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So if I have two parallel metal sheets separated by 0.1 meters with air as a dielectric with a breakdown strength of 3MV/m and I'm applying a voltage pulse that goes from 0 to 100kV in few microseconds (so we shouldn't have an arc between the two plates as the electric field is below the breakdown of air) Now assume that one of the plates got a rough surface hence a lot of sharp spikes, so there is an amplification of the electric field by a factor of 10 or even 100x that causes a breakdown to occur between the two plates through the air. How would you find how much force was exerted on the 2nd plate due to the huge electric field? Basically trying to find out if the surface of the 2nd plate would bend.

BTW I don't care about exact numbers, just looking for the general way to think about this problem. I know for a parallel plate capacitor with uniform electric field the force is F = (Permativity * Voltage ^2 * surface area)/(2*distance^2) But this isn't uniform electric field as the spikes on the surface will have much higher fields.

Also is it fair to say that when the electric field got amplified by 10x, the potential difference between this spike and the other ground plane (2nd plate) was also amplified by 10x but it wasn't a sustainable potential difference as the arc/spark connected the two metals together and got rid of this potential difference?

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In the limit that edge effects don't matter and that the surface roughness is small compared to the plate separation, the net force on the plates is the same as if there was no surface roughness. Think about it from an energy perspective: If you move the rough-surfaced plate closer to the other one, all the charge on the plate has moved in the electric potential of the other plate by an amount that is totally independent of the surface roughness. Thus the energy change is the same as if there were no surface roughness, ergo the force (which is the spatial gradient of the energy) is the same in the two cases.

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  • $\begingroup$ In an actual experiment performed, I got a spark between the two surfaces because of a spike/sharp edge on the surface. For the spark to occur, the breakdown strength of air must be exceeded. However, the voltage applied itself isn't enough to cause the breakdown, so it's mainly due to the effect of the amplified electric field from the sharp edge. Would you say that the amplified electric field caused a high enough potential difference for the spark to occur? 3MV/m breakdown strength of air and 0.1-meter gap distance means the potential difference exceeded 0.3MV >> 100kv charging voltage. $\endgroup$
    – DC Med
    Commented Dec 1, 2018 at 8:14

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