Electrons in a cathode ray tube

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Homework Help Overview

The discussion revolves around a problem involving a cathode ray tube, specifically focusing on the calculation of the number of electrons arriving at the screen per second and the force exerted on the screen by these electrons upon collision. The subject area includes concepts of electric current, kinetic energy, and momentum in the context of electron behavior in electric fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the number of electrons using current density and the charge of an electron, while expressing confusion about determining the velocity of the electrons upon impact. Some participants suggest considering the work-energy principle to find the velocity and relate it to the force calculation.

Discussion Status

Participants are exploring different methods to derive the velocity of electrons and the resulting force. Some guidance has been offered regarding the relationship between kinetic energy and voltage, indicating a potential direction for resolving the calculations. However, there is no explicit consensus on the correctness of the approaches discussed.

Contextual Notes

There is mention of confusion regarding the unit of electron volts (eV) and its application in the calculations, as well as the need for clarity on the distance traveled by electrons, which remains unspecified.

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 traveled but that's unkown too.
 
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Robin04 said:
what's the velocity of the electrons when they arrive
Think about work.
 
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haruspex said:
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?
 
Robin04 said:
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.
 
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ZapperZ said:
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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