Question

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

Answer 1

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