Fields in Physics assignment question about electron beams in a cathoe ray tube

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Homework Help Overview

The discussion revolves around a physics assignment involving a cathode ray tube, specifically focusing on calculating the speed of electrons based on provided voltage and distance data. Participants are exploring the relationship between kinetic energy and speed in the context of electron motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the speed of an electron using two different methods but arrives at different results. They express uncertainty about the correctness of their calculations and the implications of negative values in their equations.

Discussion Status

Some participants provide feedback on the calculations, noting discrepancies and questioning the use of certain variables. There is an ongoing exploration of the calculations and the underlying physics concepts, particularly regarding the interpretation of energy and charge.

Contextual Notes

Participants are discussing the implications of the negative charge of the electron and how it affects the calculations. There is also mention of potential confusion regarding the variables used in the equations.

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Homework Statement




Ok so we have been given a diagram of a cathoe ray tube, with relevant data on it. The second question asks to "calculate the speed of the electron, using data from the first question". I need to make sure the answer is correct as it leads into many different questions.

I am unable to attach the picture, but here are all details:
Distance between metal plates: 5cm
Length of metal plates: 10cm
Charge on upper plate: 500V
No charge on lower plate
Charge on anode: 1000V


Homework Equations



The first question was to calculate the kinetic energy gained by an electron traveling from cathode to anode. What i calculated was:

E_k=qV=C×V
-1.6×〖10〗^(-19)×1000
E_k=-1.6×〖10〗^(-16) N


The Attempt at a Solution



I used two different methods to work this out, but got two different answers:


F=E_k

E_k=1/2 mv^2

v=√((2×E_k)/m)

v=√((2×-1.6×〖10〗^(-16))/(9.11×〖10〗^(-31) ))

v=-1.87 〖ms〗^(-1)



OR


s=((2×C×V))/m

s=((2×-1.6×〖10〗^(-16)×500))/(9.11×〖10〗^(-31) )

s= -1.76 〖ms〗^(-48)
 
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Welcome to PF!
The first speed calc is correct except it works out to 1.87 x 10^7 on my calculator.
I don't understand the second - what is C? I don't recognize the formula.
 
Thank you!
I have put it into my calculator again a few times but still getting the same answer: 1.87x10^-24 ... I put it all in separately, so work out what 2x 1.6x10^-16 then divide it...then work out the square root. would doing this alter my answer?

In the second equation the C represents the charge on the electron. I found it on a website so I'm not totally sure it is suitable for this problem.

Also what do i do about the fact that the electron has a negative charge on it? because I'm unable to square root a negative number.
 
There is no negative number - the energy is positive. Technically, the polarity is oriented so the V is negative, too, so the qV is positive.

v=√((2×-1.6×10^(-16))/(9.11×10^(-31) ))
Say you separate the numbers and powers: 2x1.6/9.11 = 0.35
The power of 10 is -16-(-31) = 15. So under the square root, 0.35 x 10^15
Square root of that is 1.87 x 10^7.
 

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