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

[electrodacus]

Veritasium's scope screen looks sick.

That screen shows how much energy left the battery and how much energy arrived at the load (resistor).
Since current in that loop will be the same in battery and in resistor is clear to see that much more energy is delivered by the battery than dissipated on the resistor.
Where is that extra energy that battery delivered ?
Part of it is stored in the transmission line capacitance and inductance and part of it is lost as heat.
As stored energy is irrelevant as it did no work the lost energy will be in the form of heat.
All you need is a thermal camera and you can see that wires (copper pipe in this case) is what delivered the energy from battery to load.

That initial small current seen trough the load is due to energy being stored in the transmission line.
Not understanding what energy and energy storage is made Derek to come to a wrong conclusion about how energy is delivered from battery to load.

[bsfeechannel]

He used your confirmation bias against you, he only repeated his funny claim that "the energy is in the fields," giving no proof of the definition he chose (obviously).
He added an argument from authority from Rick Hartley, who himself only used another argument from authority to claim the same thing.
(It's bogus all the way down for some reason)

He also made mistakes:
- "electrons don't go to the battery" except they do it pretty quickly since they move at 1000km/s
- "charges contract radially on a wire" except you'd have a charged core of the wire, and you don't (at DC)
- he confused the 14 mW given by the capacitive coupling with the ~ mW given by the antennas (and I think this is a generous value).

All of this is irrelevant. He did the experiment and energy reached the load before it could travel the distance along the wires. End of story

Wires are not pipes.

[electrodacus]

All of this is irrelevant. He did the experiment and energy reached the load before it could travel the distance along the wires. End of story

Wires are not pipes.

25x less energy compared to the point in time where electron wave reached the load traveling trough the wire.  There is a fairly sharp transition between the two. 

[Sredni]

Veritasium's scope screen looks sick.
The green trace shows that there is a weak plus & minus  0.5V by  9.5 ns signal in his "battery" before he closes the switch (ie before his scope sends the "pulse")(or sends a signal or whatever).
The yellow trace shows that there is a similar plus & minus  1.1V by  9.5 ns signal in his "bulb".
...
What is causing these spurious initial signals?

1/9.5ns = 105 MHz.
Could it be an FM radio?
https://radio-locator.com/cgi-bin/locate?select=city&city=Pasadena&state=CA
KKGO and KPWR?

[Sredni]

That screen shows how much energy left the battery and how much energy arrived at the load (resistor).
Since current in that loop will be the same in battery and in resistor

You think that KCL holds, in the first few nanoseconds?
Oh dear, not another Kirchhoff battle...

[pepsi]

All of this is irrelevant. He did the experiment and energy reached the load before it could travel the distance along the wires. End of story

Wires are not pipes.

Is this for real? Tell this to the national electricity grid. Let's get rid of poles and wires and beam MW of power to customers 

[Syndicate]

All of this is irrelevant. He did the experiment and energy reached the load before it could travel the distance along the wires. End of story

Wires are not pipes.

What?  The existence of some wireless transfer of power was never the issue.  The problem is that the lamp doesn't turn on at "1/c seconds", and his explanations are off and misleading at best.  "Wires are not pipes/Energy doesn't flow through wires" is good example of faulty conceptualization.  And since the entire video rests on correcting conceptualization and the question of when the lamp turns on, that is a big problem for the video.

[electrodacus]

You think that KCL holds, in the first few nanoseconds?
Oh dear, not another Kirchhoff battle...

Has significantly more energy left the battery than got to the load/lamp ?
Do you agree that energy conservation can not be violated ?
If so then any difference should have ended up as heat and or stored in some form.

If you will have a sensitive enough thermal camera you will see that all energy delivered to the Load/lamp was trough the wires (not outside the wires) as you will be able to see the IR losses (yes for those first few nanoseconds).
The load/lamp is basically in series with some capacitors (the long and most likely intentionally thick transmission line).

[aetherist]

Veritasium's scope screen looks sick.

