An electron falls through a distance of $1.5 \; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \; N C^{-1}$ [Figure $(a)$]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)$]. Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

(N/A) In Figure $(a)$,the field is upward,so the negatively charged electron experiences a downward force of magnitude $eE$,where $E$ is the magnitude of the electric field. The acceleration of the electron is $a_{e} = eE / m_{e}$,where $m_{e}$ is the mass of the electron.
Starting from rest,the time required by the electron to fall through a distance $h$ is given by $t_{e} = \sqrt{\frac{2h}{a_{e}}} = \sqrt{\frac{2hm_{e}}{eE}}$.
For $e = 1.6 \times 10^{-19} \; C$,$m_{e} = 9.11 \times 10^{-31} \; kg$,$E = 2.0 \times 10^{4} \; N C^{-1}$,and $h = 1.5 \times 10^{-2} \; m$,we get $t_{e} \approx 2.9 \times 10^{-9} \; s$.
In Figure $(b)$,the field is downward,and the positively charged proton experiences a downward force of magnitude $eE$. The acceleration of the proton is $a_{p} = eE / m_{p}$,where $m_{p} = 1.67 \times 10^{-27} \; kg$ is the mass of the proton.
The time of fall for the proton is $t_{p} = \sqrt{\frac{2h}{a_{p}}} = \sqrt{\frac{2hm_{p}}{eE}} \approx 1.3 \times 10^{-7} \; s$.
Thus,the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of 'free fall under gravity' where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity. The acceleration of the proton is $a_{p} = \frac{eE}{m_{p}} = \frac{(1.6 \times 10^{-19} \; C) \times (2.0 \times 10^{4} \; N C^{-1})}{1.67 \times 10^{-27} \; kg} \approx 1.9 \times 10^{12} \; m s^{-2}$,which is enormous compared to $g = 9.8 \; m s^{-2}$. Thus,the effect of gravity can be ignored.