Veritasium "How Electricity Actually Works"

[rfeecs]

If so: HOW MUCH ENERGY does travel along the path?

If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path.
Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B.  The amount of energy is the energy converted to heat.  So it depends on which hole it goes down.

This is equivalent to say that the path the energy follows depends on where we arbitrarily choose to set the zero reference.

I don't think this is a problem if you consider the energy is relative to a reference.  So zero relative energy may still be nonzero "absolute" energy.

Instead of assuming the energy is located in the rock, you can eliminate the problem perhaps by using the gravitational potential: https://en.wikipedia.org/wiki/Gravitational_potential

If the gravitational potential is a property of location, then when you move the mass you are not "transporting energy".  As Naej said, you are only converting energy at the two end points.

We have heard the same point of view applied to electromagnetic systems.

[rfeecs]

The unasked question so far is:

Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.

What if you only connected pipe at one end.  The wire is connected at both ends.  Then the current only flows through the wire.

But since the pipe is at the same voltage as the wire, you have no electric field around the wire.  You moved the electric field to the outside of the pipe.

Now according to the Poynting vector, the energy flow is on the outside of the pipe, not even at the surface of the wire.

The current flows through the wire, but the energy flows on the outside of the pipe.  You could make the pipe 1 meter in diameter and the energy flow would be half a meter away from the current flow.

Yet another odd result of the Poynting vector applied to DC.

[aetherist]

2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.

If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.

Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.
If the switch is closer to charged capacitor / source then the wire after the switch forms another capacitor with the wire of the load thus effectively the small current Derek's is seeing is the one needed to charge that extra capacitor (the transmission line).

With the switch far away you also have a capacitor made by the transmission wires (that is fully charged before starting the test) and what the switch is doing is shorting that capacitor but from the far end of the capacitor thus electron wave that move the energy trough wires needs time to travel that distance.

Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).

The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light.
This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns.

Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity.
If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 35 ns, nearly the same as for the early (air) signal.

If the wire is insulated then this 35 ns would be 50% greater, ie  52.5 ns.  But the  34 ns for the early signal would remain at  34 ns.

Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.

[electrodacus]

Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).
The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns.
Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 36 ns, nearly the same as for the early (air) signal.

Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.

There will be no "early weak fast signal" with the switch 10m away from the source and from the load as mentioned.
The small constant electric field already exists on the transmission line going to the switch as that is just a low value capacitor and it is charged at V_i
The electric field will be converted to magnetic field and some IR heat loss as the electron wave travels along the wire and the electron wave will arrive first then you will have the electric field thus the energy is transferred by the electron wave moving inside the wire and not by any field outside the wire.

Derek (Veritasium) has no understanding of what energy is else he will not make such a claim that energy is transferred by a field.
Any of the 3 points I made are sufficient to disprove Derek's claim and all 3 of them are correct unless you have proof that they are not since I only used tested and accepted equations and theories to make those points.
Link to that post: https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462

Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions.

The most common mistake is to think that charge is conserved if you move the plates of an isolated capacitor when energy is the one that is conserved.
The explanation they have for this is also wrong as they think that since there is a force from the electric field thus you put energy in to move the plates further apart that is the reason charge is conserved thus they show final voltage as 2x initial voltage if you double the distance between plates when the correct answer is 1.414x and conserved energy not conserved charge.
The energy you put in to move the plates will be recovered as soon as you let go of the plates so it is potential kinetic energy storage.
It is like pulling two permanent magnets apart. You do not do any work you store some kinetic energy that you can get back by letting go of the magnets.

[aetherist]

Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).
The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns.
Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 36 ns, nearly the same as for the early (air) signal.

Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.

There will be no "early weak fast signal" with the switch 10m away from the source and from the load as mentioned.
The small constant electric field already exists on the transmission line going to the switch as that is just a low value capacitor and it is charged at V_i
The electric field will be converted to magnetic field and some IR heat loss as the electron wave travels along the wire and the electron wave will arrive first then you will have the electric field thus the energy is transferred by the electron wave moving inside the wire and not by any field outside the wire.

Derek (Veritasium) has no understanding of what energy is else he will not make such a claim that energy is transferred by a field.
Any of the 3 points I made are sufficient to disprove Derek's claim and all 3 of them are correct unless you have proof that they are not since I only used tested and accepted equations and theories to make those points.
Link to that post: https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462

Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions.

