Let us say that air breaks down at a field strength of $E_b$. For a capacitor to arc, does there need to be a path between the two conductors such that for every point on that path the electric field is greater then $E_b$, or can we still have arcing even if the field is not greater than $E_b$ all the way between the plates?
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asked Aug 10, 2015 at 14:29
Quantum spaghettificationQuantum spaghettification
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$\begingroup$ It's not very clear what are you asking, can you please draw some diagrams to illustrate your question? $\endgroup$– ChaositAug 10, 2015 at 16:05
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$\begingroup$ I fixed the punctuation here. Please take the time to use proper English punctuation etc. on Physics Stack Exchange. It makes the questions easier to understand and answer, and it saves other people the time of fixing it up for you :-) $\endgroup$– DanielSankAug 10, 2015 at 18:31
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1 Answer
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You refer to plates, so I assume we are talking about a parallel plate capacitor. In this case (to a good approximation) the field is the same everywhere inside the capacitor so if the field is bigger than $E_b$ somewhere inside the capacitor then it is bigger than $E_b$ everywhere.
But if the capacitor is not a parallel plate capacitor all bets are off. If one surface of the capacitor has a sharp point then the field will be stronger there. If the breakdown field is exceeded around the tip but not all the way to the other electrode then this can cause local breakdown at the tip but not elsewhere. Then we get what is called a corona discharge where a small current flows into the ionized ("broken down") air and then out into the un-ionized air which acts as a good resistor. If you increase the field further to the point where there is a path to the other electrode with $E>E_b$ along the whole path then the air will ionize along the whole path, will be a good conductor and a large current (the "spark") will flow.
