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An electron accelerated through 30kV in a cathode ray oscilloscope, enters a system of deflecting plates. The deflecting field between plates is 24kV/m. the length of deflecting plates is 0.06m and the total deflection produced in the path of the electron on a screen is 9mm. What is the distance of the screen from the near end of the plates?

7 years ago


Answers : (1)



given length of plate = 0.06 m , so field across the plates at end is : 24*0.06 = 1.44kV

so deviation in the electron due to fields in both directions is tan x = 1.44/30

let L be distance from near end of plate to screen

so tan x = 9*10-3 / (x+0.06)

on solving gives : x = 12.75 cms


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