Question

A charged particle of mass m and charge $q$ is released from rest in an electric field of constant magnitude $E$. The K.E. of the particle after time $t$ is:

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

Answer 1

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

Acceleration of the charged particle is
$a=\frac{F}{m}=\frac{qE}{m}$
Velocity of the charged particle after time $t$ is
$v=u+at=0+\frac{qE}{m}t=\frac{qEt}{m}$
So the K.E. of the charged particle after time t is
$K.E.=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{qEt}{m}\right)}^2=\frac{E^2{q}^2{t}^2}{2m}$