Electrical breakdown occurs when the potential difference between a conductor and the medium surrounding exceeds the breakdown voltage. In such scenario, the electric field at a boundary point is strong enough for an electron at the boundary to ionize atoms in the medium, which releases more electrons, causing a chain reaction.

My question is, where does the first electron come from? Is it an electron from the surface of the conductor? Do electrons from the conductor continually ionize atoms in the surrounding medium causing the conductor to lose electrons and thus charge until the potential difference between the conductor and the medium does not exceed the breakdown voltage?

  • $\begingroup$ Yes, below the breakdown threshold, there will be a continuous corona discharge. In air, this creates ozone, the smell of photocopiers. $\endgroup$ – Pieter Mar 22 '19 at 21:56
  • $\begingroup$ @Pieter, so just to be clear, the conductor will be losing charge during discharge? $\endgroup$ – Andrew Paul Mar 22 '19 at 22:14
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    $\begingroup$ Well, that depends. When it is connected to a high-voltage supply (as in the case of the corona wires in a copier), there is also a current coming in. $\endgroup$ – Pieter Mar 22 '19 at 22:30
  • $\begingroup$ Corona currents are used to actively control voltage on high voltage accelerators, such as Van de Graaffs. This provides a shorter time constant than changing the current going up to the terminal. $\endgroup$ – Jon Custer Mar 22 '19 at 23:39

Let's say we're talking about something like a spark plug. Two electrodes, separated by a small air gap.

How do you produce a large potential difference between the two electrodes? You force a large number of electrons onto one of them. If you force enough on, you produce a strong enough electric field in the gap to produce breakdown, and some of those electrons start to flow across the gap.

But if you don't continue to supply electrons to the cathode (negative electrode), the breakdown won't be sustained. Likely the potential difference required to sustain breakdown after it's started is less than was required to start it, but even so if you don't keep supplying electrons to the cathode, the breakdown will quickly be quenched.

So the cathode never really loses electrons. You have to push as many in from your generator as are leaving through the breakdown path, or else the breakdown behavior won't continue.

  • $\begingroup$ Yes indeed. Anything you apply voltage to has capacitance. Thus it has excess charge. $\endgroup$ – Jon Custer Mar 22 '19 at 23:37

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