Question
Thu June 23, 2011 By: Prashant Jain

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?

Expert Reply
Fri June 24, 2011
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 t/ (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
 
 
 
 
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