rightarrow v=\frac{qE}{m}t$

So, kinetic energy of charge will be

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m(\frac{qE}{m}t{)}^2$

$\Rightarrow KE=\frac{q^2{E}^2{t}^2}{2m}$

Hence, option $\left(b\right)$ is correct.