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(C) When a charge $q$ is released in a uniform electric field $E$,the force acting on it is $F = qE$.According to Newton's second law,the acceleration

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$A$ 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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