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This question (What is the capacitance between the Earth and Moon?) on EESE makes me wonder: How can a vacuum have a breakdown voltage?

If electrons can find the shortest path through a vacuum, how is that possible? With nothing to carry the charge, I can't understand how the charge can traverse a vacuum.

The value for dielectric strength of a vacuum comes from Wikipedia

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    $\begingroup$ See en.wikipedia.org/wiki/Vacuum_arc $\endgroup$
    – user6972
    Commented May 23, 2014 at 19:44
  • $\begingroup$ @user6972 that should be an answer...a good one too $\endgroup$
    – David Wilkins
    Commented May 24, 2014 at 0:02
  • $\begingroup$ You may find it interesting that even a single very highly charged body emits it charges due to repulsion between the free charge carriers without the availability of any other charged body in perfect vacuum. $\endgroup$
    – Rijul Gupta
    Commented May 24, 2014 at 14:19

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The electrons (torn from one surface (electrode) through, say, electron field emission, just move in vacuum in electric field to the other surface (electrode).

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    $\begingroup$ To what extent is this a property of the vacuum, rather than a property of the work function of the electrode surfaces? $\endgroup$
    – rob ♦
    Commented May 23, 2014 at 19:47
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    $\begingroup$ I'd say the surfaces play the leading role, the vacuum's role being rather passive (the electrons do move in vacuum, and the electric field exists in vacuum). $\endgroup$
    – akhmeteli
    Commented May 23, 2014 at 20:42
  • $\begingroup$ The above is true for an electric field of reasonable strength and for a relatively long vacuum gap, when vacuum polarization is not important. $\endgroup$
    – akhmeteli
    Commented May 23, 2014 at 20:47
  • $\begingroup$ I thought electromagnetic fields might have something to do with it, thanks. $\endgroup$
    – David Wilkins
    Commented May 23, 2014 at 23:58

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