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Q.

A charged particle of mass 'm' and charge 'q' initially at rest is released in an electric field of magnitude E. Its kinetic energy after time 't' will be:

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a

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

b

$\frac{Eqm}{t}$

c

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

d

$\frac{{Eq}^{2} m}{2 t^{2}}$

answer is B.

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Detailed Solution

$KE = \frac{1}{2} {mv}^{2} = \frac{1}{2} m {(at)}^{2} = \frac{1}{2} m {(\frac{qE}{m} t)}^{2} = \frac{q^{2} E^{2} t^{2}}{2 m}$

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