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With a nine volt battery, touching the two terminals together (or using a faulty terminal) will cause a spark roughly where I would want it to be.

How is this possible? Is it ionizing only a very small portion of air surrounding the wires when this happens and it is just more visible? I believe at an extremely small distance, ~300v is the breakdown point of air (often, for example according to Paschen's law) so I do not understand how the battery can do this.

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    \$\begingroup\$ You may find it useful to look up "intrinsically safe" and "intrinsic safety" as this deals very specifically with the limits of such an ability. \$\endgroup\$ – Russell McMahon Oct 10 '11 at 5:55
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As the contact is being broken, a connection is made through very small pieces of metal (microscopic features), which have enough current through them to vaporize, the ions of which then support a current through the air briefly.

While lower voltages will not, in general, jump a gap that is present before the voltage is applied, interrupting an existing current flow often produces a low-voltage spark or arc. As the contacts are separated, a few small points of contact become the last to separate. The current becomes constricted to these small hot spots, causing them to become incandescent, so that they emit electrons (through thermionic emission). Even a small 9 V battery can spark noticeably by this mechanism in a darkened room. The ionized air and metal vapour (from the contacts) form plasma, which temporarily bridges the widening gap.

Wikipedia: High voltage § Sparks in air

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    \$\begingroup\$ You can get a short spark when making a contact too, because those same features contact before the bulk of the metal. \$\endgroup\$ – Optimal Cynic Oct 10 '11 at 11:11
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    \$\begingroup\$ @OptimalCynic don't forget contact bounce which will lead to a established circuit getting broken leading to a spark \$\endgroup\$ – ratchet freak Oct 10 '11 at 17:48
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    \$\begingroup\$ This forms an amazingly clear understanding, thank you endolith, \$\endgroup\$ – Hobbyist Oct 10 '11 at 20:17
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Back EMF only occurs with an inductive or capacitive circuit,you dont have this with resistive circuit. the spark is because at last instant of contact the metal vapourises as previously described.If the voltage is sufficicent over 20 volts, the spark can become an arc,and can reach a length of several inches,the current still flows,until the separation becomes too great.If the circuit is broken to an inductance,the back emf from the coil will intensify the arc, and will assist to maintain the arc. A flow of electric current is difficult stop,and this is the beuty of DC, (but can be a nuisance) With AC, there is no net flow of current,and this flow stops and starts,so arcing is not an issue with AC,hence switches are primitive.

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    \$\begingroup\$ No, you can't have an arc of several inches with just 20 volts under standard conditions. \$\endgroup\$ – Dmitry Grigoryev Apr 13 '16 at 5:55
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To answer this question, you'll need to know Ohm's Law: V=IR, as well as inductance which "stores" current or rather, resists changes in current.

What this means is that once a wire connection is made across the battery terminals current starts flowing through the wire. The current 'I' is equal to V/R, which is the battery voltage (9V) divided by the the resistance of the wire and battery. Now, remember the inductance of the system is going to try and maintain that current. When you disconnect the wire, even for fractions of sections, the inductance tries to hold 'I' constant. The act of breaking the connection makes 'R' go from very low to very high. Now if 'I' is constant and 'R' approaches infinity, then 'V' must also approaches infinity to balance the V=IR equation. That's how you get the voltage high enough to ionize gas and spark or burn a very small amount of remaining metal contact. Of course the voltage doesn't hit infinity because once it goes high enough to arc then current flows and discharges the inductance.

Earlier in this thread someone mentioned that when the connection is made for the first time only through a few small pieces of metal which causes all the current to flow through and burn it. That's actually incorrect since the few pieces of metal have a very high resistance, which won't allow enough current though anyways. It's only when the connection is broken that the system inductance forces the current higher than the resistance alone would allow.

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  • \$\begingroup\$ This answer is plain wrong. The inductance of a shorted battery is microscopic, and pretty much negated by its (also microscopic) capacitance. \$\endgroup\$ – Dmitry Grigoryev Apr 13 '16 at 5:49

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