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(C) When a charge $q$ is released in a uniform electric field $E$,the force acting on it is $F = qE.$Since $F = ma$,the acceleration of the particle i

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

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(C) When a charge $q$ is released in a uniform electric field $E$,the force acting on it is $F = qE.$Since $F = ma$,the acceleration of the particle is $a = \frac{qE}{m}.$Since the particle starts from rest $(u = 0)$,its velocity $v$ after time $t$ is given by $v = u + at = 0 + \left(\frac{qE}{m}\right)t = \frac{qEt}{m}.$The kinetic energy $(KE)$ of the particle is given by $KE = \frac{1}{2}mv^2.$Substituting the value of $v$,we get $KE = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{1}{2}m\left(\frac{q^2E^2t^2}{m^2}\right) = \frac{q^2E^2t^2}{2m}.$

Element	$M^{3+}/M$	$M^{+}/M$
$Al$	$-1.66$	$+0.55$
$Tl$	$+1.26$	$-0.34$

$A$ 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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