The distance between the plates of condenser is 2 cm.Starting from rest an electron taken 1.5*10^-8 seconds to reach from negative to positive plates.What is the potential difference between the 2 plates?

Let the potential difference between plates is V.

 

Then, electric field, E= V/0.02 volt/m

 

Force on the electron, F = q E

 

The acceleration, a = F / me = q E / me = q V / (0.02 x me)

 

Now using the second equation of motion

 

s = u t  + (1/2 ) a t² 

 

0.02 = 0 + (1/2 ) q V t² / (0.02 x me)   

 

0.02 = q V t²/ (2 x 0.02 x me)

 

V = 0.02 x 2 x 0.02 x me / q t²

 

V = 8 x 10^-4 x 9.1 x 10^-31 / 1.6 x 10^-19 x 1.5 x 1.5 x 10^-16

 

V = 8  x 9.1 / 1.6 x 1.5 x 1.5 

 

V= 20.22 volts

 

 

 

the speed of the electron = 0.02 / 1.5 x 10^-8 = (4/3) x 10⁶ m/s