The most common mistake is to think that charge is conserved if you move the plates of an isolated capacitor when energy is the one that is conserved.
The explanation they have for this is also wrong as they think that since there is a force from the electric field thus you put energy in to move the plates further apart that is the reason charge is conserved thus they show final voltage as 2x initial voltage if you double the distance between plates when the correct answer is 1.414x and conserved energy not conserved charge.
The energy you put in to move the plates will be recovered as soon as you let go of the plates so it is potential kinetic energy storage.
It is like pulling two permanent magnets apart. You do not do any work you store some kinetic energy that you can get back by letting go of the magnets.

Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very)  weak fast signal.

It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns.  And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.

I have been looking at the arguments re capacitors, & everyone has been wrong in every way.

[electrodacus]

Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very)  weak fast signal.

It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns.  And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.

I have been looking at the arguments re capacitors, & everyone has been wrong in every way.

I specifically mentioned that circuit is symmetrical. In this case both the charged and discharged capacitors are at the example same distance from the switch.
The so called small current when circuit is not symmetrical is the current to charge the capacitor made by the transmission line.  In that case as it was in Derek's test your lamp is just an indicator to show the current used to charge that capacitor made from the transmission line wires.

As another example you can have the two capacitors circuit that I mentioned and add a lamp in series with the switch just next to the switch and you will see a current trough that lamp as soon as you close the switch but that current is provided by the energy stored in the transmission line so if you open the switch immediately after you closed it the charged in the transmission line will redistribute and the energy lamp used will be provided some nano seconds later by the main energy storage so that charged capacitor in this case.
Thus energy travels trough wires.

As long as your lamp is next to the switch it does not matter how far the battery is the lamp will see a small current immediately but that energy traveled trough wires and from wires (capacitor).
There is no evidence of energy traveling outside the wire just a wrong interpretation of the test result.

[aetherist]

Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very)  weak fast signal.

It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns.  And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.

I have been looking at the arguments re capacitors, & everyone has been wrong in every way.

I specifically mentioned that circuit is symmetrical. In this case both the charged and discharged capacitors are at the example same distance from the switch.
The so called small current when circuit is not symmetrical is the current to charge the capacitor made by the transmission line.  In that case as it was in Derek's test your lamp is just an indicator to show the current used to charge that capacitor made from the transmission line wires.

As another example you can have the two capacitors circuit that I mentioned and add a lamp in series with the switch just next to the switch and you will see a current trough that lamp as soon as you close the switch but that current is provided by the energy stored in the transmission line so if you open the switch immediately after you closed it the charged in the transmission line will redistribute and the energy lamp used will be provided some nano seconds later by the main energy storage so that charged capacitor in this case.
Thus energy travels trough wires.

As long as your lamp is next to the switch it does not matter how far the battery is the lamp will see a small current immediately but that energy traveled trough wires and from wires (capacitor).
There is no evidence of energy traveling outside the wire just a wrong interpretation of the test result.

Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's  3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.

However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.

U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.

In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.

[electrodacus]

Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's  3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.

However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.

U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.

In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.

All energy transfer from source to load/lamp is done done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.

[aetherist]

Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's  3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.

However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.

U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.

In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.

All energy transfer from source to load/lamp is done done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.

The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).

[Naej]

The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc².
If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat.

In the quasistatic idealization I had in mind, the mass takes an infinite time to travel from point A to point B. So, velocity is basically zero along the path and only changes during 'generation' (the machine raises the weight) and 'dissipation' (the weight does work against the gravitational field and it all becomes heat).

In my view the energy is in the gravitational field of the system planet + weight.  It is being added to the system during 'generation' and it is extracted from the system during 'dissipation'. The system occupies all space, so does it even make sense to ask if the energy is traveling?

It does not occupy all space, because potential energy is the sum of -Gm1m2/r.[

quote author=bsfeechannel link=topic=322795.msg4156429#msg4156429 date=1651721925]
I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.

The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.
[/quote]
Isn't it interesting how every argument for 'energy in vacuum' is an argument from authority? 

Also, Maxwell wrote:
https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/102
" We are thus led to a very remarkable consequence of the theory which we are examining, namely, that the motions of electricity are like those of an incompressible fluid, so that the total quantity within an imaginary fixed closed surface remains always the same."

[electrodacus]

The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).

What do you mean by "full electricity"?
Do you understand what energy is ?  And do you understand the conservation of energy ?

