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Take a battery and connect a small led bulb across it with the help of two wires. The bulb will glow, but if I cut a small piece of wire from any part of the connecting wires,the circuit will not work which implies that there is no current and hence no flow of electrons. But if you consider the part of the wire which is exposed to air on one end(due to the cut part) and to negative potential,then the electrons should have travelled from the negative end to the almost zero potential end(air gap) but it does not occur. Why?

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    $\begingroup$ The voltage isn't high enough. $\endgroup$
    – Hot Licks
    Commented Feb 15, 2019 at 23:27

3 Answers 3

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But if you consider the part of the wire which is exposed to air on one end(due to the cut part) and to negative potential,then the electrons should have travelled from the negative end to the almost zero potential end(air gap) but it does not occur.

They will travel from the negative potential source to the cut end. But it will only take a few nanoseconds or so for enough electrons to build up at the cut end so that they repel any further electrons from moving there. Because this motion is so brief, we normally just ignore it when talking about how the gap behaves in a low-speed circuit.

The electrons can't travel across the air gap because it takes a substantial energy (called the work function) for an electron to exit the metal material and into the air in the gap. If you were to make the gap small enough and the voltage across the cut ends large enough, you could create an arc that carries current across the gap. This is how the spark plugs in an internal combustion engine work.

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  • $\begingroup$ Good answer. I'd just add that I like to say that, when electrons "first go" to towards the negative pole, the accumulation of elcetrons creates another electric field, which goes towards the neg. pole as well; and that pushes the electrons back towards the "possitive" pole. The motion reaches equilibrium very fast. Both electric fields (potential difference & accumulation of charges) cancel each other quickly. $\endgroup$
    – FGSUZ
    Commented Feb 15, 2019 at 22:55
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The flow of current moves electrical charge. At endpoints of a 'broken circuit', charge would arrive and... sit.

But, like charges repel; very rapidly, the accumulation of electrons creates an electric field around that endpoint that pushes away any newcoming electrons.

As a rule of thumb, this process occurs with such a small amount of charge, and 'electric current' as we employ it is such a large flow, that we can consider the cessation of current to be immediate.

As to why the air gap does not participate in moving charges, that relates to the necessity (in order to create a spark) to make the air molecules into charged ions (so that the motion of those molecules allows current to flow). That's a chemical change in the air molecules, and CAN happen, but only with very high voltages and carefully shaped electrodes (like 40,000 volts applied to a platinum-tipped spark plug), and it wastes a lot of energy (which is why the spark starts that little fire in the engine cylinder).

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One reason is due to air having a very high dielectric breakdown voltage, which is the amount of energy that an electron in the air has while it constitutes a current through air. So if you had a very large voltage close to the breakdown voltage of air, you will see charge flow (in the form of a spark). It dies down after that.

The reason why you don't see spark with normal wires - at relatively small voltages ($\approx$ 12 V) is because of cancelling electric fields. That is, the electrons at the negative pole would accumulate in large numbers when the circuit suddenly becomes "open" - and the incoming charges at the other ends of the wire are slowed down rapidly by the electric field due to those accumalation of charges in the negative pole.

However, in normal life, you can see "spark effects" which I mentioned in the first para in the case of inductors. In case you don't know about them, these devices oppose changes to currents (and forces them on for just a while, despite the accumulation of charges- this still builds up voltages quickly and dies fast enough - during that period, we can see this spark).

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