Thomson Experement, Derive deltaY between E=0 and E=E0

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Homework Statement:

Imagine that you set up an apparatus to reproduce Thomson's experiment, as shown in the figure (Figure 1). In an highly evacuated glass tube, a beam of electrons, each moving with speed v0, passes between two parallel plates and strikes a screen at a distance L from the end of the plates.

First, you observe the point where the beam strikes the screen when there is no electric field between the plates. Then, you observe the point where the beam strikes the screen when a uniform electric field of magnitude E0 is established between the plates. Call the distance between these two points Δy.

B) What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.

Relevant Equations:

Fnet = q(E + v x B)
There are more parts to this problem but I can't get to those until I finish this one.

I have attached figure.1 and some my work so far, including my answer which the system rejected. (I had to copy my work from my original sheet as it contained sensitive information I didn't want to upload)

There was another thread from 2013 on here about this problem but like most of the posts I've seen on here, the student never saw it to conclusion. Though, from what I could gather from it, I was moving in the right direction.
 

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  • #2
haruspex
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Homework Statement: Imagine that you set up an apparatus to reproduce Thomson's experiment, as shown in the figure (Figure 1). In an highly evacuated glass tube, a beam of electrons, each moving with speed v0, passes between two parallel plates and strikes a screen at a distance L from the end of the plates.

First, you observe the point where the beam strikes the screen when there is no electric field between the plates. Then, you observe the point where the beam strikes the screen when a uniform electric field of magnitude E0 is established between the plates. Call the distance between these two points Δy.

B) What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Homework Equations: Fnet = q(E + v x B)

There are more parts to this problem but I can't get to those until I finish this one.

I have attached figure.1 and some my work so far, including my answer which the system rejected. (I had to copy my work from my original sheet as it contained sensitive information I didn't want to upload)

There was another thread from 2013 on here about this problem but like most of the posts I've seen on here, the student never saw it to conclusion. Though, from what I could gather from it, I was moving in the right direction.
You found the displacement at the point where the electron emerges from the plates, but it still has distance L to go.
 
  • #3
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You found the displacement at the point where the electron emerges from the plates, but it still has distance L to go.
Yes, then I tried using the angle of the triangle formed by the displacement where the electron emerges from the plates and the length of the plates (d) to calculate of the larger triangle of sides (delta y) and (d+L) but that aparently does not work.

I know that the electron only experiences a force and thus acceleration between the plates so I was trying to take that displacement and use it to find the larger displacement after it leaves the plates. and travels the remaining distance L.

Can you give me something to work with for calculating that?
 
  • #4
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You found the displacement at the point where the electron emerges from the plates, but it still has distance L to go.
some more thoughts....

Can I use my constant acceleration equations after the particle leaves the loop (F=0 so a=0) using my time d/V0 to calculate the vertical velocity of the particle as it leaves the E field and then calculate that as it moves over the remaining distance L over a time of, im thinking (L/V0)?
 
  • #5
Delta2
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I read your work at the attached pdf file and I believe your mistake is that you focus on the wrong angle ##\theta##.
Try to calculate the angle from the velocity triangle, i.e the angle that the velocity vector ##\vec{v}## at the exit point does with the horizontal, cause that is the angle that the electron will move after its exit, since its velocity ##\vec{v}## will remain constant as there are no forces (neglecting gravity and possibly air resistance) acting on the electron after its exit.
 
  • #6
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I read your work at the attached pdf file and I believe your mistake is that you focus on the wrong angle ##\theta##.
Try to calculate the angle from the velocity triangle, i.e the angle that the velocity vector ##\vec{v}## at the exit point does with the horizontal, cause that is the angle that the electron will move after its exit, since its velocity ##\vec{v}## will remain constant as there are no forces (neglecting gravity and possibly air resistance) acting on the electron after its exit.
I had no idea I could do a velocity triangle. I'll try that later after I have the spare time. I did solve the problem by splitting the problem up into two stages. The first stage with a constant, positive acceleration and a second stage with a=0 and V0y = Vy from stage 1. After I get some rest I'll read the guidlines and post my solution if it's allowed. Thank you for your help. I'll give it a try later on.
 
  • #7
Delta2
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I had no idea I could do a velocity triangle. I'll try that later after I have the spare time. I did solve the problem by splitting the problem up into two stages. The first stage with a constant, positive acceleration and a second stage with a=0 and V0y = Vy from stage 1. After I get some rest I'll read the guidlines and post my solution if it's allowed. Thank you for your help. I'll give it a try later on.
You ll have to split the problem in two stages as well with the velocity triangle approach. Though you have not shown us your work I believe that essentially is the same with what you did in solving the problem. The velocity triangle is valid only for the second stage of the problem (F=0,a=0) because in the first stage the velocity vector of the electron is not constant (neither in direction nor in magnitude) .
 
  • #8
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You ll have to split the problem in two stages as well with the velocity triangle approach. Though you have not shown us your work I believe that essentially is the same with what you did in solving the problem. The velocity triangle is valid only for the second stage of the problem (F=0,a=0) because in the first stage the velocity vector of the electron is not constant (neither in direction nor in magnitude) .
I checked the rules and, they mention that I can post my solution so I,LL go ahead and do that. Mods have mercy.
 

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  • #9
PeroK
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I checked the rules and, they mention that I can post my solution so I,LL go ahead and do that. Mods have mercy.
I agree with your answer.
 

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