1. Electric Charges and Fields
medium

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

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

B

$\frac{{2{E^2}{t^2}}}{{mq}}$

C

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

D

$\frac{{Eqm}}{t}$

Solution

(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$

$==>$ $KE = \frac{1}{2}m{v^2} = \frac{1}{2}m\,{\left( {\frac{{qE}}{m}t} \right)^2} = \frac{{{q^2}{E^2}{t^2}}}{{2m}}$

Standard 12
Physics

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