Electric field

Question

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

Moderate

Solution

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

As $F=ma$

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

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.

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