Main claim made by Darek and that is false is that "Energy doesn't flow in wires"
All 3 points that I made using the science accepted and used by everyone shows that is not true. Even just one of those points alone is enough to dismiss the claim Derek made.

And unless you are referring to IR from the wires no energy from the source travels outside the wire.

Electrons moving trough the wire did all the work.  I think the two parallel capacitors explain best what happens and that all energy flows in the wire.
All is required is to understand what energy is how energy in a capacitors is calculated (the discharged capacitor is like a energy recording device) then all energy will be accounted for and the results shows very clearly that no energy flowed outside the wire other than in this example half of the total energy at the start that ended as heat due to wire resistance.

[aetherist]

The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).

What do you mean by "full electricity"?
Do you understand what energy is ?  And do you understand the conservation of energy ?

Main claim made by Darek and that is false is that "Energy doesn't flow in wires"
All 3 points that I made using the science accepted and used by everyone shows that is not true. Even just one of those points alone is enough to dismiss the claim Derek made.

And unless you are referring to IR from the wires no energy from the source travels outside the wire.

Electrons moving trough the wire did all the work.  I think the two parallel capacitors explain best what happens and that all energy flows in the wire.
All is required is to understand what energy is how energy in a capacitors is calculated (the discharged capacitor is like a energy recording device) then all energy will be accounted for and the results shows very clearly that no energy flowed outside the wire other than in this example half of the total energy at the start that ended as heat due to wire resistance.

Me myself i reckon that the energy flows on the surface of the wires (electons)(& electrons), not in the wires (electrons).
Plus i reckon that there is a small amount of energy in the em field around a wire, but not enuff to justify some silly kind of Poynting field source.
In addition the em field around a wire transmits (can transmit) the energy of the electricity on the wire.
If electrons drift inside a wire then that is merely a drag effect, robbing energy. Energy duznt flow in wires.

The workings of capacitors can be explained by my electricity, but can't be explained by silly drifting electrons.

The full electricity arrives when the electons arrive.

[electrodacus]

Me myself i reckon that the energy flows on the surface of the wires (electons)(& electrons), not in the wires (electrons).
Plus i reckon that there is a small amount of energy in the em field around a wire, but not enuff to justify some silly kind of Poynting field source.
In addition the em field around a wire transmits (can transmit) the energy of the electricity on the wire.
If electrons drift inside a wire then that is merely a drag effect, robbing energy. Energy duznt flow in wires.

The workings of capacitors can be explained by my electricity, but can't be explained by silly drifting electrons.

The full electricity arrives when the electons arrive.

You did not provide an answer to what "full electricity" means.
You seems to have guesses about the subject but no concrete mathematical or experimental proof.
This is not unknown science. Is just Derek that comes with ridiculous false claims thanks in large part due to his inability to understand what energy is and he exposes how many have the same gap in understanding (he has millions of views on his videos and multiple other creators make videos mostly agreeing with his points).

Energy flows in to a capacitor not trough a capacitor.

[aetherist]

Me myself i reckon that the energy flows on the surface of the wires (electons)(& electrons), not in the wires (electrons).
Plus i reckon that there is a small amount of energy in the em field around a wire, but not enuff to justify some silly kind of Poynting field source.
In addition the em field around a wire transmits (can transmit) the energy of the electricity on the wire.
If electrons drift inside a wire then that is merely a drag effect, robbing energy. Energy duznt flow in wires.

The workings of capacitors can be explained by my electricity, but can't be explained by silly drifting electrons.

The full electricity arrives when the electons arrive.

You did not provide an answer to what "full electricity" means.
You seems to have guesses about the subject but no concrete mathematical or experimental proof.
This is not unknown science. Is just Derek that comes with ridiculous false claims thanks in large part due to his inability to understand what energy is and he exposes how many have the same gap in understanding (he has millions of views on his videos and multiple other creators make videos mostly agreeing with his points).

Energy flows in to a capacitor not trough a capacitor.

Full electricity is the arrival of my electons (photons hugging the surface of the wire). If u prefer to believe in old electricity (ie drifting electrons) then it is the arrival of the wavefront of the electron to electron push-wave.

I agree that energy flows in to a capacitor when it is charging.
I reckon that energy duz flow through a capacitor – by virtue of induction across the gap.

