What is the difference between voltage and the poynting vector?

1. Sep 4, 2009

user111_23

This is the article that explains it: http://amasci.com/elect/poynt/poynt.html

If the energy in a circuit is really just in the form of electromagnetic fields, what's so important about voltage? If I already know the wattage in the circuit, why should I know the potential energy per charge if the energy is flowing and not stationary? Voltage seems a useless concept if this theory is true.

Also, this was stated in another topic here, but (neglecting the above) what really happens to the potential energy of a Coulomb when it passes through a circuit? Is it converted into kinetic energy? Because when electrons pass through a light bulb filament, it is the FRICTION that causes the resistor to glow, so wouldn't that be from the kinetic energy of the charges?

2. Sep 4, 2009

mikelepore

The voltage is the reason the charges flow. For something to have potential energy means it is in a position where it has the ability to begin moving in the future, like a stretched spring has the ability to return to its original shape, or a mass at a high altitude has the ability to fall. The charge at one side of the filament of the bulb has potential energy relative to the other side of the filament, because it has the ability to move spontaneously through the filament to the other side. The charges drop to lower potential energy as they go through the filament, and the energy becomes heat.

In answer to this, I see no reason to mention an electromagnetic field or the Poynting vector.

3. Sep 4, 2009

user111_23

So the energy arises when the charges can spontaneously move through the circuit...interesting.

I mentioned the Poynting vector because I wanted to know what the difference was between electromagnetic energy and potential energy of electrons. Are they related in some way or not?

4. Sep 4, 2009

mikelepore

The Poynting vector is a feature of an electromagnetic wave, which can travel through empty space without the need for wires. If you have a circuit to power a light bulb, the analysis of the circuit isn't performed by talking about electromagnetic waves. The hot filament produces light, which consists of electromagnetic waves, but you don't have to consider that during the circuit analysis.

5. Sep 4, 2009

user111_23

Ok that makes sense. I made these topics before but none of the replies in those topics are as straightforward as yours. Thanks for the help!

6. Sep 4, 2009

mikelepore

Are you still unclear about what voltage is?

From Halliday's physics textbook, the example of a person who keeps lifting up a lot of bowling balls and putting them on an elevated platform, and then they keep rolling down a ramp to their starting point. A battery is like the person who keeps lifting them up to the top. The light bulb or other resistor is like the ramp which is their path to fall back down. The difference is a circuit is based on charge responding to electric force, not mass responding to gravitational force.

Do you remember the gravity problems? A mass at height h has gravitational potential energy mgh. The mass m is multiplied by its location's grav potential gh to express it's grav PE mgh. Similarly, in a circuit, a charge q is multiplied by its location's electrical potential V to express its electrical potential energy qV.

So the potential difference between two points is the cause that compels the current to flow.

7. Sep 4, 2009

user111_23

I heard of those analogies before. I like using those analogies only when telling someone about voltage or trying to understand it in a simple way.

I understand Voltage as the amount of energy needed to move a charge across a resistance. The difference arises (in a battery) when one terminal has more electrons than the other side. When a circuit is made, electrons go against the negative terminal gaining PE, and then get pulled by an external E-field parallel to the conductor towards the positive terminal. The work is useful because the amount of work done is translated into an equal amount of PE to the electron(s).

Not sure if I have holes in my understanding, but it seems to work with me! :tongue:

8. Sep 4, 2009

Bob S

The Poynting vector S = E x H is a very important concept at high frequencies where energy is flowing through empty space, but it also applies to microwave power inside waveguides, where currents in the waveguide walls help confine the energy flow. At even lower frequencies, wires are used to confine the energy flow. At 60 Hz, there is a voltage between the wires that produces electric field lines, and a current in the wires that produces magnetic field lines. The Poynting vector S represents the power flowing along between the wires. At some future date, power lines might be superconducting, in which case there is no currrent in the wires (it is on the surface) and no assoicated resistive power loss. So where is the actual power flow? All of the power is flowing between the wires in the form of E and H fields.

[Added] Knowing either the current or the voltage or the current is insufficient to determine the power flow. If the load is reactive I=I0 sin (wt) and V = V0 cos(wt)), then the one-cycle average of S is zero, so there is no power flow. If either the instantaneous electric field E or the instantaneous magnetic field H changes polarity, the power is flowing in the opposite direction.

Last edited: Sep 4, 2009
9. Nov 16, 2011

Omegatron

The whole point of the original linked article is that wires do not confine the energy flow. Even in a DC circuit, the energy is entirely in the fields. Do you disagree with the article?

10. Nov 16, 2011

nsaspook

The energy is in the fields but not in a classic EM wave. At 60hz the rear field at less than 1/4 is totally reactive and the fields are decoupled because the impedance of space at that frequency for any wire under a 2000 miles so much larger than the impedance of the wire confining the power flowing on the surface of it.

As posted in the original link, Feynman knows it's a fools mission to waste time with Poynting theorem with utility power transmission.

11. Nov 17, 2011

Per Oni

nsaspook:
Imagine the next scenario which doesn’t occur in real life but it could be built and made to work.

Say I want to run a DC, 10 KW electrical motor supplied by 20 KV and ½ amp in a lab. According to you all the energy (power) flows on the surface of the connecting wires between source and load. But do I really need a wire for the whole length? I could insert say a fluorescent light tube in line with the wire. As long as I make sure that the rated current for the tube is not exceeded all should work fine.

Very basically, these tubes are filled with some low pressure inert gasses aided with a drop of mercury. Therefore I can honestly not see how any meaning full amount of energy can be transported that way. Not only that but I cannot see which surface the electrical power would use.
This (and other considerations) leaves me with the conclusion that the link of the OP is close to the truth.

Also: if all this energy runs on the surface of the conductor then according to the Poynting vector the E and B fields should be disproportionally higher there then say a little outside or just inside the conductor.

12. Nov 21, 2011

Omegatron

I'm not sure what you're talking about. This is about the energy in DC circuits flowing in the fields surrounding the wires. The energy is traveling entirely outside the conductors. This is only for twin lead connections, though. For coax, the energy flow is contained entirely in the dielectric, and not outside of the cable, because the magnetic fields of inner and outer conductors cancel. https://en.wikipedia.org/wiki/Poynting_vector#In_a_coaxial_cable There is no energy flowing in the conductors. Energy flux in the conductors and up to their surface is zero.

13. Nov 24, 2011

Per Oni

Not sure this is entirely correct.

Say we have a resistive load connected with wires to a DC source. Then most of the energy flux flows towards the load but since wires are also having a finite amount of resistance it follows that some of the energy flux will go to the conductors.

The site you quoted is very educational but is not clear on that point either.

14. Nov 24, 2011

Omegatron

Right, but the energy is not flowing from the source to the load inside the conductors. When the Poynting vector points toward the resistive conductor, the energy is flowing from the source into the conductor itself. It is then dissipated. The resistance of the conductors is converting that energy into heat.

15. Nov 25, 2011

Per Oni

In that case we're on the same wavelength.