An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric

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

Accepted Answer

Using $h = \frac{1}{2}at^2$ where $a = \frac{eE}{m}$, we get $h = \frac{1}{2} \frac{eE}{m} t^2$, so $t = \sqrt{\frac{2hm}{eE}}$. Since $t \propto \sqrt{m}$ and the mass of an electron is smaller than the mass of a proton, the electron takes less time to fall.

Step-by-Step Solution

Practice All general Questions