That screen shows how much energy left the battery and how much energy arrived at the load (resistor).
Since current in that loop will be the same in battery and in resistor is clear to see that much more energy is delivered by the battery than dissipated on the resistor.
Where is that extra energy that battery delivered ?
Part of it is stored in the transmission line capacitance and inductance and part of it is lost as heat.
As stored energy is irrelevant as it did no work the lost energy will be in the form of heat.
All you need is a thermal camera and you can see that wires (copper pipe in this case) is what delivered the energy from battery to load.

That initial small current seen trough the load is due to energy being stored in the transmission line.
Not understanding what energy and energy storage is made Derek to come to a wrong conclusion about how energy is delivered from battery to load.

I aint an EE.
Q1.  Once the current is steady, i guess that the area under the yellow V trace is less than the area under the input green V trace. Q1A. Duz the diff in area tell us the amount of heat lost? Q1B. Or duz the diff in area include energy making or sustaining em radiation?

Q2.  Why did the rise in the green trace from 0.4V to 18.6V take  8.6 ns?  Q2A. Why didn’t their (costly new u beaut) scope do it in say  1.0 ns, or even 0.1 ns? 

Q3.  Why didn’t Veritasium show us the initial pulses/rises with a 1.0 ns/div horizontal scale (ie as well as the 50 ns/div)?  I suspect that the scope could do better than 0.1 ns/div (i think that the scope can give at least 20 GHz, which is better than 0.1 ns/div). This would have better shown us the  3.3 ns delay.

Q4.  If the green rise took 8.6 ns, why did the yellow rise take  17.0 ns (ie from 4.1 to 21.2).  Or  20.6 ns if u prefer (from 4.1 to 24.8 )?

Q5.  Why did the green trace reach a steady state of  19.7V at say  300 ns, which is  1.0V lower than the yellow steady state of  20.7V at  300 ns?  Even tho, early on (before 0.0 ns), the green trace sat at  0.4V while it was the yellow that sat  0.3V lower at  0.1V.

Q6.  Why did the yellow trace start its main rise at  63.0 ns, when the speed of electricity along the 21 m Cu tube (10 m out plus 1 m spacing plus 10 m back) is  3.34 ns/m in air which demands that the rise should have been at 70.1 ns?  A delay of 63.0 ns suggests a tube Cu length of only  18.9 m (2.1 m too short).  Q6A. Why was the speed of electricity  10% faster than  c?

[electrodacus]

I aint an EE.
Q1.  Once the current is steady, i guess that the area under the yellow V trace is less than the area under the input green V trace. Q1A. Duz the diff in area tell us the amount of heat lost? Q1B. Or duz the diff in area include energy making or sustaining em radiation?

Q2.  Why did the rise in the green trace from 0.4V to 18.6V take  8.6 ns?  Q2A. Why didn’t their (costly new u beaut) scope do it in say  1.0 ns, or even 0.1 ns? 

Q3.  Why didn’t Veritasium show us the initial pulses/rises with a 1.0 ns/div horizontal scale (ie as well as the 50 ns/div)?  I suspect that the scope could do better than 0.1 ns/div (i think that the scope can give at least 20 GHz, which is better than 0.1 ns/div). This would have better shown us the  3.3 ns delay.

Q4.  If the green rise took 8.6 ns, why did the yellow rise take  17.0 ns (ie from 4.1 to 21.2).  Or  20.6 ns if u prefer (from 4.1 to 24.?

Q5.  Why did the green trace reach a steady state of  19.7V at say  300 ns, which is  1.0V lower than the yellow steady state of  20.7V at  300 ns?  Even tho, early on (before 0.0 ns), the green trace sat at  0.4V while it was the yellow that sat  0.3V lower at  0.1V.

Q6.  Why did the yellow trace start its main rise at  63.0 ns, when the speed of electricity along the 21 m Cu tube (10 m out plus 1 m spacing plus 10 m back) is  3.34 ns/m in air which demands that the rise should have been at 70.1 ns?  A delay of 63.0 ns suggests a tube Cu length of only  18.9 m (2.1 m too short).  Q6A. Why was the speed of electricity  10% faster than  c?

Q1. The graph shows voltage not power.  You will need to calculate the similar graphs for power as the difference there will be way more significant about 25x vs just 5x
Then on those graphs the area under represents the energy.

Q2. They likely used a solid state switch and that has capacitance and can not close instantly. Not an oscilloscope problem.

