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I come to this question after a few thoughts on capacitors.

A capacitor essentially transfers (and stores) energy using an electric field through an insulator. The plates of a capacitor are not connected, in the sense that the electricity is not passed through using a true conductive path.

Consider a simple, two-plate capacitor. This capacitor has a charge, a capacitance, and a distance separating the plates. If we increase the charge, the capacitance increases. If we decrease the distance, capacitance increases. Therefor, if we increase the distance, we will lower the capacitance, but if we comparatively increase the charge, the capacitance will remain the same. The insulator would simply be the air.

So if we have two highly-charged plates, separated by a distance, shouldn't it be possible, in theory, to have a single-wire connection rather than a whole circuit?

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    \$\begingroup\$ Capacitance is not normally dependent on voltage - voltage is usually independent of capacitance. Maybe you should rephrase your question with a better understanding of what a capacitor is. As it stands the basic errors in the assumptions make it pointless answering. Consider also EM waves - you can transfer power with zero wires and ditto magnetism. Having 1 wire is barely no more attractive than having two wires. Having zero wires on the other hand seems far more attractive. \$\endgroup\$ – Andy aka Jun 3 '13 at 9:34
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    \$\begingroup\$ You could just fly a kite in a storm. \$\endgroup\$ – RedGrittyBrick Jun 3 '13 at 9:57
  • \$\begingroup\$ Current needs a closed loop, but the loop does not have to be all wire. In your case, your capacitor is a component that closes the loop. Note that you can provide power using zero wires, for instance using a (varying) (electro-) magnetic field. \$\endgroup\$ – Wouter van Ooijen Jun 3 '13 at 11:16
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    \$\begingroup\$ @WoutervanOoijen, to have a current doesn't require a closed loop. A circuit requires a closed loop but not current. Consider, for example, the charge "sloshing" back and forth in an antenna. \$\endgroup\$ – Alfred Centauri Jun 3 '13 at 11:22
  • \$\begingroup\$ If you increase charge, charge increases. Capacitance is not dependent on charge: A discharged capacitor has the same capacitance as the same capacitor when charged. Please review your concepts of capacitance, this question still stumbles on invalid concepts. \$\endgroup\$ – Anindo Ghosh Jun 3 '13 at 11:26
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Your capacitor misconceptions aside, it is possible to deliver power by effectively one wire. Driving an antenna is an example of one wire power delivery. Power is delivered to the antenna via the transmitter and the antenna radiates the energy as an electromagnetic wave.

enter image description here

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  • \$\begingroup\$ Well, I was thinking strictly in terms of an electric field, rather than EM waves, as EM waves like to 'fly off', never to be recovered. ;) So, apart from my capacitor misconceptions, should it be possible for what I described to happen? \$\endgroup\$ – SparkGap Jun 3 '13 at 11:48
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    \$\begingroup\$ @SparkGap, your description is inchoate. It isn't clear what you're describing so, until you sharpen your concepts and description, I don't have an answer. \$\endgroup\$ – Alfred Centauri Jun 3 '13 at 12:05
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It is possible using the "single wire transmission line" concept. If I understand it correctly, it only works with high-frequency AC.

In one half of the phase, the electrons travel through the load (doing useful work there) and are then stored in a capacitor that is charged this way. This does not use a normal capacitor, but the "self- and parasitic capacitance" of a large conductive object – I think it's equivalent to connecting only one wire to a normal capacitor. In the second half of the phase, the current flows in reverse through the same wire.

And there is one major way to improve this system:

"Although the self-capacitance of even large objects is rather small in ordinary terms, as Tesla himself appreciated it is possible to resonate that capacitance using a sufficiently large inductor (depending on the frequency used), in which case the large reactance of that capacitance is cancelled out. This allows a large current to flow (and a large power to be supplied to the load) without requiring an extremely high voltage source." (source)

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Although single wire power transferring is possible, it may not be applicable in this case. Current needs a closed path to travel. If you use a single wire, then the electrons that come to one end from the other end need to get back to the first end for the current to be continuously flowing. Since there is no closed loop there will be no current flowing.

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  • \$\begingroup\$ P=V*I, so if there is no current, there is no power. Could you explain why you say that power transfer over one wire is possible, but that there won't flow any current? \$\endgroup\$ – user17592 Jun 4 '13 at 17:18
  • \$\begingroup\$ @CamilStaps: The number of electrons that leave an object cannot differ much from the number that return. It is possible to transfer power by having electrons leave when it is easy for them to do so, and return when it is more difficult, or vice versa (the direction of power transfer will depend upon which side is causing the difficulty to increase or decrease, and which direction is "easy"). \$\endgroup\$ – supercat Jun 4 '13 at 17:46
  • \$\begingroup\$ @supercat but then there is a current flowing, not? You're talking about moving electrons, don't you always have a current then? \$\endgroup\$ – user17592 Jun 4 '13 at 17:47
  • \$\begingroup\$ @CamilStaps: There is an AC current, but no DC current. \$\endgroup\$ – supercat Jun 4 '13 at 18:01
  • \$\begingroup\$ @supercat and that makes the statement in the answer incorrect, what I wanted to point out. \$\endgroup\$ – user17592 Jun 4 '13 at 18:02

protected by W5VO♦ Jun 3 '13 at 13:17

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