Question

An electron moves a distance of $6cm$ when accelerated from rest by an electric field of strength $2\times {10}^4{NC}^{-1}$. Calculate the time of travel. (mass of electron $=9.1\times {10}^{-31}kg$).

Answer 1

$F=qE$

$\Rightarrow ma=qE$

$\Rightarrow a=\frac{qE}{m}$

$\Rightarrow a=\frac{1.6\times {10}^{-19}\times 2\times {10}^{-4}}{9.1\times {10}^{-31}}$

$\Rightarrow a=\frac{32\times {10}^7}{9.1}m/s^2$

Now, using second equation of motion:
$s=ut+\frac{a{t}^2}{2}$

$6\times {10}^{-2}=0t+\frac{32\times {10}^7{t}^2}{2\times 9.1}$

$\Rightarrow t^2=\frac{2\times 6\times 9.1\times {10}^{-9}}{32}$

$\Rightarrow t^2=11.375\times {10}^{-10}$

$\Rightarrow t=3.372\times {10}^{-5}s$