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Electrical breakdown of air occurs when the electric field exerted by a charged object exceeds the electrical breakdown limit of air which is $3.10^6 V/m$. Since even 1 Coulomb of charge can exert an electric field whose magnitude can be as high as $10^{10} V/m$, this amount of charge can ionize the air surrounding it.

Moreover, 1 Ampere is defined as 1 Coulomb of charge flowing in a circuit every second. Even simple DC circuits in small electronic devices have several amperes of current flowing in them. Therefore, why that amount of charge flowing in a circuit doesn't ionize the air surrounding it; however, when we place 1 Coulomb of static charge on a conductor, it generates an electric field capable of ionizing the air?

I searched about it everywhere but there seems to be little research made on the complete electrical breakdown of air.

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  • $\begingroup$ Yes, but one Coulomb would correspond to nearly $10^{19}$ free electrons, which is getting close to a mole of electrons... Moreover, the magnitude of the electric field outside such a distribution would decrease as roughly $r^{-2}$. So only the neutral particles close by (i.e., within ~100 m in your example) would be subject to the breakdown. None of this matters though since electrical circuits almost never carry more than a few kV... $\endgroup$ Jan 22, 2016 at 14:30
  • $\begingroup$ @honeste_vivere Wouldn't objects within 100 m getting subjected to this breakdown cause disasters such as arc discharges? Moreover, doesn't the magnitude of the electric field and the potential carried in a circuit correlate with each other? If the magnitude exceeds the breakdown limit, shouldn't the potential exceed the breakdown voltage? $\endgroup$
    – user57144
    Jan 22, 2016 at 14:40
  • $\begingroup$ Yes, but the last point in my comment is the relevant part. Namely that you are not going to have, under most conditions, 1 Coulomb of free charges just sitting in our atmosphere. Thunderstorms routinely exceed the breakdown limit and so do spark plugs in cars, but on much different scales... Regardless, see garyp's answer... $\endgroup$ Jan 22, 2016 at 14:46
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    $\begingroup$ The "complete electrical breakdown of air" is called plasma physics and the world has probably spent tens of billions of dollars over the years on it over the past half century. As to why you never see electric breakdown in low voltage circuits: you never have a lot of free charge around: a typical arrangement of wires has a capacitance of 100pF, charged to e.g. 12V it carries a charge of no more than 1.2nC. $\endgroup$
    – CuriousOne
    Jan 22, 2016 at 15:56

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Wires in a circuit are electrically neutral. They have as many positive charges as negative charges, so the net charge is zero.

There is a spatial distribution of charges within a wire. The charge is not uniform across the cross section. But that variation in charge density is fairly small. Given the overall neutrality, you'd only have to move a short distance from the wire before the field is zero.

Furthermore, if ionization were to occur, where would the liberated electrons go? Unless there was a conductor nearby at higher potential, they would snap right back onto their ions due to Coulomb attraction.

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  • $\begingroup$ So if an object is not electrically neutral and stores an amount of 1 Coulomb of charge on it, it can ionize the air surrounding it since its field will not be zero a short distance away from the object. Right? $\endgroup$
    – user57144
    Jan 22, 2016 at 14:46
  • $\begingroup$ If the object is small enough so that its surface charge density is large, then 1 C can ionize air. But what happens next depends on factors that you have left unspecified. If there is no object "nearby" at higher potential, then the electrons will snap back onto the object. This is not electrical breakdown. If there is a "nearby" object, then breakdown, and a large current (spark, lightning) can occur. (I put nearby in quotes because the details of electrical breakdown are complicated enough that there is no simple answer to what nearby is.) $\endgroup$
    – garyp
    Jan 22, 2016 at 15:01
  • $\begingroup$ I suspect some of the confusion is between electromotive force in the circuit, measured in volts, and electric field strength, measured in $\frac{V}{m}$ . You can't get an arc until the field strength is high. $\endgroup$ Jan 22, 2016 at 15:32

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