Veritasium "How Electricity Actually Works"

[hamster_nz]

You say "There is no electric field before closing the switch". I don't think this is true. Here is the same diagram with two negative charges electrons sitting between the plates of your capacitor(s). What way do they move? and why do they move, and what provides the energy for them to move?


  ________________ common plate
  +  +  +  +  +  +  +

     -                 -

  _______/  ________
  -----------

Also the positive charges are way too close together. Their mutual repulsion will cause them to spread out over the plate pretty much uniformly.

Also the unconnected wire will have potentials at either end, as one end is closer to a large static charge.

That is incorrect and I noticed the same wrong explanation in Derek's video. The electrons and holes will be equal and they will extend just up to the switch even on the common plate.
Only after the switch is closed electrons and holes will move symmetrically on top and bottom plate.
Keep in mind the drawing is nowhere near to scale. There may be just 0.1mm between plates and plates may be a few meters long for the charged capacitor and then for the discharged capacitor that represents the transmission line the distance between plates may be order of magnitude higher but even if the same 0.1mm there charge will end at the switch both on top and bottom plates.

Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...

You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).

[electrodacus]

Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...

You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).

Yes that is exactly what I'm saying and I'm fairly sure that I'm correct. They will move across the switch when switch is closed and symmetrically on the top plate.
After the right side is charged they will be also uniformly distributed across the entire new capacitor that has now higher capacity.
This is assuming that distance between plates is the same on the left side and right side else if say on the right side the distance between plates is larger (lower capacity) there will be proportionally less electrons on that side.

[hamster_nz]

Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...

You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).

Yes that is exactly what I'm saying and I'm fairly sure that I'm correct. They will move across the switch when switch is closed and symmetrically on the top plate.
After the right side is charged they will be also uniformly distributed across the entire new capacitor that has now higher capacity.
This is assuming that distance between plates is the same on the left side and right side else if say on the right side the distance between plates is larger (lower capacity) there will be proportionally less electrons on that side.

No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.

 ________________ common plate
 +++++++

     -                 -   << These guys which way do the move and why.

  _______/  ________
  -----------

If they are free to move, which way to they go and why?

[electrodacus]

No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.

 ________________ common plate
 +++++++

     -                 -   << These guys which way do the move and why.

  _______/  ________
  -----------

If they are free to move, which way to they go and why?

Not sure why you had to draw those ones as they are already on the diagram. And yes they are incorrectly drawn under the bottom plate (limitations of this CAD tool ).

The ++++ and ------ symbols in my diagram represents holes and electrons in the plates/wires as that is what is of interest in this discussion.
They will move when switch is closed and they will move symmetrically on the top and bottom plate's.

To maybe help you understand that there is symmetry between the charges on the top and bottom plate imagine there is also a switch on the top plate exactly above the one on the bottom and now thing what will happen if you start with both switches open and only close the top one.
Do you think there will be any current trough that closed top switch ? If not that means charges remain arranged the same as they where before you closed that switch.

[hamster_nz]

No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.

 ________________ common plate
 +++++++

     -                 -   << These guys which way do the move and why.

  _______/  ________
  -----------

If they are free to move, which way to they go and why?

Not sure why you had to draw those ones as they are already on the diagram. And yes they are incorrectly drawn under the bottom plate (limitations of this CAD tool ).

The ++++ and ------ symbols in my diagram represents holes and electrons in the plates/wires as that is what is of interest in this discussion.
They will move when switch is closed and they will move symmetrically on the top and bottom plate's.

To maybe help you understand that there is symmetry between the charges on the top and bottom plate imagine there is also a switch on the top plate exactly above the one on the bottom and now thing what will happen if you start with both switches open and only close the top one.
Do you think there will be any current trough that closed top switch ? If not that means charges remain arranged the same as they where before you closed that switch.

They will not move symmetrically, because there is no symmetry to begin with.

Electrostatics is relatively easy so I wrote a quick solver for distribution of 50 + charges and 50 - charges on 10m and 5m wires one 1m apart. (so a 10:1 aspect ratio).

You can make what you want of this, but the net horizontal force on each charge (other than those at the ends of the 'wires', where they are constrained) is zero.

[electrodacus]

They will not move symmetrically, because there is no symmetry to begin with.

Electrostatics is relatively easy so I wrote a quick solver for distribution of 50 + charges and 50 - charges on 10m and 5m wires one 1m apart. (so a 10:1 aspect ratio).

You can make what you want of this, but the net horizontal force on each charge (other than those at the ends of the 'wires', where they are constrained) is zero.

You do not call that symmetry ?
Do you understand that my drawings where not done at scale?
Make the same again with 5m and 10m plates but 1mm apart (is even much lower than 1mm for any typical capacitor).
Or if you want to be even more realistic make the first 5m that overlap 0.1mm apart and then the remaining 5m 0.5m from the center as that will be the transmission line.

Still I feel that you try to look at other aspects only not to talk about the main one. No electron movement no current.
When you close the bottom switch is when the electrons will start to move and you will have energy transfer from the source (charged capacitor) to the load in this case a discharged capacitor that is the equivalent of the transmission line (so yes it also have some inductance and resistance).


Edit: looking closer at your charge density distribution it is still no where close to accurate for those dimensions.
Edit 2: and just to be clear there are free electrons in any conductor so even a discharged capacitor will have free electrons in both plates the difference is that a discharged capacitor will have the exact same density of free electrons in both plates.
In a charged capacitor what you will represent is just the excess on one plate and deficit on the other plate and they will be symmetrical in all aspects.
The electrons and holes want to be as close as possible to each other.

[hamster_nz]

Yes, one capacitor can charge another, and there is energy that can be taken from the system, one way or another (e.g. heating of components). Does it have any utility here? No, not that I can see.

