Question

A charged particle of mass $m$ and charge $q$ is released from rest in an electric field of uniform strength $E$. The kinetic energy of the particle after a time $t$ will be

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

Answer 1

Answer: (C)

$\frac{q^2{E}^2{t}^2}{2m}$

Electrostatic force ,$F=qE\Rightarrow m\frac{dv}{dt}=qE$

Integrating, we will get, $v=\frac{qE}{m}t+c$ where $c=$ integraing constant.

at $t=0,v=0$ , so $c=0$

now, $v=\frac{qEt}{m}$

$\therefore K.E=\frac{1}{2}m{v}^2=\frac{q^2{E}^2{t}^2}{2m}$

Ans:(C)