Question

1. An electron is traveling horizontally at a speed of 3.75x10⁶ m/s and enters a uniform electric field. a) What is the magnitude and direction of the electric field that uniformly stops the electron at a distance of 28 mm? b) How much time does it take for the electron to stop? Hint: Use the kinematic equations and Newton’s laws.

Answer 1

Given:

$v_i=3.75*10^6\:\rm m/s$

$v_f=0\:\rm m/s$

$d=0.028\;\rm m$

$e=1.6*10^{-19}\:\rm C$

The energy-work theorem says

$$\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=W=eEd$$ $$0-\frac{mv_i^2}{2}=eEd$$

(a) The electric field strength

$$E=\frac{mv_i^2}{2ed}=\frac{9.1*10^{-31}*(3.75*10^6)^2}{2*1.6^{-19}*0.028}=1428\:\rm V/m$$

Direction is the same as velocity of electron.

(b) The time of motion

$$t=\frac{2d}{v_i+v_f}=\frac{2*0.028}{3.75*10^6+0}=15*10^{-9}\:\rm s=15\: ns$$