# Electric current transmitted over great distances.

1. Dec 30, 2011

### siddharth5129

Kind of a basic doubt here, but it has been nagging me for a while now. The electric field generated by a power station is a non-conservative one, and it varies as 1/r with respect to the distance of a point from the source.( I am assuming this statement is accurate.) So what I don't get is how electricity is transmitted over such great distances. I mean, they way I see it, a greater distance means a smaller electric field, which would set up a smaller current. So, over thousands of kilometers, the electric field should be of a magnitude too small to set up a significant current. I am definitely missing something crucial here, so any sort of clarification would be greatly appreciated.

2. Dec 31, 2011

### Staff: Mentor

Power stations don't transmit an electric field, they physically push electrons through wires.

3. Dec 31, 2011

### Born2bwire

The power is transmitted as a guided electromagnetic wave.

4. Dec 31, 2011

### fonz

Imagine if the power station was just a huge battery. The electric field exists only inside this battery and it creates a difference in potential at each terminal. This potential difference causes a current to flow through the wires which connect at each terminal of the battery (via your house). The size of this current depends on the load in your house and the potential difference created by the battery.

5. Dec 31, 2011

### Tea Jay

I suppose if the field radiated by a station was sufficient, we could do away with all of those high maintenance power lines and rights of way, etc.

There are some devices out now that DO use the magnetic fields to induce electrical current, without wires, so for example you could re-charge your phone by laying in the device, and the device is not plugged in, but getting its power from local magnetic fields.

Wires of course make everything more efficient for long distance transmissions.

The wires themselves have electric and magnetic fields, I measure the magnetic fields radiating perpendicular to the flow through the wires once in a while for work.

In a three phase system, the fields look a bit like a puffy 3 leaf clover.

6. Dec 31, 2011

### siddharth5129

Thanks for all the replies, but am still a little hazy here.

What do you mean by physically push? How do electrons drift unless there is a local electric field to bias the haphazard motion. Isn't your physical push the same as an electric field?

Could you elaborate? I don't understand this at all.

Ok, lets model the entire scenario with a battery, connecting wires, and a load. The potential difference sets up an electric field. By Ohm's law ,the local current density is directly proportional to the strength of the electric field at that point ( J = σE ). So again, the electric field should diminish over large distances and result in a negligible current density at the load. Another way to look at it would be V = ∫E.dr , and over large distances, this would mean a small electric field and a small current density.

The way I see it, those high maintenance power lines are there for one reason. To provide a high conductance path for electron flow. J = σE. So again, over thousands of kilometers, this should mean a very small current.

7. Dec 31, 2011

### Staff: Mentor

You got it right... High conductance is exactly why those power lines are there.

Remember that the electrons are charged, so as they move down the wire they carry charge with them, and that charge generates an electric field. The strength of that field doesn't depend on the distance from the power station; it depends on the distance from the electron, and that electron can be way closer than the power station.

With an ideal superconducting transmission line (resistance equals zero) you'll end up with exactly the same potential at both ends of the line; the electrons will space themselves out equally across the entire length so every point on the line will be equally close to the same number of electrons.

With a real transmission line (non-zero resistance) the potential difference between the two ends of the line will be given by Ohm's law V=IR; some potential difference is needed to push the electrons against the resistance.

The resistance is proportional to the length of the transmission line, so you're right that if the line is long enough you won't be able to get any current through it. But like you said, this is a "high conductance" path, meaning that the constant of proportionality is very small, so it takes a very long line before you lose a lot of current flow to resistance. And if you increase the diameter of the line, you reduce the resistance per unit length, so proportionately reduce the total resistance of the line - regardless of it's length.

8. Dec 31, 2011

### siddharth5129

But at any given time, the wire is electrically neutral, so isn't it incorrect to consider the field due to the charge on the moving electrons, which would be negated by the fixed positive charges. Alternatively, the E that appears in Ohm's law ( J=σE ) is a function of V (E =V/R), which is a fact that allows us to switch between the two forms of ohms law (J = σE and V = IR ).So shouldn't the 'field' that we are talking about be a property of the voltage generated by the source alone? If that were the case then the field does depend on the distance from the power station. Further, aren't transmission wires relatively thin?

9. Jan 1, 2012

### nsaspook

10. Jan 1, 2012

### Staff: Mentor

If you have a long hose, completely filled with ping-pong balls and you push on one at one end, another pops out the other end. How did you push the one at the far end if your finger is only three inches long? The one you pushed on pushes on the next one, which pushes on the next one, etc.

Someone said electricity is a guided electromagnetic wave. Well -- the ping-pong ball example is a guided mechanical (sound) wave.
No, the physical push is a disturbance in the electric field. Every electron in the wire going from the power plant to your house already has its own electric field, whether the power plant is generating or not. I think that may be what you're missing here. What is transmitted from the power plant isn't the electric field of the coils in the generator, but what they do to the electrons in the wire nearby. Each electron is pushed by the electron behind it, similar to the ping pong balls. The "local electric field" is created by the electrons themselves.
No, the field only has to travel as far as the next ping-pong ball....er, electron in order to make it to your house.

11. Jan 1, 2012

### nsaspook

The only thing I would add to the above is that the 'field' that moves energy from electron to electron is in a particle other than the electron and the field disturbance flows at light speed so it must be massless.

http://en.wikipedia.org/wiki/Massless_particle

12. Jan 1, 2012

### siddharth5129

Thanks. That really cleared it up for me. Just two more things though. You say electricity is a guided electromagnetic wave. So shouldn't it's intensity vary as the inverse square of the distance from the generating source, much like light? And as I understand it, the intensity of an electromagnetic wave is proportional to the electric field strength at that point. Secondly, if we invoke simple circuit theory, shouldn't the length of the transmission wiring result in a resistance too high to allow a significant current to flow? It doesn't add up. I'm still missing something crucial here.

13. Jan 1, 2012

### nsaspook

The electromagnetic wave is confined near the very low impedance path of the wire (a short length of wire in terms of a 60hz wavelength) not in free space, so there is no inverse square loss.

Invoke simple circuit theory:
How much current does it take for 1 million watts if the voltage is 500,000 volts?
What resistance is required for 2 amps of current at 500,000 volts?
90% of the power is delivered to the load at the end of the transmission lines, what is the total wire resistance (to the load and back)?