Question

An electron with speed 2.85×10⁷ m/s is traveling parallel to a uniform electric field of magnitude 1.20×10⁴ N/C .

How far will the electron travel before it stops?

How much time will elapse before it returns to its starting point?

Answer 1

a) Let us first find the acceleration of the electron,

Fnet=m*a

qE=m*a

(-1.6*10^-19*1.2*10^4)=(9.1*10^-31)*a                => a= - 2.11*10^15 m/s^2

Now when the electron stops Vf=0m/s

Use equation,

Vf^2=Vi^2+2*a*d

0^2=(2.85*10^7)^2-2*(2.11*10^15)*d                =>   d= 0.1924m = 19.24cm

b) Since for returning trip acceleration is same as above, speed at the starting point will be same as speed at initial point and

Hence here Vi=0m/s , Vf= -2.85*10^7m/s (-ve sign for direction) and a= -2.11*10^15m/s^2

Use equation,

Vf=Vi+a*t

-2.85*10^7=0-(2.11*10^15)*t                    =>   t= 1.35*10^-7 s =

total time = 2*t = 2*135ns= 270ns