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Question

(A) The force acting on the particle is $F = qE$.According to Newton's second law,the acceleration of the particle is $a = \frac{F}{m} = \frac{qE}{m}$

Vedclass · Accepted Answer

$qEy$

Vedclass · Answer

(A) The force acting on the particle is $F = qE$.According to Newton's second law,the acceleration of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.Using the kinematic equation $v^2 = u^2 + 2ay$,where initial velocity $u = 0$:$v^2 = 0 + 2 \left(\frac{qE}{m}ight)y = \frac{2qEy}{m}$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.Substituting the value of $v^2$:$K = \frac{1}{2} 	imes m 	imes \left(\frac{2qEy}{m}ight) = qEy$.Therefore,the correct option is $A$.

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

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