Q3. It will not have been relevant. He anyway did not understood what it is actually seeing there.

Q4. speed electron wave is as fast as it is allowed in this universe.

Q5. Oscilloscopes are only accurate in time domain not that great as voltmeters.

Q6. I have not looked at the graph that closely and I do not know their exact setup and how accurate their length measurements where. But is also irrelevant.

[Syndicate]

25x less energy compared to the point in time where electron wave reached the load traveling trough the wire.  There is a fairly sharp transition between the two.

It is 25x less power.  Energy-wise it is far worse because the power is transitory in the DC system.

[antenna]

Q1:  Why is everyone so fixated on the ringing in the voltages? Have you tried making a fast rising edge square pulse with absolutely no ringing? On top of feeding it through connectors, a test lead, then another transition to a slightly different complex impedance....  Have fun with that!

Q2:  Why is everyone so fixated on the difference in rise times? It's called frequency dispersion.  It happens when TDR pulses are used to check underground cables all the time. 

Q3: With all the complaints about the speed of light, has anyone included the meter of probe cable? Or its velocity factor? Just wondering...

Q4: Why is everyone picking on him for using such a "slow" rise time while having such fancy equipment at his disposal when that increases the higher frequency content and thus attenuation and dispersion?  Do you think he might have wanted to limit the nonsensical attacks speeding it up would bring?

Q5: Why are people questioning the "steady state" voltages when the screenshot only shows a portion of the response that is riddled with reflections that superimpose?

Q6: Why are there markers pointing out the times on the "unusual" peaks that any TDR operator would call an inductive discontinuity? Do you really expect them to perfectly match the impedances at the transition between transmission line and load, or the probe cable and transmission line etc? Come on now, you're just bored and pissed off at this point....

Q7: will I get the spam-hammer for offending "the elites" with my stupid questions now?

[electrodacus]

It is 25x less power.  Energy-wise it is far worse because the power is transitory in the DC system.

I was referring to just the initial phase before the electron wave travels the transmission line.
The 25x average lower power supplied to lamp/load (power on resistor) vs power provided by battery.
Large majority of the energy delivered from battery (or whatever they used as supply) was stored in the transmission line then a smaller part ended up as heat as all energy was delivered by the wire/pipe including the initial few ns and the only reason lamp/load received any energy was because it was in series with the battery and the energy storage device (transmission line).
By the time the energy storage is charged electron wave gets to Load/lamp and so energy delivered to Lamp/load is not delivered outside the wire at any point in time.

[EEVblog]

Derek made it very clear that he didn't make things very clear in the original video.  (Have any of us never made a meal of expressing an idea?)

It was very apparent to me that when he was talking about a light illuminating "at any amount of current" in the first video, he was NOT including the extreme case of leakage current - but current that results from closing the switch.  If it wasn't, then why even have a switch?

Yes, also I assumed the switch meant that leakage current should not be included. That seemed very obvious and was the only genuine way to approach the problem.
Not that the problem was designed to be analysed by EE's in the first place as Derek alluded to, and from talking to him can confirm that. He readily admits is was a poor example.

[Naej]

He used your confirmation bias against you, he only repeated his funny claim that "the energy is in the fields," giving no proof of the definition he chose (obviously).
He added an argument from authority from Rick Hartley, who himself only used another argument from authority to claim the same thing.
(It's bogus all the way down for some reason)

He also made mistakes:
- "electrons don't go to the battery" except they do it pretty quickly since they move at 1000km/s
- "charges contract radially on a wire" except you'd have a charged core of the wire, and you don't (at DC)
- he confused the 14 mW given by the capacitive coupling with the ~ mW given by the antennas (and I think this is a generous value).

All of this is irrelevant. He did the experiment and energy reached the load before it could travel the distance along the wires. End of story

Wires are not pipes.

If you want to do scipop with antenna then you a) use the word antenna b) don't take as the only example unintended antennae.
Why are you so fixated on the magical words "the energy is in the fields" when it's only science-babble (not even wrong) ? How many mistakes are ok if you utter the magical words? (I forgot the part where resistance is explained, which is completely wrong)
And: wires are pipes, they are pipes for current and pipes for energy.

