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

Solution :
Electric force on charged particle is given by $$F = qE$$
Kinetic energy attained by particle = work done
$$\eqalign{ & = {\text{force}} \times {\text{ displacement}} \cr & = qE \times y \cr} $$
Alternative
Force on charged particle in a uniform electric field is $$F = ma = Eq$$
or $$a = \frac{{Eq}}{m}\,......\left( {\text{i}} \right)$$
From the equation of motion, we have
$$\eqalign{ & {v^2} = {u^2} + 2ay \cr & = 0 + 2 \times \frac{{Eq}}{m} \times y\,\,\left[ {u = 0} \right] \cr & = \frac{{2Eqy}}{m} \cr} $$
Now, kinetic energy of the particle
$$\eqalign{ & K = \frac{1}{2}m{v^2} \cr & = \frac{m}{2} \times \frac{{2Eqy}}{m} \cr & = qEy \cr} $$