# Deflection of an Electron Beam by an Electric Field

## Homework Statement

Electrons which have been accelerated from rest through a potential of 500 V pass between two parallel plates that are 6.0 cm long and 2.0 cm apart, with a potential of 150 V between them. The screen is 20 cm beyond the end of the charged plates. What is the size of the deflection as seen on the screen?

## The Attempt at a Solution

I have calculated that the deflection of the electrons while passing through the region between the plates is 0.014 m. This is correct, same answer as the answer key.
To calculate the size of the deflection as seen on the screen, I would say that
D/0.014 = (20+6) / 6
as there are similar triangles
D/0.014 = (20+3) / 3
Why does it only take half of the length of the charged plate to do this proportionality between similar triangles and not the whole side?
I know that the deflection is not significant until it reaches maybe half the plate. Is this the reason why it only takes half?
Any help really, really, greatly appreciated.

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You should make a drawing of the trajectory.
Neglecting fringe fields, when the electron is leaving the plate region it has a certain position and a certain velocity.
You have to calculate a straight line trajectory.
The final displacement is not proportional to the displacement obtained between the plates.
The final displacement is the sum of:

- the displacement obtained between the plates
- the displacement after the plate determined by the velocity when leaving the plate gap (two components)

Trajectory calculations always start from intitial conditions: position and initial velocity.
Forces modify the velocity.

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I have uploaded a diagram http://img263.imageshack.us/my.php?image=crtsn3.jpg".

Thank you for your help lalbatros. I know your method is easier.

D/0.014 = (20+3) / 3
You can say that the triangle formed underneath the parallel plates (base is y) is similar to the whole big triangle (base is D).
Then D / y = (L+l) / l
I know y = 0.014 m
L = 20 cm
l = 6 cm
So D / 0.014 = (20+6) / 6
D/0.014 = (20+3) / 3
Why does it only take half of the length of the parallel plates when they do this proportionality?

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No husky88, this is not true:

You can say that the triangle formed underneath the parallel plates (base is y) is similar to the whole big triangle (base is D).
The trajectory underneath the plates is NOT a straight line, it is a parabola.
Therefore, there is no way to solve that using similar triangles, because underneath the plates the trajectory is NOT straigth.

Here is a picture of a part of the trajectory:

http://www.geocities.com/l.albatros/pictures/beam.JPG

It is clear from this picture that you cannot calculate the final displacement from similar triangles.
I did this picture by calculating the trajactory one point every 0.5 ns approximatively, using the familiar formulas for accelerated motion.

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Thank you for the reply again.

The Physics course I am taking does not involve calculus, so I guess the key just assumed it a straight line. So the key and me are both wrong.
I understand your suggested method and that's the way I am going to solve it.

Not much calculus involved.
Only the familiar laws for uniformly accelerated motion needed in the plate region.
Outside the plate region, velocity is constant -> uniform motion.

This problems seems a little more complex to me. The 500V potential (presumably, in relation to the source?) on the plate is responsible for the horizontal component of motion of the electron, correct? But, from the information given in the diagram what is the distance that Ve is accelerating the electrons??? How did you calculate a horizontal velocity for an electron entering the parallel plate region?

Also, I suppose it is assumed that the Ve plate produces no electric field in the direction of the parallel plate region; just between the source and the plate, right? Otherwise, that Ve plate will serve to decelerate electrons in the horizontal direction after they pass beyond the plate. Nevertheless, assuming that's not the case, my main concern still stands: if we can't calculate the electron's velocity as it exits the Ve plate and enters the parallel plate region, how are we to determine the deflection within that region, let alone the deflection at the screen? Or, am I missing something and just being stupid (which happens more frequently than I'd care to discuss) jf

From the diagram provided by husky88 I guessed that the electrons enter the plate region with their full speed obtained by an acceleration by 500 V. The energy of the electrons is then simply mv²/2 = eV, from which the velocity is easy to calculate. For the fun, you will check that this velocity is about 4% of the speed of light. Therefore small relativistic corrections could be considered to as a second exercice! Of course these relativistic effect would be much smaller than all the parasitic fields effects. It is clear that the geometry of the electric field will be more complicated than assumed.

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Right. It hit me while I was in the shower. The distance from plate to source doesn't matter. The kinetic energy at the plate must be electron charge times the potential difference between the plate and the source.

Thank you so much lalbatros (and jackiefrost) for such a good explanation.

The problem is not so complex. The electric field only acts where there is voltage applied. And I guess it is assumed that all the energy it receives goes into kinetic energy.

I know it should be a parabola, because the force is perpendicular to the speed. I just had no idea how the answer key gets the formula
D/0.014 = (20+3) / 3
(but I guess it is wrong (?), although the numerical result is alike with the real one)

So does this look ok? (I am not referring to numerical values because the calculations are long, but I was wondering if the method seems ok.:) )
I found out the deflection while passing through the plates (0.014 m). Then the vertical speed after it leaves the plates (6 000 000 m/s) and the time it takes to get to the screen from the plates (1.5 * 10^-8 s).
The deflection after the plates is the vertical speed times the time = 0.09 m.
0.09 + 0.014 = 0.10 m

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Here are three points on the trajectory I calculated:

t = 0.0 ns, x = 0.00 m, y = 0.000 m
t = 4.5 ns, x = 0.06 m , y = 0.014 m
t = 19.6 ns, x = 0.26 m, y = 0.104 m

I think you obtained the same results, right?

Happily yes, thank you. Then I think you could calculate other point of the trajectory as well and make a graphic.