2
$\begingroup$

If I have two spheres of the same size and one sphere has a small amount of charge compared to the other that has a lot more charge, then clearly the sphere with the larger charge has a larger voltage (relative to the ground). My question is do high voltage power lines have a lot more charge that is placed on them? Is that what gives them the high voltage? I think I have a grasp of the step up stations that use transformers to kick up the voltage from power plants.

This question seems almost silly to me but I have been struggling with this for a long time. I’ve done several searches online and I am not able to find answers. If there is a link that someone can provide, I much appreciate it.

$\endgroup$
2
$\begingroup$

For an analogy, you should consider voltage more like a "pressure" than an amount of charge. It just so happens that in your sphere example, with a fixed number of charges (electrons), you get this nice correlation between charge and voltage.

But in a conductor, the pressure can come from far away so to speak - from the step-up transformer in your example.

Put another way, the voltage is defined as the potential to conduct current through a resistance. This property increases when you "push" more in the other end. In the spheres, the push comes from the increased number of charges in a confined space, and in the high-power wires, the push comes from/through the transformer.

The actual reason why a domestic current-distribution network uses high voltages as opposed to low voltages is another question.

$\endgroup$
  • 1
    $\begingroup$ To be fair, if you ignore the OP's commentary and just answer the question title, that may help a lot of lurkers. The answer of course is that power P= V * I but wires dislike carrying a lot of I so bumping up the V delivers more power with less physical wire :-) $\endgroup$ – Carl Witthoft Nov 18 '13 at 18:17
  • $\begingroup$ @Bjorn Wesen: Let me try to be more clearer. A high voltage line means it is measured relative to the ground. If true, can I think of a high voltage line as an "equipotential?" This is tricky because it's ac, not dc, so I really do not know enough at the moment to say so. I follow when you ask to think of voltage more like pressure, but what sets up the "pressure difference" in the wires? Are you telling me it is not charge? $\endgroup$ – Jesus Nov 18 '13 at 22:20
  • $\begingroup$ @Carl Witthoft: I was not asking about why the voltage is high and the current is low in power lines. I feel that I understand that very well. $\endgroup$ – Jesus Nov 18 '13 at 22:24
1
$\begingroup$

They do carry a bit of extra charge, but it's sort of a side effect. Every conductor (such as a high voltage wire) has some capacitance. The capacitance of an object can be defined as the amount of charge added, per unit change in voltage on that object (keeping all other voltages constant). When we energize the high voltage line (say, increasing its voltage from zero to 1 megavolt), therefore, we must add some charge.

In the case of the high voltage line the capacitance is probably something on the order of 10 picofarads per meter (this is an educated guess based on approximate dimensions). This means that, in order to charge up a 100 km length of power line to 1 megavolt, we need to add about 1 coulomb of charge.

$\endgroup$
  • $\begingroup$ Doesn't this extra charge help to create the high voltage of the power line? $\endgroup$ – Jesus Nov 18 '13 at 22:26
  • $\begingroup$ In some sense, yes. These charges are (locally) what its holding the potential energy of the electrons 1 mega-electronvolt lower than in the surroundings. However, the capacitance is quite geometrical in nature and the figure of 10 pF/meter could easily change to 20 pF/meter if there were more grounded objects closeby to the high voltage line. Or, if I somehow made my high voltage line very very thin, the capacitance would be less. $\endgroup$ – Nanite Nov 19 '13 at 8:22
  • 1
    $\begingroup$ To make a hydraulic analogy, where voltage on a wire becomes the pressure in a water-carrying pipe, the capacitance becomes the flexibility of the walls of the pipe, and the tiny bit of extra charge we are talking about becomes the tiny bit of extra water being carried by the pressurized pipe because the pipe has expanded slightly. The extra bit of water is certainly "creating" the high pressure locally inside the pipe because this excess of water strains against the walls of the pipe. However the amount of extra water is a bit incidental since we could choose a more or less rigid pipe. $\endgroup$ – Nanite Nov 19 '13 at 8:27
  • $\begingroup$ these comments are really helpful. Thanks. Please see my comments below. $\endgroup$ – Jesus Nov 19 '13 at 18:04
1
$\begingroup$

The voltage in electric cables has almost nothing to do with the amount of electric charge on or in some portion of the cable.

The voltage is a measure of electric potential.

Charge carriers are present in all conductors, even those that have no voltage across them. Applying a high voltage does not alter the number of charge carriers in any section of cable.

The presence of voltage (i.e. an electric field) is what causes the randomly moving charge carriers to, on average, slowly drift in one direction - but the voltage is not produced by the presence of charge carriers in the conductor.

There are smaller effects but these are distractions from the fundamentals.

Most power lines on land are AC, but most undersea power lines are DC, so AC effects can be disregarded when considering the fundamentals of what is meant by voltage or high voltage and it's relation to charge.

$\endgroup$
  • $\begingroup$ If no charge is added to the high voltage power lines, does the high voltage come from the amplitude of the ac voltage waveform that goes through the wire that shakes the electrons back and forth then? If this is true, then a 500,000 V line would have a peak voltage at the crest/trough locations and zero voltage at the “nodes.” Is this correct thinking? $\endgroup$ – Jesus Nov 19 '13 at 18:05
  • $\begingroup$ @Jesus: The crest/trough and nodes are time-dependent, not position-dependent. Apart from that you are correct. $\endgroup$ – RedGrittyBrick Nov 19 '13 at 18:47
  • $\begingroup$ this was very helpful! $\endgroup$ – Jesus Nov 19 '13 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.