[aetherist]

Thanx for your answers to my questions.  Some have suggested that Veritasium Pt2-X is a triumph. Some have suggested that Veritasium Pt2-X (rise times etc)(sensitivity & screen scale etc)(reflexions etc)  is severely constrained by practical switching problems & practical probing problems.  I reckon that u/we can do a search on this topic for "Howardlong".  See especially my reply#1048. He did his own little table-top X, using a puny 4 ft of ladder antenna. He had no such problems. And, his scope screen scale was 50 ps/div, compared to Veritasium's 50 ns/div.
Howardlong used a 20 GHz scope. I am thinking that Veritasium's scope was just as good or better.

I detect a continuing love affair around here for lumped element transmission line models.  I wish to point out that there has never been any good correlation tween such a model & experiment for a DC transient of the Veritasium gedanken kind, albeit using 1000 m of Cu (AlphaPhoenix) or 42 m of Cu (Veritasium) or 8 ft of Cu (Howardlong) or any m or ft of Cu.

I see much chatter around here re crosstalk due to radio waves, & crosstalk due to capacitance tween parallel wires, & crosstalk due to induction tween parallel wires.  I wish to advise that these 3 effects are (basically) the same effect.  Certainly in the case of electricity. Here the cause is em radiation.
And we can add a 4th kind of crosstalk, ie induction due to static charge. Alltho here the cause is not em radiation, the cause is the electric field (there being zero magnetic field).
So, what i am saying is that there is one basic kind of crosstalk. Induction.

If the cause of the crosstalk is due to electrons (or ions) crossing a gap (rather than simply em radiation) then i am not sure whether i would call that a kind of induction. I would have to have a think. Some other day.

If the cause of the crosstalk is due to photons (eg gamma rays), or some other exotic powerful subatomic rays, then that probly aint induction.

While i'm hot. All electric radiation is em radiation. An electric field is simply em radiation (ie an em field) where the magnetic field part is nett zero due to 2 opposing magnetic fields negating.  Hence in a sense charge too is em radiation. But i don’t want to start any arguments about that today. This contradicts what i said above, but i wont bother to edit that wording.

[aetherist]

HERE IS A COPY OF MY REPLY#1048.
START #####################################################################################################
Howardlong messaged me the following info.
Rise time 10-90% at the scope is 36 ps. By the time it gets to the feedpoint, it'll be about 45 ps due to dispersion in the coaxial feed.
Fall time looks similar visually but I didn't take a measurement. Pulse width is 608 ps.

Howardlong has already mentioned that his signal crosses (first reaches) to the opposite wire in  80 ps which he says accords with the speed of light for the  24 mm distance tween the pair of wires in his  450 Ohm antenna ladder line. Howardlong in effect says that this supports Veritasium's expectation that Veritasium's bulb can possibly light (start to light) in 1/c seconds (ie 3.3 ns for Veritasium's 1000 mm spacing).

These kinds of transients have at least say 4 stages.
I wanted to have a closer look at Howardlong's experiments to look at the first stage, stage-1 of his transient. But i will come back to that another day.
Today i will jump ahead & look at stage-2 of his transient.

Howardlong X using 4 ft of ladder antenna line (wires 24 mm apart). He got 12 mV, with 58 mV in the other wire, which is 20.7% (20 GHz scope).
Schantz X using 100 ft of 300 ohm twin lead antenna line (wires 7 mm apart). He got 60 mV, with 340 mV in the other wire, which is 17.6% (100 MHz scope).
AlphaPhoenix X using 1000ft of 24AWG  enameled copper wire (wires 250 mm apart). He got 0.2 V, which climbed to 1.7 V, which is 11.8%(100 MHz scope). Actually his source is 5.0 V, so 0.2 V is 4.0%.
Silicon Soup (youtube) does a Finite-Difference Time-Domain simulation (1000 mm), gets a 0.3 mA signal from a 1.47 mA current, which is 2.0% 20.4%, for a mini-version of the Veritasium circuit. I don’t know how his pseudo-signal happens (its something to do with Maxwell)(displacement current perhaps).

All of the above percentages are astonishingly high. But i think i know what happens.

