I have always wondered how it travels. One thing I do know is magnets are used but I am not sure. The type of devices I mean are ultility poles that carry electricity from the powerplant to a house usually. So I mostly want to know how it travels on ultility poles.


The simplest explanation is the analogy most textbooks used. The powerplant generates a potential difference between the hot and the neutral line. The electricity, i.e. electrons, wants to travel in such a way that it reduces the potential energy. The force experienced by the electrons is caused by the potential difference. The analogy to this is a ball rolling down an inclined plane.

  • $\begingroup$ Yes. Reduce the potential energy. Thank you. Original answer edited. $\endgroup$ – Lost in Knowledge Oct 14 '15 at 20:39

I ... want to know how [electricity] travels on utility poles.

This is an interesting question and many aspects of the answer are surprising the first time you hear/read them.

Firstly, nothing really travels through the wires on the poles all the way from the power-plant to your home.

What does move physically is not "electricity" but charged particles, in the case of metal wires on utility poles these are electrons that are not strongly bound to the metal atoms. These are called "free electrons" because they can move around more easily.

The electrons quickly jiggle and bounce around a lot because the metal has a temperature hundreds of degrees above absolute zero. By which I mean normal outdoor temperature.

The power plant applies a strong force to the bouncing electrons which causes them to, on average, very slowly drift in one direction for about 0.02 seconds and then causes them to drift back in the opposite direction for 0.02 seconds. This is called alternating current (AC)

The movement of charge carriers (in this case electrons) is called electric current. We measure electric current in units called Amps. A current of 1 Amp is defined to mean the current from 6,241,000,000,000,000,000 electrons going past every second. However in the case of our AC current, the net number of electrons going past is harder to measure because after one second of drifting back and forth they are mostly back where they started (approximately speaking). So we average the motion out by calculating the root mean square (RMS) current - which makes subsequent arithmetic a little easier.

The force that causes the charge carriers to slowly drift is a force produced by what is called an electric field. We can measure the strength of this field as an electric potential measured in volts (so what is measured is called a voltage).

The upshot is that the power plant converts some form of energy such as chemical energy or gravitational potential energy into electrical energy and the utility poles cause this energy to be made available in your home (but don't think the energy always travels inside the wires - this also is a misconception)

  • $\begingroup$ I sometimes use a freight train as an analogy: The track is like the wire, the cars are like the free electrons, the force between one car and the next is like voltage, etc. When the engine starts pulling a mile long train, the "news" that the train is moving takes about a second to reach the last car---orders of magnitude faster than the actual speed of the engine at that moment in time. $\endgroup$ – Solomon Slow Oct 15 '15 at 13:31
  • $\begingroup$ Nice answer, I have a few additions, to further stress your points: 1. Even if the power lines were DC the slow drift velocity of the electrons (~mm/s) means that the electron from the power plant would need on the scale of a year (assuming a distance of 100km) to reach your home. 2. So what electric wires actually do is to "guide" the electromagnetic field. The power transfer from the power plant to your home is actually achieved through all the space around us (as one can verify by considering the Poynting vector – for a Ohmic load it points inwards, for a battery it points outwards!). $\endgroup$ – Sebastian Riese Oct 16 '15 at 1:09

Power-plant generates potential difference. This potential causes a electric field. Now we know that every conductor have free electrons. So this free electrons of the conductor experience a force $(F=eE)$, which is called electromotive force. This force gives a drift velocity to the free electrons and electrons move through the conductor wire. We know current flows in opposite direction to moving electrons. So a current flows due to this movement of electron in potential difference.

In this process electric current flows through conductor from one place to another due to potential difference generated by power-plant.

  • $\begingroup$ "We know current flows in opposite direction to moving electrons." It were better to say: "People 150 years ago, who did not know current in metals was carried by negatively charged particles, defined current to have the opposite direction from the electron's movements." In other words: The direction of current we define is completely arbitrary, the actual current are the moving electrons. $\endgroup$ – Sebastian Riese Oct 15 '15 at 0:38
  • $\begingroup$ yeah, you are right. $\endgroup$ – Rajesh Sardar Oct 15 '15 at 4:05
  • $\begingroup$ "we know that every conductor have free electrons" - this is true for commonplace metallic conductors, but there are many types of conductors where the charge carriers are not free electrons. So your statement is untrue. I suggest changing "every" to "metal". $\endgroup$ – RedGrittyBrick Oct 15 '15 at 10:50
  • $\begingroup$ I think usually the word "conductor" is used to define metallic conductor not a semi-conductor or super-conductor. And I also use the word "conductor" to define metallic conductor. You should note that here the question is "how current travel form powerplant to our house. So it is a matter of electrical and in electrical we generally use word "conductor" to indicate metallic current carrying conductor. @ RedGrittyBrick $\endgroup$ – Rajesh Sardar Oct 15 '15 at 14:04
  • $\begingroup$ So are metals the only materials with mobile charge carriers in their atom/ions conduction band? $\endgroup$ – Ubaid Hassan Apr 13 at 17:56

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