Veritasium "How Electricity Actually Works"

[electrodacus]

Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.

The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.

If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.

Pipes are essential to the system working, even though they carry no usable energy.

The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.

I guess I need to be more explicit as you did not understood any of my compressed air analogies.

Air molecules are the electrons and compressed air cylinder with with two chambers represents the capacitor.
A discharged capacitor is like an air cylinder that has the same pressure in both chambers say atmospheric pressure but it can be any value as long as the same number of air particles are found in both chambers and inside the hose's.

Now you can charge the capacitor using say mechanical energy to pump air from one chamber to the other one. You can use the same pump in reverse to do work with the stored energy.
Since now there are more air molecules (electrons) in one chamber (capacitor plate) than the other one you have stored energy.
If you connect the two chambers through a pipe air molecules (electrons) will flow (electric current) from the higher pressure chamber to the other one until pressure (voltage) is equalized. 

[IanB]

I see. So if you have three capacitors in series, the energy to charge the middle capacitor flows through the outer capacitors. If energy does not flow through the outer capacitors the middle capacitor will not be charged. If energy does flow through the outer capacitors the middle capacitor will be charged.

At no point in time any energy flows through the capacitor.

What I wrote was not a question. It follows directly from what you wrote. If you contradict it, you contradict yourself.

[Sredni]

Now, I am well aware that after the Faraday vs. Kirchhoff threads I am the last person who should point this out but...

This short video summarize perfectly this thread:



As for the ongoing discussion about the role of the wires, I would like to point out the role of surface charges.
To simplify: we need surface charge of different polarities at the resistor ends. The surface charge distribution will cause conduction electrons inside the resistor to be accelerated and then collide with the ion lattice, releasing heat (classical ED). Now, why don't the electrons that arrive on the other side of the resistor cancel the positive charge on the surface on that side? Because we have the wire to scoop them away (and to replenish on the other side before they enter).

The role of the wires is that to provide the boundary conditions that will keep the surface charge distribution at the resistor. And what makes the charge distribute on the surface of the conductors and resistor? The fields all around them. The electric field inside is the result of the surface charge distribution.

[hamster_nz]

Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.

The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.

If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.

Pipes are essential to the system working, even though they carry no usable energy.

The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.

I guess I need to be more explicit as you did not understood any of my compressed air analogies.

Air molecules are the electrons and compressed air cylinder with with two chambers represents the capacitor.
A discharged capacitor is like an air cylinder that has the same pressure in both chambers say atmospheric pressure but it can be any value as long as the same number of air particles are found in both chambers and inside the hose's.

Now you can charge the capacitor using say mechanical energy to pump air from one chamber to the other one. You can use the same pump in reverse to do work with the stored energy.
Since now there are more air molecules (electrons) in one chamber (capacitor plate) than the other one you have stored energy.
If you connect the two chambers through a pipe air molecules (electrons) will flow (electric current) from the higher pressure chamber to the other one until pressure (voltage) is equalized.

What? This was in reply to your post:

It is like saying that energy from a compressed air tank is delivered to a compressed air tool outside the hose.

I was pointing out a system where a vacuum tank and pipes can deliver energy to an air-powered tool, using the air outside of the hose, by literally delivering nothing in the pipes.

You just don't like it because the open air molecules have no 'special' added energy, and they also have no 'special' added energy when in the vacuum pipes on the way back to the tank & pump, pipes are required, those pipes are empty, and yet somehow still deliver energy to the tool.

[electrodacus]

What I wrote was not a question. It follows directly from what you wrote. If you contradict it, you contradict yourself.

It can not follow from what I wrote as I never wrote that energy flows through a capacitor but energy flows in or out of a capacitor.
Is like you are unable to see a certain color (that color will be called energy storage).

[electrodacus]

I was pointing out a system where a vacuum tank and pipes can deliver energy to an air-powered tool, using the air outside of the hose, by literally delivering nothing in the pipes.

You just don't like it because the open air molecules have no 'special' added energy, and they also have no 'special' added energy when in the vacuum pipes on the way back to the tank & pump, pipes are required, those pipes are empty, and yet somehow still deliver energy to the tool.

