Why does the breakdown voltage increase when the separation between the electrodes is decreased beyond a particular value?

Question

When I was learning about Paschen's law from Wikipedia, I came across the following statement:

Paschen studied the breakdown voltage of various gases between parallel metal plates as the gas pressure and gap distance were varied:

• With a constant gap length, the voltage necessary to arc across the gap decreased as the pressure was reduced and then increased gradually, exceeding its original value.

• With a constant pressure, the voltage needed to cause an arc reduced as the gap size was reduced but only to a point. As the gap was reduced further, the voltage required to cause an arc began to rise and again exceeded its original value.

The first point was easy to understand intuitively. The conduction of gas in the discharge tube increases at first with decrease in pressure as we'd reduce the rate of recombination of ions. Beyond a certain limit, it increases due to lack of charge carriers.

However, I'm unable to understand the second point in a similar way. When the pressure is fixed, it is said that initially the voltage required for dielectric breakdown of air decreases with decrease in the distance between the two electrodes. This is understandable. But the next statement seems counter-intuitive to me:

As the gap was reduced further, the voltage required to cause an arc began to rise and again exceeded its original value.

I expect that the potential difference for observing an electric arc must decrease with decrease in the gap size. But I don't understand why it must increase beyond a particular separation between the electrodes. It would be helpful if you could explain: Why does the breakdown voltage increase when the separation between the electrodes is decreased beyond a particular value?

Answer 1

Breakdown depends on an 'Avalanche' taking place: positive ions formed by electrons being knocked out of molecules, and negative ions formed by the knocked-out electrons sticking to neutral molecules, are accelerated by the electric field. If they gain enough KE before colliding with other molecules they will cause more ions to be formed. These will also be accelerated, cause more ionisation and so on.

If the gas density is too high the ions won't gain enough KE before colliding to cause more ionisation unless the pd is raised.

If we keep the density constant, a smaller gap will need a smaller pd to get an electric field strong enough for ions to pick up enough KE between collisions in order to ionise. But if the gap is smaller than a few mean free paths, the voltage will have to be increased in order to make collisions more energetic and cause more ionisation per collision, to compensate for the short and less effective avalanche.

Answer 2

the initiation of breakdown depends on the availability of ultraviolet photons. by placing the electrodes ever closer together, it gets hard to get any UV photons into the gap to trigger the discharge, and the breakdown voltage rises.