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Suppose I have 2 pointed stainless steel paper pins. I also have a convex lens and super strong sunlight. The pointed head is painted black. Sunlight is focussed on pointed head of one pin. The pin will get heated. So electrons will come out due to photoelectric and thermoionic emmisions. Using this experiment, can I generate enough potential difference for dielectric breakdown of air, assuming the experiment is done indoors and non-windy air? If no, what can I do to cause dielectric breakdown?

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    $\begingroup$ Why do you think that emitting a few electrons will lead to dielectric breakdown? $\endgroup$ – Jon Custer Jan 7 at 14:35
  • $\begingroup$ Even if you had used the best electropositive metal in place of stainless steel, you need to create at least 3 kilovolt/mm potential difference. Stopping potential over typical metals is about $ 1V $. And with two small paper pins. I don't think there is any chance. $\endgroup$ – KV18 Jan 7 at 14:42
  • $\begingroup$ I don't think you'll get many photoelectrons from stainless steel using visible light. $\endgroup$ – PM 2Ring Jan 7 at 14:47
  • $\begingroup$ Use some photo-elements and a voltage multiplier. $\endgroup$ – Vladimir Kalitvianski Jan 7 at 14:47
  • $\begingroup$ Playing with insulators of different sorts is usually easy. Never got electric breakdown opening a door or greeting a friend? $\endgroup$ – Alchimista Jan 8 at 13:08
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If you have two pins and you focus sunlight onto one of the pins, photoelectric effect might occur but that is just not enough. Have you ever thought about a lightening? So now you should realize the dielectric strength of air. A very high electric field of 3,000,000V/m is required for dielectric breakdown of air. Which means that you have to create a potential difference of 3,000,000 V within 1 metre . Connecting the 2 pins to a battery might help but even that is not enough. A device called Van de Graff generator is used to produce such high voltage. So try connecting the pin to this generator . https://en.m.wikipedia.org/wiki/Van_de_Graaff_generator

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Yes, probably you can. Whatever the heat source is, if one pin is too hot and the other is too cold, the termoelectic emission will charge the other pin either until air breakdown (if the distance is too small) or untill the "saturation" is established (the repulsing potential of the other pin and the positively charged first pin will prevent the other pin charging), or the other pin gets hot enought to produce the same eletron flux towards the first pin.

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