All analogies have limitations and if I used an analogy then it was used to explain just a single aspect and you can not go with the analogy further.
That vacuum pipe will carry air molecules that are taken from ambient so you just reversed where the source of air particle is the polarity.
There is a delta in pressure between the vacuum (absence of air molecules) and ambient. So you vacuum cylinder will be filled with air molecules at same concentration as ambient when all energy is used up.
The vacuum analogy is not great for a capacitor as you can never remove all electrons from a plate not even close.
So your empty pipe is no longer empty when it delivers energy.

[SiliconWizard]

It's been two threads and things are still running in circles.

While I do not agree with everything electrodacus has written, I understand his question here: I don't think I've really seen a proper answer (except possibly in the early stages of the first thread), neither on here or in Veritasium's videos, about the role of wires and why we actually needed them (but we do for sure.)

Wires are wave guides.

Fields emanate from charges - charges live in conductors and the field disturbances carry the energy which influences nearby charges or charges in other conductors. I posted a video a few replies ago that illustrates this. And Kraus in Chapter 10 makes this explicit when he describes the Poynting Vector in circuit theory terms in a chapter about wave guides (including showing what happens when an infinite conducting sheet is placed between battery and load and how this changes the shape of the Poynting vector field).

And Feynman includes a lecture on this:
https://www.feynmanlectures.caltech.edu/II_24.html

IIRC, I did post a link to this very lecture quite a while ago.

I suggest anyone not fricking tired of it already to go read the 80 pages of the other thread and then this one just to 1/ learn a few things and 2/ realize how this has been running in circles for ages now. I'm sure it can keep going almost forever though. It's almost as though perpetual motion has finally been resolved.

For a wire carrying a DC current it's a slightly different story.

It is, and as we discussed a long time ago already, the unfortunate part of the whole Veritasium example (not the theory itself) is that the transient phenomenon upon closing the switch and the operation at steady state have been blurred somewhat. I don't think this point has been fully addressed in the second video. But the problem is that having a transient phase is the only straightforward way of making the point. *Insert coin*

[electrodacus]

It is, and as we discussed a long time ago already, the unfortunate part of the whole Veritasium example (not the theory itself) is that the transient phenomenon upon closing the switch and the operation at steady state have been blurred somewhat. I don't think this point has been fully addressed in the second video. But the problem is that having a transient phase is the only straightforward way of making the point. *Insert coin*

The claim Derek (Veritasium) made that "energy doesn't travel through wires" is more than ridiculous.
His so called "evidence" was to show that some energy arrives at the lamp before the electron wave has the time to travel the entire wire distance.
For anyone that properly understand what a capacitor is the obvious answer to that small amount energy that arrives earlier is the loss in wires/lamp while charging that capacitance and it is accurately described and calculated by the lumped model.
You can get rid of this capacitance charging through the lamp by shielding just one side of the circuit bypassing the lamp then it can be seen that no energy arrives at the lamp before the electron wave gets there.
This will show that capacitance charging is what produces an electron flow through the lamp and also shows that energy is delivered through wires else no energy could arrive at the lamp as it is shielded from electric field.

I need to apologise for being a bit rude in some of my replayes as most of you rely on you learned from others and unfortunately the way this is teached in schools is not up to current understanding.
I do not think it is OK for Maxwell to still be relevant when at his time the structure of the atom was not discovered. Yes the equations still work but the conclusions that can be drawn from them can be very misleading.
To be fair to schools I do not think any engineering school teaches that energy travels outside the wires but it also not makes that clear enough so that a youtube celebrity can not convince them the opposite is true.

[PlainName]

You can get rid of this capacitance charging through the lamp by shielding just one side of the circuit bypassing the lamp then it can be seen that no energy arrives at the lamp before the electron wave gets there.

You mean "then it should show". Unless you've actually done this then it is supposition, but you continue to pretend things like this are fact. That's one of the things that rubs people up the wrong way.

[snarkysparky]

""It can not follow from what I wrote as I never wrote that energy flows through a capacitor but energy flows in or out of a capacitor.""

So can energy both flow in and out of a capacitor at the same time ?

[electrodacus]

You mean "then it should show". Unless you've actually done this then it is supposition, but you continue to pretend things like this are fact. That's one of the things that rubs people up the wrong way.

I have not done this but I do not need to do something if I understand how it works. You can do accurate predictions about a system if you understand how it works.
You only need to shield one of the two sides of the circuit either left or right so in the Derek example just 10m of shielding from the corner (where the pipe is bent 90 degree) up to that 1.1k Resistor but you connect the shield on the other side of the resistor not on the same side you shielded.
Then there will still be a capacitance between the shield and the lower wire/pipe but the current to charge that capacitance will bypass the resistor as you connect the shield on the other side of the resistor.
So energy will start flowing through resistor but only after about 65ns needed for electron wave to reach the resistor through wire.

