The diagram shows a schematic of a TV tube. Electrons are accelerated from the c

Question

The diagram shows a schematic of a TV tube. Electrons are accelerated from the cathode at the rear of the tube (on the left side of the drawing) which is at -10 kV, rightwards through a hole m the anode which is at 0 V, and are then deflected by the plates. The potential difference between the plates is 1500 Y, their separation is 1.2 cm, and the plates are 5 cm long. What is the speed of the electrons as they pass through the anode ? How long do the electrons take to pass between the plates ? What is the field between the plates ? What is the vertical acceleration of the elections between the plates ?

Explanation / Answer

a)

  Use energy conservation and assume the electron starts at zero kinetic energy (at rest).

qV = (1/2)mv^2

-> v = sqrt(2qV/m)

=sqrt(2*1.6*10-19*10*103/9.31*10-31)

=sqrt(3.437*1015)

=0.586*108m/s

b)

s=0.5at2

t=sqrt(0.05/0.5*214.82*108)

=2.15*10-6seconds

c)

field between plates

E=V/d

=1500/1.2*10-2

=1250v/m

d)

F=qE

ma=qE

a=1.6*10-19*1250/9.31*10-31

=214.82*108m/s2

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