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"Veritasium" (YT) - "The Big Misconception About Electricity" ?

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electrodacus:

--- Quote from: Sredni on May 01, 2022, 02:27:56 pm ---
it's going into building the fields that will make the steady-state power transfer possible.
Or, if you dislike the idea of fields storing energy, it is going into separating the surface charge that will create the electric field inside the good conducting wires and the badly conducting load and that are responsible for the the local dissipation of energy in situ.

--- End quote ---

The creation of the electric field outside the wire due to line capacitance (storing energy) is what makes the initial small current flowing trough the lamp/load.
That initial field will collapse (being discharged) if the circuit is closed (ends not opened).
With ends open that field after is created remains there but since there is no longer any change in the field no energy is transferred.
With the ends closed the electron wave powers the lamp/load and not that initial field outside the wire that is not even there anymore (there is some but orders of magnitude smaller and constant (assuming load is constant)).
   


--- Quote from: Sredni on May 01, 2022, 02:27:56 pm ---No, you will see the resistor of the lamp getting hotter while the wires are cold. And even in steady state you would see the lamp filament very hot and the wires not even lukewarm. You can't see the 'transfer' of energy: you can only see where it is dissipated.

--- End quote ---

lukewarm :)
There are 20m or so of thick pipe representing the wire so not only much lower resistance than the resistor/lamp filament but also much lower surface area to dissipate that heat.
You can make the wires so much thinner that they get warmer than the lamp filament. The thinner the wires the higher the wire temperature and without the wire there will be no energy transfer to the lamp/load.

The main claim is that the field outside the wire is what transfers the energy to the lamp and that is just so obviously not true when we are talking about DC.
The transmission line model is a perfect finite model approximation of what happens.

A constant electric field can not transfer any energy same way as a constant magnetic field can not transfer any energy (do any work).
The transfer of energy happens when create the electric field (charge the capacitor) or when you are discharging it.
The field inside an isolated charged capacitor is doing no work (excluding the small amount of leakage).

Sredni:

--- Quote from: electrodacus on May 01, 2022, 04:13:43 pm ---
--- Quote from: Sredni on May 01, 2022, 02:27:56 pm ---it's going into building the fields that will make the steady-state power transfer possible.
Or, if you dislike the idea of fields storing energy, it is going into separating the surface charge that will create the electric field inside the good conducting wires and the badly conducting load and that are responsible for the the local dissipation of energy in situ.

--- End quote ---
The creation of the electric field outside the wire due to line capacitance (storing energy) is what makes the initial small current flowing trough the lamp/load.
That initial field will collapse (being discharged) if the circuit is closed (ends not opened).

--- End quote ---

You seem to think that the current we are talking about is the displacement current that - so to speak - is moving from the lower leg to the upper leg, let's say 'vertically'.
No, the current I am talking about is flowing inside the resistor (lamp) and nearby wires in the upper leg, and it is 'horizontal'.

It is caused by the surface charge that has been induced by the electric field disturbance that is propagating in space between the two legs of the circuit. Classically, power dissipation happens locally inside the resistor due to the great acceleration imparted to the electrons there by the 'strong' electric field that is associated with the charge displaced at the resistors ends. Before the perturbation in surface charge has traveled along the wires to the moon and back, there is only a fraction of the charge that will be there in steady state, and its spatial distribution on the surface of the conductors is also not yet final . But still, you will have power dissipated in the resistor due to current flowing INSIDE IT.

The surface charge distribution can be maintained (and subsequently reinforced) only if the circuit is closed: after a few back and forth you get the final configuration where the surface charge is such that there is a small, almost negligible, electric field directed longitudinally along the wires and a very strong electric field INSIDE THE RESISTOR that (classically) accelerates the electrons entering it, imparting them locally a lot of energy that is locally dissipated by means of collisions with the resistive material lattice.

The role of the battery and the wires is that to keep the separation of charge at the resistor extremes, so that electrons that arrive there as pacifists will be turned into a warmongering hoard of wrecking Ralphs that will make the lattice red hot. You need the wires to get, and keep, the right configuration of electric and magnetic field that will make power come out of the resistor.
But the energy is not carried by the electrons traveling into the wires; it is imparted to them by the field that they find there when they arrive.

Naej:

--- Quote from: bsfeechannel on May 01, 2022, 03:48:47 pm ---
--- Quote ---And: wires are pipes, they are pipes for current and pipes for energy.

--- End quote ---
Your theory lost. Nature won. Learn to live with that.

--- End quote ---
Nah, the antennae successfully worked, and the capacitors charged as predicted. Looks to me like another win for Maxwell's equations, but you are free to give a participation trophy to Poynting instead.

electrodacus:

--- Quote from: Sredni on May 01, 2022, 08:17:01 pm ---You seem to think that the current we are talking about is the displacement current that - so to speak - is moving from the lower leg to the upper leg, let's say 'vertically'.
No, the current I am talking about is flowing inside the wires and resistor (lamp) in the upper leg, and it is 'horizontal'.

It is caused by the surface charge that has been induced by the electric field disturbance that is propagating in space between the two legs of the circuit. Classically, power dissipation happens locally inside the resistor due to the great acceleration imparted to the electrons there by the 'strong' electric field that is associated with the charge displaced at the resistors ends. Before the perturbation in surface charge has traveled along the wires to the moon and back, there is only a fraction of the charge that will be there in steady state, and its spatial distribution on the surface of the conductors is not yet final . But still, you will have power dissipated in the resistor due to current INSIDE IT.

The surface charge distribution can be maintained (and subsequently reinforced) only if the circuit is closed: after a few back and forth you get the final configuration where the surface charge is such that there is a small, almost negligible, electric field directed longitudinally along the wires and a very strong electric field INSIDE THE RESISTOR that (classically) accelerate the electrons entering it, imparting them locally a lot of energy that is locally dissipated by means of collision with the resistive material lattice.

The role of the battery and the wires is that to keep the separation of charge at the resistor extremes, so that electrons that arrive there as pacifists will be turned into a warmongering hoard of wrecking Ralphs that will make the lattice red hot. You need the wires to get the right configuration of electric and magnetic field that will make power come out of the resistor.
But the energy is not carried by the electrons traveling into the wires; it is imparted to them by the field that they found there when they arrive.

--- End quote ---


Not sure how much you understand a battery so is best to replace that with a charged capacitor as it is simpler to understand than a battery.

----------------[RESISTOR]--------------------
-------------------{-CAP+}--s/ ------------------

The open loop above is just a charged capacitor "CAP" not connected to anything if the switch is open (ignoring the super small switch capacitance).
As soon as you close the switch "s/"  you are paralleling the charged "CAP" with the two series capacitors formed by the lines on each side and those caps are in series with the resistor but that is not very relevant (it is just like having a wire there).

So what you have when the switch is closed is a closed loop made up of 3 capacitors in series. You can consider those two discharged capacitors in series as a single capacitor and then simplification will be a charged capacitor in parallel with a discharged capacitor.
 _________I I__________ 
I                                         I
I                                         I
I                                         I
I_________I I__________I

There is no current flowing trough the capacitor dielectric and yes an electric field will be formed there as the capacitor charges but that is due to the electrons moving from the charged capacitor. There will not be a field at the discharged capacitor before the electrons from the charged capacitor get there.

Sredni:
You talk about the dielectric, while I talk about what happens INSIDE the plates.

Here: https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant
I put some references in this answer. A good deal of that is freely available on the net. Try to read at least the essay by Chabay and Sherwood.

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