How Does Thomson's Experiment Calculate Electron Deflection?

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In summary, you set up an apparatus to reproduce Thomson's experiment and observed the point where the electron beam strikes the screen when there is no electric field or when a uniform electric field of magnitude is established between the plates. You found that the distance, delta y, between the two points that you observe is (sr of qEr/m)*((d+L)/v)-1/2g((d+L)/v)^2.
  • #1
xinlan
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Homework Statement


Imagine that you set up an apparatus to reproduce Thomson's experiment, as shown in the figure. In an highly evacuated glass tube, a beam of electrons, each moving with speed vo, 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 is established between the plates. Call the distance between these two points delta y .

What is the distance, delta y, 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 ,vo ,L , and Eo .


Homework Equations





The Attempt at a Solution



I really have no idea how to begin with it.
I assume I need to find the distance when the beam is not deflected and then the distance after the beam is deflected. then substract them.

Please help me..
 

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  • #2
Hint: Which way does the force act? How long does it act for? (How long is the electron in the field?)
 
  • #3
the force acts on the electron.
F = q(VxB).
about how long is it? does it depend on the length of the plate?
 
  • #4
xinlan said:
the force acts on the electron.
F = q(VxB).
It's an electric field, not a magnetic field. :wink: What direction does the force act?
 
  • #5
ups..
F=qE
about the force, do you mean it is perpendicular to the electric field?
 
  • #6
The force is parallel to the field, not perpendicular.
 
  • #7
can you give me another hint? I'm really clueless..
 
  • #8
Which way does the field point? (Look at the diagram!) Which way does the force act?
 
  • #9
the E points downward, so the F will point downward too..
 
  • #10
Careful. What's the sign of the electron's charge?
 
  • #11
the sign is negative charge.
 
  • #12
Right. So the force points opposite to the field. (That's why the diagram shows the electron being deflected up.)
 
  • #13
okay. the foce is going to point upward since the electron has a negative charge.
I look up my textbook, but there is not any equation for finding a distance.
 
  • #14
You have the force. Figure out the acceleration.

Hint to find the time that the electron is in the field: Does the field affect the horizontal motion of the electron?

You want to find the direction that the electron is moving in once it gets out of that field.
 
  • #15
F= ma
a=F/m
= qE/m
v^2/r = qE/m
V=sr of qEr/m

then use kinematic equation = delta y = vot-1/2gt^2
delta y = (sr of qEr/m)t-1/2gt^2

t = (d+L)/v
delta Y = (sr of qEr/m)*((d+L)/v)-1/2g((d+L)/v)^2

is that correct?
 
  • #16
xinlan said:
F= ma
a=F/m
= qE/m
OK.
v^2/r = qE/m
Why introduce centripetal acceleration? The acceleration is upward, not radial.

Once the electron leaves the field, what's its acceleration?

What is the transit time for the electron through the field?
 
  • #17
I am really confused.
to find the time is the distance / the velocity.
I still don't get it
 
  • #18
That's right. The time = distance/velocity, for constant speed motion. You have the horizontal distance through the field and the horizontal speed. So what's the time?

Then you can find the deflection while in the field and the final velocity as it leaves.
 
  • #19
the horizontal distance is (d+L) right?
then the time is (d+L)/v
how to find the deflection? is it using delta y = Vot+1/2at^2 ?
 
  • #20
xinlan said:
the horizontal distance is (d+L) right?
then the time is (d+L)/v
That's the total horizontal distance and time. Better break this problem into two parts: the accelerated part within the plates, and the non-accelerated part after it leaves the plates.

How long is it between the plates? What's the vertical deflection when it leaves the plates? What direction is it moving in? (You'll need the vertical speed it gains while in the field.)
 
  • #21
I have a similar problem where I am told to find the total deflection
I have derived the following equations

delta Y = Y(in capacitor) + Y (in the distance after the capicator to the screen)
= (.5*(eE/me)(Lof cap/V0)2)+((eE/me)*(Lof cap/V0)(Dto screen/V0))

ok now assuming that, that is correct...
I am only given V0, Lof capacitor, Seperation between plates, Distance to screen and the potential difference between the plates.

How would I go about finding the Efield between the plates. I haven't been told the area either just the length.

Thank You
 
  • #22
avenkat0 said:
I have a similar problem where I am told to find the total deflection
I have derived the following equations

delta Y = Y(in capacitor) + Y (in the distance after the capicator to the screen)
= (.5*(eE/me)(Lof cap/V0)2)+((eE/me)*(Lof cap/V0)(Dto screen/V0))
Looks OK.

ok now assuming that, that is correct...
I am only given V0, Lof capacitor, Seperation between plates, Distance to screen and the potential difference between the plates.

How would I go about finding the Efield between the plates.
All you need to find the field is the potential difference and the separation between the plates. How are those quantities related?
 

Related to How Does Thomson's Experiment Calculate Electron Deflection?

What is a Thompson experiment?

A Thompson experiment is a scientific experiment that was conducted by British physicist J. J. Thompson in the late 19th century. It involved using a cathode ray tube to study the properties of electrons and led to the discovery of the electron as a fundamental particle.

What was the purpose of the Thompson experiment?

The purpose of the Thompson experiment was to investigate the nature of cathode rays and determine whether they were made up of particles or waves. This ultimately led to the discovery of the electron and revolutionized our understanding of atomic structure.

How did Thompson conduct the experiment?

Thompson used a cathode ray tube, which is a sealed glass tube with metal electrodes at either end. The tube was filled with a low-pressure gas and an electric current was passed through it. Thompson then used magnets to deflect the rays and measure their properties.

What were the results of the Thompson experiment?

The results of the Thompson experiment showed that cathode rays were made up of negatively charged particles, which were later identified as electrons. This disproved the prevailing theory at the time that cathode rays were a form of light or waves.

How did the Thompson experiment impact science?

The Thompson experiment had a significant impact on science as it provided evidence for the existence of electrons and helped pave the way for the development of atomic and quantum theory. It also led to the development of new technologies such as cathode ray tubes, which were used in early television sets and computer monitors.

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