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Can the air disruptive field (about 3*10^6 V/m) be calculated?

  1. Mar 9, 2013 #1
    My book says that the air disruptive field (about 3*10^6 V/m). Is this value empirical or can be deduced from the physical properties of O2 and N2 molecules in the atmosphere?

    I tried to calculate this disruptive field assuming it to be the field necessary to pull an eletron out of an nitrogen atom. The periodic tables says that the nitrogen atom radius is 60 pm and it's effective atomic number is 3.9 (I assume that the atractive field made by the nucleus is equavalent to that produced by a charge of 3.9 times the carge of the eletron) so the eletric field produced by the nucleus is about 1.6*10^12 N/C. This is way above the disruptive field! How shall I do my math?
  2. jcsd
  3. Mar 9, 2013 #2


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    Staff: Mentor

    With your approach, you would calculate the value where no air is possible at all.

    There are always some free electrons/ions in the air. They get accelerated along the electric field. If they meet an ion/electron, they can recombine. If they hit an atom with sufficient energy, they can ionize this. So the basic question is: What happens more frequently? If recombination dominates, everything is fine. If ionization dominates, you get more and more electrons and ions and the air begins to conduct.
    With the mean free path length, it should be possible to get some approximation.
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