Question

An electron traveling with initial velocity 4000 m/s, enters aregion of uniform electric field 1 × 10−8 N/C. After 3 sec, what is thespeed of the electron () in m/s?

Answer 1

Given data

*The given initial velocity is v = 4000 m/s

*The given uniform electric field is E = 1 × 10^-8 N/C

*The given time is t = 3 s

The formula for the acceleration of the electron in the electric field is given as

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

Substitute the known values in the above expression as

$\begin{gathered} a=\frac{(1.6\times10^{-19})(1\times10^{-8})}{(9.11\times10^{-31})} \\ =1.75\times10^3m/s^2 \end{gathered}$

The formula for the speed of the electron in m/s is given as

$v_r=u+at$

Substitute the known values in the above expression as

$\begin{gathered} v_r=(4000)+(1.75\times10^3)(3) \\ =9250\text{ m/s} \\ \approx9300\text{ m/s} \end{gathered}$