An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

Force experienced by a charged particle in an electric field, $F=qE$

As $F=m a$

$\therefore ma=eE\Rightarrow a=\frac{eE}{m} ..................\left(i\right)$

As electron and proton both fall from same height at rest. Then initial velocity=0

$\text{ From }s=ut+\frac{1}{2}a{t}^2 (\because u=0)$

$\therefore h=\frac{1}{2}a{t}^2\Rightarrow h=\frac{1}{2}\frac{eE}{m}{t}^2 [\text{ Using (i)]}$

$\therefore t=\sqrt{\frac{2hm}{eE}}\Rightarrow t\propto \sqrt{m}$  as ‘ q’ is same for electron and proton.

$\therefore$ Electron has smaller mass so it will take smaller time.