[bsfeechannel]

Point to me where I have posted a wrong equation.

Ex falso quodlibet, after a contradiction you can conclude whatever.

However, equations are not the question, here. Maxwell debunked the hydraulic analogy for the flow of energy based on experimental data. He noticed that a stretch of wire through which a current was flowing would tend to show a spike of voltage if interrupted at the point of the interruption, just like a pipe experiences sudden pressure at the point where a valve has interrupted the flow. However what hinted him that the energy didn't flow like a fluid in a pipe is that if you coil up the wire the voltage increases, while with a coiled up pipe you do not have this effect. Also, if you put a piece of iron inside the wire coil, the voltage will increase even more, but not the pressure for the pipe coil. Electricity flowing in wires affect other wires nearby not connected to each other either by attracting or repelling them if they too are conducting current, or inducing voltages if the currents vary.

Derek repeated the experiment Maxwell observed, only that, instead of opening the switch, he closed it. And he observed that, unlike a pipe, where the energy would never arrive at 1m/c, the electric energy does, and that the hydraulic analogy is what it is, a (should I say poor) analogy, not a description of How Electricity Actually Works.

So the experimental data debunks whatever theory you may have to sustain your misconception.

[bsfeechannel]

Isn't it interesting how every argument for 'energy in vacuum' is an argument from authority? 

Also, Maxwell wrote:

Contradiction, thy name is Naej.

https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/102
" We are thus led to a very remarkable consequence of the theory which we are examining, namely, that the motions of electricity are like those of an incompressible fluid, so that the total quantity within an imaginary fixed closed surface remains always the same."

If you didn't take snippets of texts out of context like every pseudo-sciencer you would turn the page and find the following snippet:

https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/103

The peculiar features of the theory as we have now developed them are: -

That the energy of electrification resides in the dielectric medium, whether that medium be solid, liquid, or gaseous, dense or rare, or even deprived of ordinary gross matter, provided it be still capable of transmitting electrical action.

[electrodacus]

Point to me where I have posted a wrong equation.

Ex falso quodlibet, after a contradiction you can conclude whatever.

However, equations are not the question, here. Maxwell debunked the hydraulic analogy for the flow of energy based on experimental data. He noticed that a stretch of wire through which a current was flowing would tend to show a spike of voltage if interrupted at the point of the interruption, just like a pipe experiences sudden pressure at the point where a valve has interrupted the flow. However what hinted him that the energy didn't flow like a fluid in a pipe is that if you coil up the wire the voltage increases, while with a coiled up pipe you do not have this effect. Also, if you put a piece of iron inside the wire coil, the voltage will increase even more, but not the pressure for the pipe coil. Electricity flowing in wires affect other wires nearby not connected to each other either by attracting or repelling them if they too are conducting current, or inducing voltages if the currents vary.

Derek repeated the experiment Maxwell observed, only that, instead of opening the switch, he closed it. And he observed that, unlike a pipe, where the energy would never arrive at 1m/c, the electric energy does, and that the hydraulic analogy is what it is, a (should I say poor) analogy, not a description of How Electricity Actually Works.

So the experimental data debunks whatever theory you may have to sustain your misconception.

As any analogy there will be limitations but I was not mentioning the hydraulic model in my demonstration. https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462

Derek clearly has no understanding of what energy is and is not just based on this two videos related to "how electricity actually works".

Do you agree with this simple statement:
Energy flows in or out of a capacitor and not trough a capacitor.

I say and is easy to demonstrate that energy flows in and out of a capacitor while Derek that has no clue what energy is basically saying that energy flows trough a capacitor.

[T3sl4co1l]

Why not both?

Stick a 10mF capacitor in series with an AC load and tell me it doesn't transmit energy.

Tim

[electrodacus]

Why not both?

Stick a 10mF capacitor in series with an AC load and tell me it doesn't transmit energy.

Tim

It will not.  You will have a lot of heat loss in the conductors due to repeatably charging and discharging that capacitor.
No energy will get across the dielectric.
That is why there is no current with DC after the capacitor is charged and gets to same voltage as your source and if you disconnect the source and short the capacitor you will get all that energy you put in back as heat.


Edit: And if you attempt to do this test make sure to use a non polarized capacitor (likely much smaller value than 10mF as you do not need that). Use a resistor to limit the current.

[T3sl4co1l]

Whelp... time for another experiment I guess.

