What happens to electrons in an open circuit?

In the Physics classes, the professor did an experiment using de Van de Graaff generator, by which he held a neon tube radially outward to the V d Graaff dome, and the neon lit up. I understood that this was because there was a potential difference between the 2 ends when placed radially which caused the electrons to flow, therefore lighting up the neon. My question is, since it was not a closed circuit, couldn't the electrons "run out"? I mean, that while under the electrical influence, there wasn't a way for them to return to the other end of the tube, and so, wouldn't eventually all of them be transferred to the other side?

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Hi andreas.vitikan - we prefer to have only one question per post here. I've edited out everything but the first question you asked; feel free to post your other two questions as separate posts. – David Z Aug 27 '12 at 21:26
I'm sorry I didn't know that, and thanks! – andreas.vitikan Aug 28 '12 at 3:46
Are you sure it was Van de Graaff generator (en.wikipedia.org/wiki/Van_de_Graaff_generator) and not Tesla coil or something? (en.wikipedia.org/wiki/Tesla_coil) – Yrogirg Sep 3 '12 at 12:14

1 Answer

1. The circuit likely was closed by his body, or by a grounding wire he was holding. At least that's one way the demonstration has been done. I assume the other end was in contact with the generator. Another way is to suspend the tube so it is not in electrical contact with anything, and swing it around, so that you observe a momentary discharge when the tube is orthogonal to the field. In this case the discharge shuts itself off because there is no closed circuit path as you observe.

2. Yes.

3. Think of V as like the height of a hill for a positive charge. E wants to push the charge down the hill, in the direction of lower V. Hence the minus sign.

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Since we like to have only one question per post, I've edited the original post to remove all but the first question. You may want to edit your answer accordingly. – David Z Aug 27 '12 at 21:26