In a given CRT, electrons are accelerated horizontally by 7.4 kV . They then pass through a uniform electric field E for a distance of 2.6 cm which deflects them upward so they reach the top of the screen 20 cm away, 13 cm above the center.Estimate value of E
First finding the velocity of the electron from the kinetic and potential energy
v =Sqrt(2qV/m) =Sqrt(2*1.6*10-19*7.4*103/(9.11*10-31) =5.098*107m/s
now the time by the electron to cross the electric field is
t =distance /velocity =2.6*10-2m/5.098*107m/s=0.5100*10-9s =0.510ns
Now the vertical deflection of the electron in the electic field is
y =(1/2)at2
Now the force acting on the electon in an elctric field is given by
F =qE ===>ma =qE
then a =qE/m =
Now from the above equations we get electric field is E =2ym/qt2 =2*0.25*9.11*10-31/(1.6*10-19)(0.5100*10-9s)2=10.945*106V/m
In a given CRT, electrons are accelerated horizontally by 7.4 kV . They then pass through...
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