Q. A charged particle of mass m and charge q is released from rest in an uniform electric field E neglecting the effect of gravity,the kinetic energy of the charged particle after t second is
A) $\frac{2E^{2}t^{2}}{mq}$
B) $\frac{Eq^{2}m}{2t^{2}}$
C) $\frac{Eqm}{t}$
D) $\frac{E^{2}q^{2}t^{2}}{2m}$
Explanation
(d) Electrostatic force on charged particle
$\ \ \ \ \ \ \ \ \ \ \ \ F = qE$
So, by Newton’s law of motion
$\ \ \ \ \ \ \ \ \ F = ma = qE\ or\ \ \ a = \frac{qE}{m}$
Velocity attained by particle $v = 0 + \frac{qE}{m}\ t$
So, kinetic energy $K = \frac{1}{2}mv^{2}$
$K = \frac{1}{2}\frac{mq^{2}E^{2}}{m^{2}}\ t^{2} = \frac{E^{2}q^{2}t^{2}}{2m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.