An electrostatic CRT has parallel deflection plates 1.5 cm long and 6mm apart. The screen is 55cm from the centre of the deflecting plates. The deflecting and accelerating potentials are 150 V and 2.5 kilovolts respectively. Determine the:

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 Ld = length of deflecting plates, m

 d = distance between deflecting plates,m
m = mass of the electron ,kg
L = distance between screen and the centre of the deflecting plates, m
Ea = accelerating voltage to the plates, volts
Ed = deflecting potential of the plates

(I) Velocity of the electron
Soln
Velocity of electron,  Vox = √{(2eEa)/(m)}
Ea = 2500 v
Ed = 150v
Ld = 1.5cm
L = 55cm
d = 6 mm
m = 9.1094 x 10^-31 kg
e = 1.6022 x 10^ -19 Coulombs

= {√ (2 x 1.6022 x 10^-19 x 2500)/(9.1094x 10^-31)}
= √ 8.794x 10^ 14
= 29655037.64 m/s

(II) Amount of deflection

D = {((L x Ld x Ed)/(2 x d x Ea)) x (m)}
    = {((0.55 x 0.015 x 150)/(2 x 6 x 10^-3 x 2500))}
    = 0.04125 m

(III) Deflection sensitivity
  
          = { ( Ld x L )/(2 x d x Ea)}
          = { ( 1.5 x 10^ -2 x 0.55)/(2 x 6 x 10^ -3 x 2500)}
          = 2.75 x 10^ -4 m/ volt
(IV) Deflection factor
           = 1/deflection sensitivity
            = 1/ ( 2.75 x 10^ -4)
            = 3636.3636 volt/meter

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