Ed: You're not just thinking parallel (shunt), right?  You know what I mean by "series"?

Tim

[electrodacus]

Whelp... time for another experiment I guess.

Ed: You're not just thinking parallel (shunt), right?  You know what I mean by "series"?

Tim

If you want to do the experiment then in order to make things simpler for you just select the R and C so that capacitor is fully charged or almost fully charged with just half sine. That way you much easier can see what happens as measurements will be more accurate.
Yes of course resistor will be in series with capacitor in order to limit the current and be able to calculate the current by measuring the voltage drop across the resistor.

[EEVblog]

2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.

If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.

Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.

The position of the switch matters, otherwise that would break speed of light causality.
If anyone can't get their head around that then put the switch at a Derek inspired moon length. Flipping the switch at the moon can't instantly make something happen on earth, to do so obviously requires the speed of light to be infinite. This is wacko land.

[Naej]

However, equations are not the question, here. Maxwell debunked the hydraulic analogy for the flow of energy based on experimental data. He noticed that a stretch of wire through which a current was flowing would tend to show a spike of voltage if interrupted at the point of the interruption, just like a pipe experiences sudden pressure at the point where a valve has interrupted the flow. However what hinted him that the energy didn't flow like a fluid in a pipe is that if you coil up the wire the voltage increases, while with a coiled up pipe you do not have this effect. Also, if you put a piece of iron inside the wire coil, the voltage will increase even more, but not the pressure for the pipe coil. Electricity flowing in wires affect other wires nearby not connected to each other either by attracting or repelling them if they too are conducting current, or inducing voltages if the currents vary.

Yes, this is how the hydraulic analogy is inaccurate. Which is why the word 'analogy' is used: it's very similar, not identical.

Derek repeated the experiment Maxwell observed, only that, instead of opening the switch, he closed it. And he observed that, unlike a pipe, where the energy would never arrive at 1m/c, the electric energy does, and that the hydraulic analogy is what it is, a (should I say poor) analogy, not a description of How Electricity Actually Works.

So the experimental data debunks whatever theory you may have to sustain your misconception.

The energy arrives 1m/c where c is the speed of sound in air. For the same reason: opening the valve emits a sound wave, because you're violently compressing/accelerating water.
Just as in the experiment proposed, it's a tiny amount of energy.
Also there's a pressure wave propagating inside the pipe, much like there's a potential wave propagating along the wire (telegrapher's equations also work for the pipe).

[snarkysparky]

Why not both?

Stick a 10mF capacitor in series with an AC load and tell me it doesn't transmit energy.

Tim


It will not.  You will have a lot of heat loss in the conductors due to repeatably charging and discharging that capacitor.
No energy will get across the dielectric.

Yes it will.   It's called the displacement current.   

Displacement current is "flowing" anywhere ,  even in an insulator ,  where the electric field is varying with time.

https://en.wikipedia.org/wiki/Displacement_current

[electrodacus]

The position of the switch matters, otherwise that would break speed of light causality.
If anyone can't get their head around that then put the switch at a Derek inspired moon length. Flipping the switch at the moon can't instantly make something happen on earth, to do so obviously requires the speed of light to be infinite. This is wacko land.

Dave, I do understand that.  The point is that source (battery or charged capacitor) is 1m from the lamp/discharged capacitor on earth while switch is at the moon.
The transmission line is just a very long capacitor and it is charged as same potential like the the battery/charged capacitor. So when you connect a multimeter across the open switch at the moon you will not wait the time it takes electron wave propagation from earth to moon to measure the voltage it will be basically instant as the transmission line/capacitor is fully charged.
At the moment you close the switch first electron from one side of the switch will move on the other side collapsing the electric field associated with that electron imbalance. This will continue at the speed of light so if you where able to visualize the electric field line you will see how they collapse starting from the switch on the moon all the way down to earth at basically the speed of light and all that energy contained in the transmission line is radiated as infrared photos in the vacuum of space so ends up as heat.
Up to this point no energy left the battery/charged capacitor and no energy was delivered to bulb/discharged capacitor.

So when switch is closed the transmission line is a long capacitor with significant inductance and resistance.
No energy will flow trough the capacitor / transmission line but it will flow in or out.  So all energy delivered to load flows in the wire and not outside of it unless you count the radiated heat in the form of infrared photons due to wire resistance and it is not even that if you are using a superconductor.