In: Physics
Electrons in a cathode ray tube start from rest and are accelerated through a potential difference of 11.6 kV. They are moving in the +x-direction when they enter the space between the plates of a parallel plate capacitor. There is a potential difference of 320 V between the plates. The plates have length 8.93 cm and are separated by 1.10 cm. The electron beam is deflected in the negative y-direction by the electric field between the plates.
where P = 8.93 and Q = 320.
What is the value of Δy, the vertical deflection?
Through what potential difference do the electrons move while between the plates?
What is the kinetic energy of the electrons as they leave the plates?
Initially electron is accelerated by 11. 6 kV
Hence energy gained by electron = 11.6 keV = 11.6
1000
1.6
10-19
kinetic energy of electron = 1.856
10-15 J ....... (1)
velocity of electron in x-direction is obtained from
(1/2)mvx2 = 1.856
10-15
where m = 9.1
10-31 kg , is mass of electron
hence velocity of electron, vx = { 2
1.856
10-15 / ( 9.1
10-31 ) }1/2
velocity of electron, vx = 6.387
108 m/s ...............(2)
Acceleration a experienced by electron in vertical plates is given by ,
a = qE/m
where q is the charge of electron and E is electric field
a = 1.6
10-19
( 320 / 1.1
10-2 ) / [ 9.1
10-31 ] = 5.115
1015 m/s2
Time t to reach the exit of plate = distance / horizontal velocity
Time t to reach the exit of plate = 8.93
10-2 / ( 6.387
108 ) = 1.4
10-10 s
Vertical distance h travelled by electron is given by,
h = (1/2) a t2 = 0.5
5.115
1015
1.4
1.4
10-20 = 5.013
10-5 m
vertical velocity component of electron at exit of plate ,
vy = a
t = 5.115
1015
1.4
10-10 = 7.161
105 m/s .............(3)
Reultant velocity at exit of plate v = { vx2 + vy2 }1/2
if we compare the velocity component s as given in eqn.(2) and eqn.(3),
we see that vy
10-3 vx
Hence at exit of plate, velocity is nearly equal to vx = 6.387
108 m/s
Hence kinetic energy KE of electron at exit of vertical plates is given by eqn.(1)
KE = 1.856
10-15 J