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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
$\frac{{E{q^2}m}}{{2t}}$
$\frac{{2{E^2}{t^2}}}{{mq}}$
$\frac{{{E^2}{q^2}{t^2}}}{{2m}}$
$\frac{{Eqm}}{t}$
Solution
When charge $\mathrm{q}$ is released in uniform electric
field $\mathrm{E}$ then its acceleration $\mathrm{a}=\frac{\mathrm{q} \mathrm{E}}{\mathrm{m}}$ ( is constant )
So its motion will be uniformly accelerated motion and its velocity after time $\mathrm{t}$ is given by
$\mathrm{v}=\mathrm{at}=\frac{\mathrm{qE}}{\mathrm{m}} \mathrm{t}$
$\Rightarrow \mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2}\left(\frac{\mathrm{qE}}{\mathrm{m}} \mathrm{t}\right)^{2}=\frac{\mathrm{q}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{2 \mathrm{m}}$