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(C) The force acting on the particle in a uniform electric field is given by $F = qE$.According to Newton's second law,the acceleration $a$ of the par

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$qEy$

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(C) The force acting on the particle in a uniform electric field is given by $F = qE$.According to Newton's second law,the acceleration $a$ of the particle is $a = F/m = (qE)/m$.Using the equation of motion $v^2 = u^2 + 2ay$,where initial velocity $u = 0$ and distance $s = y$,we get $v^2 = 0 + 2(qE/m)y = 2(qE/m)y$.The kinetic energy $(KE)$ is given by $KE = (1/2)mv^2$.Substituting the value of $v^2$,we get $KE = (1/2)m[2(qE/m)y] = qEy$.

$A$ particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The $KE$ attained by the particle after moving a distance $y$ is

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