Q. A charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E.$ Neglecting the effect of gravity, the kinetic energy of the charged particle after $'t'$ second is
A) $\frac{Eq^{2}m}{2t^{2}}$
B) $\frac{2E^{2}t^{2}}{mq}$
C) $\frac{E^{2}q^{2}t^{2}}{2m}$
D) $\frac{Eqm}{t}$
Explanation
(c) When charge $q$ is released in uniform electric field $E$ then its acceleration $a = \frac{qE}{m}$ (is constant)
So its motion will be uniformly accelerated motion and its velocity after time $t$ is given by $v = at = \frac{qE}{m}t$
$\Rightarrow KE = \frac{1}{2}mv^{2} = \frac{1}{2}\left( \frac{qE}{m}t \right)^{2} = \frac{q^{2}E^{2}t^{2}}{2m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.