I was playing with some High-Voltage the other day, when a question popped into my head. Can you calculate length of an electrical arc? It probably would be proportional to :-
1. Voltage of the source
2. Spark gap
3. Relative Humidity (in air only)
4. Resistivity of Medium
5. Shape of the Electrodes
Can anyone come up with a dimensional formula from this...?

  • 1
    $\begingroup$ You look up the breakdown voltage of whatever material (like air) you want to arc thru. Then multiply that by the distance between electrodes and that's the voltage you need. $\endgroup$ Sep 7, 2014 at 12:20

1 Answer 1


Paschen already did. See Paschen's Law

The underlying physical intuition here has to do with the mean free path of electrons, electron-impact ionization and coulomb collisions.

Electric fields will separate electrons from atoms in a gas, for example, and cause them to accelerate. Those electrons will (1) lose energy (coulomb collisions) on other electrons as they accelerate and (2) may knock out other electrons (impact ionization) if they are sufficiently energetic. These are two competing terms.

If for example, the mean free path for electron-electron collisions is small compared to the gap size, and the electric fields are relatively low, then (1) will dominate and an arc is not sustainable.

On the other hand, if the mean free path is on the order or larger than the gap size, and or the electric fields are very large, then (2) will dominate and an arc is sustainable.

If you want a characteristic length, the mean free path is a good rule of thumb. It should be obvious based on the above discussion how the voltage of the source and the gap come into play (voltage / gap = electric field). The resistivity of the medium in the traditional sense (the cold matter resistivity) doesn't matter. Once you have the arc, what matters is the plasma resistivity which depends largely on the mean free path for coulomb collisions. The shape of the electrodes can affect the electric fields, especially near the electrodes.

  • $\begingroup$ Basically this is a link-only answer. Would you consider adding a few more details in your post? Otherwise it is only valuable as a comment rather than an answer. $\endgroup$
    – Bernhard
    Sep 7, 2014 at 12:07
  • $\begingroup$ @Bernhard I added details despite my unwillingness to do so. I don't think that we should be duplicating textbooks or Wikipedia. It's one thing to go research and read something on your own, not understand it, or lack insight, and then come to Physics SE with a reasonable question. That's why this site exists. This was not one of those questions. The answer is self-contained in the link. Either the OP didn't bother searching the web or other resources or couldn't find it in their search for whatever reason. $\endgroup$ Sep 10, 2014 at 17:42
  • $\begingroup$ I see your point, but in that case, don't answer, but leave it as a comment, downvote the question, and flag it for closing. That is the way it should work on SE. $\endgroup$
    – Bernhard
    Sep 10, 2014 at 20:23
  • $\begingroup$ @Bernhard fair enough $\endgroup$ Sep 10, 2014 at 21:12

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