Will seeing that the above is true convince you that energy travels through wires ?

[electrodacus]

So can energy both flow in and out of a capacitor at the same time ?

No.
Energy flows in capacitor while it is being charged and stays there until you discharge the capacitor then energy flows out.
Even if you have a resistor in parallel with the capacitor energy can only flow in or out of capacitor not both at the same time.
 

[PlainName]

You can do accurate predictions about a system if you understand how it works.

It's still 'should' and not 'will', particularly when it is a) an off-the-wall suggestion no-one has done before, and b) the whole point of this mega-thread is to determine how exactly what you're going on about works.

[electrodacus]

You can do accurate predictions about a system if you understand how it works.

It's still 'should' and not 'will', particularly when it is a) an off-the-wall suggestion no-one has done before, and b) the whole point of this mega-thread is to determine how exactly what you're going on about works.

Not going to claim I'm a grammar expert because I'm not. Still I will have used will to show how sure I'm about my prediction.
b) I'm not saying anything controversial unlike Derek.  I never heard the claim that "energy doesn't travel through wires" before so he needed extraordinary evidence for such an extraordinary claim. He failed to do that with his only so called evidence the small current flow through load much earlier with not even a mention about line capacitance in first video and some mention in the second but ignored.

Seeing the question snarkysparky just posted I realized that I should have been more clear on what energy in and out means and that it was not in and out as the same time since that will just mean trough.
I meant in when charging and out when discharging and this things never happen at the exact same time.
 

[snarkysparky]

How does IanB's middle capacitor get charged if no energy can flow through the outer capacitors.

[electrodacus]

How does IanB's middle capacitor get charged if no energy can flow through the outer capacitors.

Two or more capacitors in series are no different from a single capacitor.
You can imagine a two parallel plates capacitor and you insert another plate in between those plates and that plate is not connected to anything and it will not affect the functionality or even capacity of that capacitor if plate is thin enough compared to the gap.
Inserting that unconnected plate will make that system a two capacitor in series setup and adding another plate will be equivalent with three capacitors in series.

a) -|     |- 
b) -|  |  |- 
c) -| | | |-

a) 100pF single capacitor.
b) two 200pF capacitors in series.
c) three 300pF capacitors in series.

For the unconnected plates in the middle all that will happen is that free electrons already in those plates when capacitor is charged or discharged will move from one side of the plate to the other.

[TimFox]

An interesting variation on case B is if you use double-sided aluminized Mylar for the center plate.
Back in grad school, when some of my buddies were building wire chambers for high-energy physics experiments, we considered using the tooling to build large electrostatic speakers, but the engineer in charge warned us that the double-sided Mylar film used in the chambers would burn out if we tried to use them for the membrane, since charge would have to flow through a very thin film to get from one side to the other as the membrane moved.

[electrodacus]

An interesting variation on case B is if you use double-sided aluminized Mylar for the center plate.
Back in grad school, when some of my buddies were building wire chambers for high-energy physics experiments, we considered using the tooling to build large electrostatic speakers, but the engineer in charge warned us that the double-sided Mylar film used in the chambers would burn out if we tried to use them for the membrane, since charge would have to flow through a very thin film to get from one side to the other as the membrane moved.

There will have been no reason for that not to work with low power. The wires mesh plates should have been fairly close to be an effective speaker but he should have allowed you to test this if nothing else will have been damaged other than worst case some inexpensive mylar tape.

[snarkysparky]

""For the unconnected plates in the middle all that will happen is that free electrons already in those plates when capacitor is charged or discharged will move from one side of the plate to the other.""

Nope.  Not the same.  The unconnected plates will have no net change in charge levels.

The plates on the middle capacitor will gain charge on one side and lose charge on the other.  The proof is that you can remove the middle capacitor after applying voltage to the series set and it will have a charge on  it.

How did that energy get there.

[electrodacus]

""For the unconnected plates in the middle all that will happen is that free electrons already in those plates when capacitor is charged or discharged will move from one side of the plate to the other.""

Nope.  Not the same.  The unconnected plates will have no net change in charge levels.

