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(C) The force acting on a charged particle in a uniform electric field is given by $F = qE$.According to Newton's second law,the acceleration $a$ of t

Vedclass · Accepted Answer

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

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(C) The force acting on a charged particle in a uniform electric field is given by $F = qE$.According to Newton's second law,the acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.Since the particle is released from rest,its initial velocity $u = 0$. The velocity $v$ after time $t$ is given by the equation $v = u + at = 0 + \frac{qE}{m}t = \frac{qEt}{m}$.The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.Substituting the value of $v$,we get $K = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{1}{2}m \cdot \frac{q^2 E^2 t^2}{m^2} = \frac{E^2 q^2 t^2}{2m}$.

$A$ 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 $t$ seconds is

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