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I have a question about electron guns. I have read a lot, and all designs use high NEGATIVE voltage ($-1000$ V for example). Is possible to use high POSITIVE voltage like this drawing? Or do you think it will not work this way?

Considering that all the chamber walls are at ground voltage ($0$ V) in both cases.

enter image description here

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The two diagrams describe the same apparatus. Only the voltage difference matters.

From a practical perspective, one wants the outside surfaces of a device to be near the ground potential. Exposed high-potential surfaces risk dangerous electrical arcs. In many countries, electrical wall sockets are polarized so that, even without a dedicated ground pin, the outer surfaces of the device are connected to the low-voltage "return" line of the power cable.

I was involved years ago with the design of a neutron decay experiment where either the proton or electron detector electronics (I have forgotten which) needed to be at 30 kV relative to ground. Safely accessing the detector side of the experiment was going to take some major safety engineering — but it was doable with existing industrial equipment, rather than our team needing to invent such an engineering system from scratch.

For an electron gun, the beam target needs to be at the same potential as the anode — otherwise the electrons will accelerate or decelerate en route.

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  • $\begingroup$ Welcome to the club. I had the exact same problem once for a beamline diagnostic system. It sounded like a hard problem at first, but we solved it with some commercial isolated power supplies. 30kV is not an outrageously high voltage. A quarter inch of insulation will suffice if I remember correctly. $\endgroup$
    – FlatterMann
    Commented Aug 11 at 11:03
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Might be wrong, but I thought voltage (or potential difference) refers to the ratio between the voltages at two points; giving me the impression that these two diagrams are exactly the same?

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    $\begingroup$ It does not make sense to say that voltage is the ratio between voltages. $\endgroup$
    – G. Smith
    Commented May 26, 2019 at 3:38
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    $\begingroup$ I explained it badly but the gist is that it is to do with the difference between two points $\endgroup$
    – Randomer
    Commented May 26, 2019 at 11:00

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