But otherwise, yes, the power is transmitted in the fields. A capacitor is an electic field.Yes, that explains how power flows through a capacitor. Rings a bell.
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My counter-argument is this: If Dave's argument is true, we would not have EMI, grounding, ground-loop, signal-integrity etc. issues in our circuits, because if practicing engineers really knew and understood how the EM-fields were actually flowing in the circuits, they would not make that many troublesome designs.
Since in practice we see way too often designs that have these kind of issues, it is clear that a) practicing engineers do not really understand EM-fields, and b) it is not really sufficient to think that the energy flows only along wires.
In a world where EMI etc was the only objective in all designs, then sure, that argument would be absolutely true. Not saying it's false in its reasoning, but maybe not the only conclusion you can draw. We would probably also have to assume that all designs produced from a true understanding of EM would also meet thermal, mechanical, cost, and functional constraints... and still meet all of those constraints when the sales team decide what they actually wanted was a plastic enclosure and have it delivered the week previous. It's always a balance and most engineers are unfortunately humans: mistakes and carelessness happen, things get overlooked with different priorities, it doesn't say anything about their understanding.
Attached article provides a quantitative analysis which shows at DC condition Poynting vector has two components, one along the surface of the wire in the direction of power transfer and the other one perpendicular to the wire directed into the wire, representing power loss dissipated inside the wire due to active resistance.
Anyone in wild agreeement/disagreement with those points?
So both of these points indicate to me that it is not the E field of the battery that is of interest in deriving the correct arrangements of E/B fields to account for the energy flow in this circuit, but is rather the local E fields generated by the local charge distributions on the surface of the wires that we need to know about.
So if we want to explain clearly how the Poynting vector and subsequent energy flow arises in this DC steady state case, then we have to derive the charge distributions on the wires; it's not enough to hand-wave and point at the E field of the battery.
Anyone in wild agreeement/disagreement with those points?
Only the fields from the orthogonal wires contribute to the poynting vector
Anyone in wild agreeement/disagreement with those points?
1. A PP3 9 V battery is sufficiently small that its E field will look like a dipole to the rest of the circuit, and so drops in intensity by 1/r^3.
Near the resistance, this will be too small to account for the required intensity of the Poynting vector. If this is isn't true for a circuit with 10 m sides, then it certainly is for one with 10 km sides, yet the resistance will clearly still dissipate the same power in the latter.
The first detail he missed is that electromagnetic fields significantly decrease in strength over distance. I did a calculation that estimates the magnitude of the poynting vector going directly from source to load, and finds the energy transferred over 1/c seconds. For the calculation, I assumed a typical bulb resistance and ideal wire. Only the fields from the orthogonal wires contribute to the poynting vector, and they are so far away that I can assume they are effectively a point charge. The value for the magnitude of the poynting vector and for the energy transfer are so small, they are effectively zero. These values increase as the position of the vector moves towards the wires, and it doesn't become significant until you are right next to them. So the path of the energy does follow the wire. It doesn't just jump through thin air wherever it wants to go.
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Since EM fields are created and carried by charges, you can't have one without the other. So it's silly to say, "the energy doesn't transfer through the electrons, it travels through the EM fields." You might as well say, "My car didn't move me across town, the gas in the tank moved me." In both statements, the distinction is misleading.
I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.
You have a fundamental misconception about how power is transferred in DC. In your video you say that the direction of the Poynting vector is away from the source only in AC, and throw in for some reason the skin effect,
Anyone in wild agreeement/disagreement with those points?
Thanks for not drawing the battery at the top of the diagram, otherwise all electrons would fall out.
1. A PP3 9 V battery is sufficiently small that its E field will look like a dipole to the rest of the circuit, and so drops in intensity by 1/r^3.
This would be true (maybe, idk actually) if the battery were just floating in space with no wires attached. However, that's not the real situation here. One of those wires has a voltage 9V higher than the other, and since E = ∇ V, there an electric field between between the two wires.
I cannot stress enough that if you correctly integrate the Poynting vectors over a plane dividing the battery from the load, you will see the correct value for the power being transferred from battery to load. If you disagree, you made a mistake in the calculation
However... there is one huge, laughable error in his presentation. Observe his graphic of an EM wave fields in space. Spot the error.
My comment on youtube, buried in 46,785 other comments:
Speaking of 'wrong things', your graphic of the E and M fields for light, radio etc is wrong. One field is phase shifted from the other by 90 degrees. Eg the blue will be maximum while the red is going through zero. It's the rate of change of each one that creates the other.
Did anyone else in this long thread notice that?
If we could see the magnitude of S as a glowing substance in my circuit, it would be glowing hot near the wires, and much dimmer away from the wires (if my hand-waving is even nearly right). However hot the glow is 1 cm from the wire, it would be 1% of that intensity 10 cm away.
https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant/532550#532550
John D. Jackson, the dreadfully respected author of the bible of Classical EM, wrote a paper about the role of these surface/interface charges, where he noted that this knowledge sorely lacks in most curricula at the time (1996, but I see not much has changed).
John D. Jackson
Surface charges on circuit wires and resistors play three different roles
American Journal of Physics 64 (7), July 1996
A much much easier read might be the note from Chabay and Sherwood:
Bruce A. Sherwood, Ruth W. Chabay
A unified treatment of electrostatics and circuits
American Journal of Physics
This can be find online.
a wire sitting at ground potential will have zero E field around it if everything nearby is at ground potential. But add nearby wires at non-ground potentials, and you'll see an E field appearing between those wires. It's very much a function of the spacing between the wires and the voltage across those wires; I'd personally not put too much thought into "local charge distributions on the surface of the wires", I don't quite see how that could lead to correct computation of the E field around the wires.
If we could see the magnitude of S as a glowing substance in my circuit, it would be glowing hot near the wires, and much dimmer away from the wires (if my hand-waving is even nearly right). However hot the glow is 1 cm from the wire, it would be 1% of that intensity 10 cm away.
I fully agree with the statement above (if anything, I suspect the falloff might be more extreme than r^2 even), and the reasoning leading up to it.