A step signal (voltage)(current)(Heaviside might say that the step signal is energy current)(Dollard might say impulse current)(is say elektons) propagates say to the right along the right half of our circuit, along the say bottom wire.
The bottom wire in that half is gradually flooded with negative charge (elektons), starting at the source (at the midpoint of the circuit), the flooding progressing to the right towards the short at the end.
The growing negative charge on the surface of the bottom wire gradually repels more & more free surface elektons on (along) the top wire, some go right (to the end), & some go left (to our bulb).
The elektons pushed right (along the top wire) tend to bunch up, because they are propagating in the same direction as the propagating step (in the bottom wire).
The result is that say 50% of the escaping elektons in the top wire go left & 50% go right.
The elektons propagating left create a flow of elektons flowing left through our bulb, which manifests as a voltage drop across our bulb.
Our bulb turns on (weakly) a little after d/c seconds, ie as soon as (enough) elektons start to flow (leftwards) through the bulb on our top wire.
Our bulb glows brighter as the flow of elektons through the bulb increases.
After a short time the flow through our bulb reaches its initial maximum (say 10% of the current in the bottom wire).
[In the Veritasium gedanken (wire spacing d is 1000 mm) this would be a little after 1/c.]
Eventually the step (propagating right) in our circuit will get to the end of the bottom wire & will enter the top wire (via the short), & go to our bulb.
When the main signal reaches our bulb the bulb will achieve full brightness, ie there will be a big sudden jump step in the voltage.
[In the Veritasium gedanken the main signal would reach his bulb in 1 second (his half circuit is 1 light second long).]

The elektons escaping to the left will give a current & voltage (signal) at the midpoint of our top wire (ie at our bulb). The size of the signal will depend on the wire spacing. The signal will begin to grow as soon as the E×H radiation reaches across, ie the delay is d (metres)/c (m/s), where d is the spacing, & c is the speed of light in the medium (usually air). More exactly, the delay will depend on the location of our switch, relative to our bulb.
[In the Veritasium gedanken this switch-to-bulb distance is approx the same as the spacing tween his wires anyhow.]

I doubt that a (simple conventional) LCRX lumped element transmission line model can predict transient current, using a simple LCRX paradigm, using simple speed of light.
Any such model needs smarter components.
And truer speeds (& truer flow of surface elektons).
However i have never had any hands-on experience with transmission lines, or TL models (or the application of electricity theory of any kind).
However the repulsion of the elektons from (along) our top wire is not unlike the action of lots of little capacitors tween the bottom wire & the top wire.

Perhaps someone could do a (simple conventional) transmission line model for Howardlong's experiment.
END ################################################################################################################

[Sredni]

You think that KCL holds, in the first few nanoseconds?

Has significantly more energy left the battery than got to the load/lamp ?
Do you agree that energy conservation can not be violated ?
If so then any difference should have ended up as heat and or stored in some form.

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.

If you will have a sensitive enough thermal camera you will see that all energy delivered to the Load/lamp was trough the wires (not outside the wires) as you will be able to see the IR losses (yes for those first few nanoseconds).

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.

[bsfeechannel]

If you want to do scipop with antenna then you a) use the word antenna b) don't take as the only example unintended antennae. Why are you so fixated on the magical words "the energy is in the fields" when it's only science-babble (not even wrong) ? How many mistakes are ok if you utter the magical words? (I forgot the part where resistance is explained, which is completely wrong)

And if you conjure the RF spirits with the magic word antenna you think that you'll ward off the curse of the energy in the fields, don't you?

And: wires are pipes, they are pipes for current and pipes for energy.

Your theory lost. Nature won. Learn to live with that.

[bsfeechannel]

Is this for real? Tell this to the national electricity grid. Let's get rid of poles and wires and beam MW of power to customers 

Not a bad idea. The problem is the cost of the wave guides. If we only had some kind of guide for LF AC fields to carry the energy...

Sigh. I think we'll have to wait for the invention of wires.

[electrodacus]

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.

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)).
   

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.

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]

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.

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).

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]

And: wires are pipes, they are pipes for current and pipes for energy.

Your theory lost. Nature won. Learn to live with that.

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]

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.

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.