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CBSE Class 12-science Answered

find distance
Asked by vaibhavkalind | 28 Jun, 2011, 01:58: PM
Expert Answer
let the electric field between the two parallel conductors is E.
Then the force on proton = q E, where q is the charge of the proton
So the acceleration in the motion of proton, ap = q E / mp
let the proton meets the electron at x distance away from the positively charged plate in time t. then using the second equation of motion
 
x = 0 + (1/2) at2 = (1/2) (q E / mp )t2 ...............(1)
 
The force on the electron = - q E, q is the charge of electron
 
So the acceleration of the motion of electron, ae = q E / m
 
Again using the second equation of motion,
 
5 - x = 0 + (1/2) at2 = (1/2) (q E / m)t2 ...............(2)
 
On dividing eq. (2) by eq. (1)
 
(5 - x) / x = (mp / me) = 1835
 
5/x = 1 + 1835
 
x = 5 / 1836 = 0.0028 cm
Answered by | 29 Jun, 2011, 05:17: PM
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