Question

An electron falls through a distance of $1.5cm$ in a uniform electric field of magnitude $2\times {10}^{4}N/C.$ Now the direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. Compute the time of fall in each case, neglecting gravity.

Answer 1

ATQ,

$S=1.5cm=1.5\times {10}^{-2}m$

$u=0m/s$

$m_e$ : mass of electron $=9.10\times {10}^{-31}Kg$

$m_p$ : mass of proton $=1.672\times {10}^{-27}Kg$

from coloumb's force, $\overrightarrow{F}=q\overrightarrow{E}$

and given $\overrightarrow{E}=2\times {10}^{4}N/c$

force proton ${\overrightarrow{F}}_{(+)}=1.6\times {10}^{-19}\times 2\times {10}^4=3.2\times {10}^{-19}N$

force on electron, ${\overrightarrow{F}}_{(-)}=-1.6\times {10}^{-19}\times 2\times {10}^4=-3.2\times {10}^{-15}N$

From Newton's Second law, $\overrightarrow{F}=m\overrightarrow{a}$

$\therefore$ for proton, $a_{(+)}=\frac{{\overrightarrow{F}}_{(+)}}{m_p}=1.913\times {10}^{12}m/s^2$

for electron, $a_{(-)}=\frac{{\overrightarrow{F}}_{(-)}}{m_e}=-0.35\times {10}^{16}m/s^2$

from equ of motion, $S=ut+\frac{1}{2}{at}^2⟶\left(1\right)$

ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.

$\therefore$ Substituting values in equ (1) for proton,

$1.5\times {10}^{-2}=\frac{1}{2}\times 1.913\times {10}^{12}\times t^2$

$1.568\times {10}^{-14}=t^2$

$\Rightarrow t=1.252\times {10}^{-7}seconds$

Substituting values in equ(1) for electron,

$1.5\times {10}^{-2}=\frac{1}{2}\times 3.5\times {10}^{15}\times t^2$

$0.857\times {10}^{-17}=t^2$

$\Rightarrow t=2.927\times {10}^{-19}seconds$