Question

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

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

Answer 1

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

Initial velocity of the particle$=u=0$
let,at $t$ velocity of the particle$=v$ and acceleration$=a$.
using formula, $v=u+at$, we get , $v=at=\frac{F}{m}t$
In electrostatic, $F=qE$ so,$a=\frac{qE}{m}t$
The kinetic energy of the particle, $KE=\frac{1}{2}m{v}^2=\frac{1}{2}m(\frac{qE}{m}t{)}^2=\frac{q^2{E}^2{t}^2}{2m}$