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

• garthenar
In summary, the student was trying to use a constant acceleration equation to calculate the velocity of an electron after it leaves the plates of a high-voltage vacuum tube.
garthenar
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.

#### Attachments

• Fig1 from Thomeson Experement.PNG
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• thomson exp work so far.pdf
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garthenar said:
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.

haruspex said:
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?

haruspex said:
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, I am thinking (L/V0)?

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.

Delta2 said:
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.

garthenar said:
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) .

Delta2 said:
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.

#### Attachments

• 2019-10-29_084537.pdf
1.4 MB · Views: 294
garthenar said:
I checked the rules and, they mention that I can post my solution so I,LL go ahead and do that. Mods have mercy.

## 1. What is the Thomson Experiment?

The Thomson Experiment, also known as the Cathode Ray Experiment, was conducted by physicist J.J. Thomson in 1897 to study the nature of cathode rays and the properties of electrons.

## 2. What were the key findings of the Thomson Experiment?

The key findings of the Thomson Experiment were that cathode rays are negatively charged particles, now known as electrons, and they have a much smaller mass compared to an atom. This led to the development of the plum pudding model of the atom.

## 3. How did J.J. Thomson derive the deltaY between E=0 and E=E0?

Thomson derived the deltaY by analyzing the path of cathode rays in an electric field with varying strengths. He found that as the electric field increased, the path of the cathode rays curved more, which indicated a change in the velocity of the particles due to the electric field.

## 4. What is the significance of the deltaY between E=0 and E=E0 in the Thomson Experiment?

The deltaY between E=0 and E=E0 is significant because it provided evidence for the existence of electrons and their charge-to-mass ratio. This helped to further develop the understanding of the structure of atoms and the role of electrons in electricity.

## 5. How did the Thomson Experiment contribute to our understanding of atomic structure?

The Thomson Experiment contributed to our understanding of atomic structure by providing evidence for the existence of electrons and their role in the structure of atoms. It also led to the development of the plum pudding model, which was later replaced by the more accurate Rutherford model, and eventually the modern atomic model.

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