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There is a well known experiment to determine the specific charge of an electron like in the following picture

enter image description here

Electrons are emitted from a heating spiral and then accelerated by an acceleration voltage $U_{\mathrm{B}}$. Then the electrons describe a circular path in a magnetic field. For an actual setup see for example here.

In evaluating this experiment one usually needs to express the speed $v$ via the equation

$$ e U_{\mathrm{B}} = \frac{1}{2} m v^2 $$

I.e. the speed is assumed to be determined by the acceleration voltage. Then it is said that the electric field outside of the acceleration capacitor is zero.

However it is an electrostatic field, which must be conservative, i.e. the work done must be independent of the path. This seems to imply that the electron must loose all it's kinetic energy on its circular path until it comes back to the first capacitor plate, i.e. the starting point.

But if this were true, you wouldn't observe a circular path.

How can I resolve this apparent contradiction?

Edit After reading FlatterMann's answer, I have some refinements of the question:

  1. Do I understand it correctly, that the challenge is to make the field configuration such that the field from leaving the gun around the largest part of the circle is almost zero, such that the Integral over $\vec{E}$ up do this point will be also very small. However since the field is conservative the integral would have to gather significant contributions from the little lest of the path. So the field must be relatively strong at the end of the path, just "before hitting" the source again (in practice the electron wouldn't make the last part at all because it is stopped by some metal part).

  2. Suppose one would indeed use just a parallel plate setup as in my picture. This is highly symmetric, so if the field would be strong enough just below the the bottom plate to stop the electron, then it must be equally strong above the upper plate, but this would alter the electron speed after leaving the gun significantly, so the experiment wouldn't work as expected. Thus such a symmetrical setup is practically/technically impossible.

  3. How does the actual electric field look like in a working setup exactly? Do you have a plot?

  4. If this problem is a known technical challenge, there might be some technical papers out there, which address exactly this issue in detail. Do you have some good references?

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  • $\begingroup$ Well, sure, if you let the energetic electron impact the electron gun it will lose all of its energy. So, instead, create the energetic electrons outside of the magnetic field region, and let them circle around inside the magnetic field without hitting the gun... $\endgroup$
    – Jon Custer
    Apr 18, 2023 at 18:54
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    $\begingroup$ "...loose all it's..." $\endgroup$
    – hft
    Apr 18, 2023 at 18:54
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    $\begingroup$ This isn't a full answer to your question, but as a trivial idealized example consider a capacitor made of two spherical shells, with the outside at $V = 0$ and the inside at $V = V_0$. There is zero field outside the capacitor. An electron that starts at the inner shell and exits through the outer shell will pick up energy $q V_0$, and it will never lose any of that energy unless it somehow gets back inside the capacitor. That's an example of perfect shielding. $\endgroup$
    – knzhou
    Apr 10 at 18:19
  • $\begingroup$ I agree that the setup shown in your picture wouldn't work, but presumably it's a schematic standing in for something that does work. A lot of idealized pictures in textbooks are poor like this. $\endgroup$
    – knzhou
    Apr 10 at 18:20

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There is an electrostatic shield around the electron source. You are correct that there is a field that would stop the electrons, but it's contained inside the electron gun assembly, which we can ignore because the electrons never make it back in there.

This website has a diagram of a typical electron gun wiring: https://www.semitracks.com/reference-material/failure-and-yield-analysis/failure-analysis-die-level/scanning-electron-microscope.php. As you can see, the "accelerating electrode" aka "anode" is held on ground potential and it's the focusing electron optics (Wehnelt cylinder) and the filament that are on a negative potential. In practice the "anode" surrounds the entire assembly, which suppresses the stray field on the outside of the gun. Maybe somebody can draw or find a better diagram which is free of copyright concerns that can be included in another answer.

I had to work on a few electron sources in my life and the necessary "wiring" and shielding to make the outside field free is, indeed, a non-trivial technological problem, especially if we want a fairly high energy DC source (e.g. 60-100keV, like in one of my experiments). For acceleration voltages even higher than that, the early builders of accelerators like Van de Graafs etc. had to go through a lot of trouble to make that electrical shielding work up to millions of Volts. Thankfully AC accelerators can avoid this problem altogether.

So, no, it's not a paradox. It's a real technical challenge.

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