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Electrical breakdown occurs due to the fact that the magnitude of electric field of a charged object is above the electrical breakdown limit of the insulator that surrounds the charged object. Moreover, it can be said that the magnitude of electric field of a charged object at a point is inversely proportional to the distance between that point and the charged object. Since the magnitude of electric field of a point charge, $q$ at a point whose distance from the point charge is $d$ is; $$ |E|=\frac {q}{4\pi \epsilon_0 d} $$ And to find the magnitude of the electric field of a charged object, we can add electric fields created by each charge on the object, which can be done by integration according to the shape, volume etc. of that object. But that won't change the basic principle that magnitude of electric field is inversely proportional to the distance. Thus, we can say that magnitude of the electric field created by a charged object increases as the distance decreases.

So then if we decrease the distance between the point and the charged object to an infinitesimal value the magnitude of electric field will eventually surpass the value of electrical breakdown limit independent to the amount of charge on the object. Here is where I got confused. Does every statically charged object's electric field surpass the electrical breakdown limit of the insuator surrounding it and ionize the insulator? Because even the amount of charge is really small, at a microlength its electric field's magnitude would pass the insulator's electrical breakdown limit.

P.S: I know about corona discharge but as long as I know corona discharge occurs at sharp sides of a charged object. However, according to my question ionization of insulator would be everywhere near the charged object.

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  • $\begingroup$ If your charge is surrounded by a conductor, no breakdown field strength matters: It will already conduct the charge away. You meant a surrounding isolator. $\endgroup$
    – user73762
    Apr 5, 2015 at 19:36
  • $\begingroup$ @pyramids Yes. Sorry, I meant an insulator. $\endgroup$
    – Starior
    Apr 5, 2015 at 19:39
  • $\begingroup$ I would say the electric field is proportional to the inverse square of the distance? $\endgroup$
    – akhmeteli
    Apr 5, 2015 at 20:11
  • $\begingroup$ @akhmeteli Yes, that's true for a point charge but I didn't want to make a general statement like that because the power of the distance can change for differently shaped object due to the integration for some conditions. $\endgroup$
    – Starior
    Apr 5, 2015 at 20:28

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The wikipedia article on Paschen's law should answer your question quite well. http://en.wikipedia.org/wiki/Paschen%27s_law

Basically the breakdown field strength increases at smaller distances. That's why, when you "glue" two pieces of plastic foil together with static charge, you can easily get around 30 MV/m or more in the microscopic air gap that's in between the two foils. Which makes the foils stick together with a much greater force than would otherwise be possible.

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  • $\begingroup$ So the minimum breakdown voltage goes to infinity as the distance goes to zero. Thus, ionization of an insulator doesn't occur everywhere near a charged object. Is this right? And does Paschen's law work for other forms of matter like solid other than gas? I mean does the minimum breakdown voltage increase relative to distance if the surrounding insulator is a solid? $\endgroup$
    – Starior
    Apr 6, 2015 at 16:11
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You neither have arbitrarily small distances or arbitrarily point-like charges: Usually, even "localized" charges are distributed over an entire ion, and any other molecule cannot come arbitrarily close (without becoming absorbed and hence part of the surface, acting as a buffer to other surrounding material).

What matters in practice is not if an individual charge gets removed, but if the field is strong enough to cause dielectric breakdown: Once an atom or molecule of your isolator gets ionized, can it gain enough energy to ionize one of its neighbours simply by being accelerated in the field? If so, a large current can flow and remove your entire charge distribution. Otherwise, only individual charges will constitute a small leakage current that might not do much soon to the total charge of a macroscopically sized charged object.

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  • $\begingroup$ I am trying to understand it. So the ionization due to the charged object occurs at microdistances but is not important when we think about the entire charged object because this ionization effect is negligible and doesn't affect anything. Am I correct? $\endgroup$
    – Starior
    Apr 5, 2015 at 20:21
  • $\begingroup$ The ionization is important if it leads to a noticeable reduction in the total charge of your object. For charged everyday objects, that usually and most obviously occurs when dielectric breakdown (runaway ionization) occurs. If anything below that threshold matters depends on your specific situation. If (too) many atoms that got close to your charge would get ionized, it would matter. But that actually does take a field that is not easy to create by charging some everyday object. You just can't (easily) go to subatomic distances with atoms, or charges significantly beyond ionization energies. $\endgroup$
    – user73762
    Apr 5, 2015 at 20:33
  • $\begingroup$ "You just can't (easily) go to subatomic distances with atoms, or charges significantly beyond ionization energies." Can you please explain this statement more? So as far as I understand ionization occurs even for objects with small amount of charge but it doesn't matter since it occurs at microdistances(several molecules surrounding the object for example) $\endgroup$
    – Starior
    Apr 5, 2015 at 20:46
  • $\begingroup$ The point is that I don't think your understanding describes any situation I can readily visualize at the moment. My argument is that if you had constant ionization, you would have a significant leakage current associated with it (ionized particles would be pushed away by your charge's field) and you soon would lose even a macroscopic charge. The problem is that most isolating gas molecules do not easily get ionized, and if a simple atomic distance to a single charge would suffice more than very rarely, you would already be well above the bulk dielectric breakdown field strength in the gas. $\endgroup$
    – user73762
    Apr 5, 2015 at 20:52
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    $\begingroup$ Field ionization is an electric field strong enough to ionize an isolated atom or molecule. Breakdown occurs if an ion creates more new ions (before it recombines to become neutral again) as it collides with other atoms or molecules with enough energy from acceleration in the field to ionize them, too. $\endgroup$
    – user73762
    Apr 5, 2015 at 21:14

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