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Electrons in a cathode ray tube

  • Thread starter Robin04
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Homework Statement


The accelerating voltage in a cathode ray tube is ##1000 V##, the electron current density is ##150 \mu A##.
a) How many electrons arrive at the screen in a second?
b) What's the magnitude of the force exerted on the screen by the electrons if they stop on collision?

(Sorry for my bad English)

Homework Equations




The Attempt at a Solution


a) The definition of current is ##I=\frac{Q}{t}## so if we divide the electron current density by the charge of a single electron and multiply it by 1 second that should give the answer:
##N = \frac{1,5 \cdot 10^{-4} A}{1,6 \cdot 10^{-19} C} = 9,3 \cdot 10^{14} \frac{1}{s}##

b) This is where my confusion starts. The force should be given by ##F = N \frac{\Delta p}{\Delta t}## where ##p = mv## is the momentum of one electron but I don't know what's the velocity of the electrons when they arrive at the screen. I can calculate their acceleration from the acceleration voltage like this ##a = \frac{UQ}{sm}## where s is the distance the electron travelled but that's unkown too.
 

Answers and Replies

  • #2
haruspex
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what's the velocity of the electrons when they arrive
Think about work.
 
  • #3
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Think about work.
Hmm, so let's say the electric power is ##P = UI = N \frac{1}{2} m v^2## because the work done by the electric field in a second gives kinetic energy to ##N## electrons. So ##v = \sqrt{\frac{2UI}{Nm}} = 1,8 \cdot 10^{7} \frac{m}{s}## and then ##F = N \frac{\Delta p}{\Delta t} = N mv = 1,59 \cdot 10^{-8} N##
Am I correct?
 
  • #4
ZapperZ
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Hmm, so let's say the electric power is ##P = UI = N \frac{1}{2} m v^2## because the work done by the electric field in a second gives kinetic energy to ##N## electrons. So ##v = \sqrt{\frac{2UI}{Nm}} = 1,8 \cdot 10^{7} \frac{m}{s}## and then ##F = N \frac{\Delta p}{\Delta t} = N mv = 1,59 \cdot 10^{-8} N##
Am I correct?
This is confusing, and you're working too hard!

If an electron is accelerated by a potential of 1000V, it will gain a kinetic energy of 1000 eV! So this is equal to ½mv2!

Use the charge of an electron "e", multiply that by 1000V and you have the kinetic energy in Joules. It should be trivial for you to find the velocity of each electron hitting the anode by then.

Zz.
 
  • #5
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This is confusing, and you're working too hard!

If an electron is accelerated by a potential of 1000V, it will gain a kinetic energy of 1000 eV! So this is equal to ½mv2!

Use the charge of an electron "e", multiply that by 1000V and you have the kinetic energy in Joules. It should be trivial for you to find the velocity of each electron hitting the anode by then.

Zz.
Thank you very much! I wasn't really familiar with the eV unit, but it seems obvious. :)
 

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