The plates on the middle capacitor will gain charge on one side and lose charge on the other.  The proof is that you can remove the middle capacitor after applying voltage to the series set and it will have a charge on  it.

How did that energy get there.

Yes the unconnected plates will have no net charge but you can charge the setup c) and after you charged that you can connect a lamp between the two unconnected internal plates and discharge the middle capacitor.

So it is exactly the same thing the difference is that each unconnected plate is in case of capacitors made of two plates connected through a wire so charge moves from one plate to the other trough that wire instead of directly from one side of the plate to the other.
That group of two plates connected by a wire is the same as a single plate and there will be no net charge in either case.

[PlainName]

Not going to claim I'm a grammar expert because I'm not. Still I will have used will to show how sure I'm about my prediction.

The problem there is you are using a circular argument: my theory says X will do Y, and X doing Y proves my theory. Which is fine if you show that X does indeed do Y, but otherwise it should do Y. That's still a 100% confidence on your part, but you are accepting reality (that no-one's seen it yet, is it is supposition regardless of how entrenched the idea is).

b) I'm not saying anything controversial unlike Derek.  I never heard the claim that "energy doesn't travel through wires" before so he needed extraordinary evidence for such an extraordinary claim.

It's not that controversial - Derek is only putting forward something that has been figured out for a while. But the difference here is that he did the actual experiment and the result was as he predicted. That's a lot better than just saying "it will" and then expecting it to be taken on trust but those that are doubting your theory in the first place.

He failed to do that with his only so called evidence the small current flow through load much earlier with not even a mention about line capacitance in first video and some mention in the second but ignored.

Well, that is what this thread is supposed to figuring out. Simply bashing your opponent with a circular argument before diverting to some irrelevant stuff isn't doing a great job of disproving what he thinks he's shown.

[snarkysparky]

Just answer this.

In the three capacitor setup where did the energy come from that charged the middle capacitor.

There can be no doubt that disconnecting the circuit and taking the middle capacitor out it will have a charge.

If you remove the two middle plates they will not have a charge.  Because they were isolated.

[electrodacus]

The problem there is you are using a circular argument: my theory says X will do Y, and X doing Y proves my theory. Which is fine if you show that X does indeed do Y, but otherwise it should do Y. That's still a 100% confidence on your part, but you are accepting reality (that no-one's seen it yet, is it is supposition regardless of how entrenched the idea is).

It is not my theory (I did not come up with it). It is how things work.
Many people (including me) seen this and use this to do shielding from electric fields.

It's not that controversial - Derek is only putting forward something that has been figured out for a while. But the difference here is that he did the actual experiment and the result was as he predicted. That's a lot better than just saying "it will" and then expecting it to be taken on trust but those that are doubting your theory in the first place.

Many people including me could have predicted the result. That result is not representing what he claimed "energy doesn't travel in wires"

Well, that is what this thread is supposed to figuring out. Simply bashing your opponent with a circular argument before diverting to some irrelevant stuff isn't doing a great job of disproving what he thinks he's shown.

So you mean line capacitance is irrelevant ? Or the fact that a capacitor is an energy storage device.

To be clear are you saying that shielding the way I mentioned will not eliminate the energy through lamp/resistor in the first 65ns in Derek's setup ?
If this is demonstrated to you will you then admit that energy doesn't travel outside the wire?

[electrodacus]

Just answer this.

In the three capacitor setup where did the energy come from that charged the middle capacitor.

There can be no doubt that disconnecting the circuit and taking the middle capacitor out it will have a charge.

If you remove the two middle plates they will not have a charge.  Because they were isolated.

The two or three capacitors in series are just one capacitor.
Yes in case c if you remove the middle plates they will not have a charge but that is not what you are doing when you remove the middle capacitor.
You first isolate the charges by disconnecting the capacitor pins the remove the middle capacitor.

d) -| |_| |_| |-

So d) is the same as c) but you can disconnect one face of the plate from the other by disconnecting the wire connecting the two faces.
Then those isolated middle faces will contain 33.33% of the energy that was put in those 3 capacitors in series.

[PlainName]

To be clear are you saying that shielding the way I mentioned will not eliminate the energy through lamp/resistor in the first 65ns in Derek's setup ?

To be Frank I have no idea since I've given up following your diversions so haven't really looked at it. Also it seems a pretty pointless thing to consider since you're unlikely to replicate Derek's experiment with whatever shielding, so it just comes back to what you say would occur rather than what actually does.