In fact, I've lost track of exactly what you were trying to prove... I only joined in again as you were commenting that there was no electric field before the switch was closed, when there clearly one is present.

Can't be bothered. Enjoy whatever.

[electrodacus]

Yes, one capacitor can charge another, and there is energy that can be taken from the system, one way or another (e.g. heating of components). Does it have any utility here? No, not that I can see.

In fact, I've lost track of exactly what you were trying to prove... I only joined in again as you were commenting that there was no electric field before the switch was closed, when there clearly one is present.

Can't be bothered. Enjoy whatever.

There is electric field in the charged capacitor (the one representing the battery)
There is no electric field in the discharged capacitor.

Edit: A good visualization of the capacitor

[hamster_nz]

So I set this up with three meters, two caps and a switch on the bench.

You have the switch open. You short out all the caps so everything is at 0V.

Code: [Select]

    +------------+
    |            |
  ---- 0V      ---- 0V
  ----         ----
    |            |
Gnd +----/ ------+
You charge up the capacitor on the left hand side, Gnd to the switch side, +10V to the wire common between the two capacitors.

You now have 10v across the left hand side cap, 0V across the right hand side cap, and 10V over the switch terminals .

Code: [Select]

    +------------+
    |            |
  ---- 10V     ---- 0V
  ----         ----
    |            |
GND +----/ ------+
          10V

Q1) How did that 10V measured over the switch get there  ?  My suggestion: as no charge has moved across the capacitor, no current is moving so it can't be magnetic. So unless I want to add a new fundamental field it must be electric field across the capacitor. This has caused charges to move around to balance out the electric field, until there is zero voltage gradient over the capacitor, resulting in 10V over the opened switch.

I've measured this all on the bench - there is 10V across the switch.

You close the switch a small spark is heard and the 10V of potential difference is now gone.

Code: [Select]

    +------------+
    |            |
  ---- 5V      ---- 5V
  ----         ----
    |            |
GND +------------+
           0V
You now have 5V across both caps - and half the energy goes out of the system as heat.

What part am I missing?

[EEVblog]

Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED

[EEVblog]

Dereks's video at 21:10 Re. Rick Hartley about fields is 100% correct for high speed PCB design. But that does NOT apply at DC, not at all, not even one tiny bit.

[EEVblog]

I posted this comment on Derek's video, slighly expanded:

Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?

The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.

Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.

[bsfeechannel]

Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED

Double-posting, Dave?

https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432

Anyway, I commented there and won't repeat it here.

The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.

Think for a bit. If you are convinced that energy doesn't flow in the wires for AC (or RF for that matter), why would it for DC?. Doesn't it take more energy to push electrons to and fro than to let them go indefinitely in the same direction?

[EEVblog]

Double-posting, Dave?

https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432

Yes, because there are now two threads, and many people are not reading the other thread any more.

[T3sl4co1l]

I posted this comment on Derek's video, slighly expanded:

Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?

The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.

Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.

Easy:

1. The given problem is a dynamics problem.  Therefore we're only concerned with AC behavior.

2. He does discuss field within a conductor (though not how it gets there -- skin effect bridges the "AC" and "DC" regimes, as it turns out).  And we can solve for the energy density of that field.  It is terribly small in comparison, but nonzero.

The fact that most of the system's energy is stored in the magnetic field around the wires, hints further that "energy flows outside the wires", but this is an interpretation, and one can take either point; they're equivalent, once everything's quiescent, i.e. you can solve from one given the other (and material properties, boundary conditions, all that).

That is, the E-field inside the wire, given its resistance, tells the current density, and the wire diameter give the total current, and the wire placement is given, so the magnetic field can be solved.

Tim

[snarkysparky]

""The given problem is a dynamics problem.  Therefore we're only concerned with AC behavior.""

Does the video make this clear.  I missed it.

He does a poor job of separating the transient from the steady state conditions

[Sredni]

Double-posting, Dave?

https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432

Yes, because there are now two threads, and many people are not reading the other thread any more.

So let me double post this from the other thread, because I have yet to find a satisfying answer:

Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):

I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat.  Let's say I 'generated' 1 joule of energy.

Has this energy traveled along the 1000 meters path?

What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?

Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?

[electrodacus]

Q1) How did that 10V measured over the switch get there  ?  My suggestion: as no charge has moved across the capacitor, no current is moving so it can't be magnetic. So unless I want to add a new fundamental field it must be electric field across the capacitor. This has caused charges to move around to balance out the electric field, until there is zero voltage gradient over the capacitor, resulting in 10V over the opened switch.

You closed the switch just not the one you are concentrating at. When you connected the multimeter you basically closed a switch that is in series with 1MOhm resistor so electrons will flow from the charged capacitor to the discharged one trough your multimeter.

I've measured this all on the bench - there is 10V across the switch.

You close the switch a small spark is heard and the 10V of potential difference is now gone.

Code: [Select]

    +------------+
    |            |
  ---- 5V      ---- 5V
  ----         ----
    |            |
GND +------------+
           0V
You now have 5V across both caps - and half the energy goes out of the system as heat.

What part am I missing?

You are missing the ESR.   The electron wave can move at the speed of light and how many of them can flow per unit of time depends on resistance in this case mostly the ESR and some of your wire resistance.
The reason you do not have 7.07V at the end of the experiment so same energy is because half of the energy was dissipated as heat in the ESR + wires.
If you could measure the temperature accurate enough you will see that capacitors and wires heated up by the exact amount representing half the original energy.

[rfeecs]

Hontas Farmer is back still saying the Derek is both right and wrong according to QFT/QED

She says QED is in fact simpler than classical field theory?

What if you need to actually calculate something for a macro system like this?  Given the dimensions of the wires (diameter, spacing), the input waveform (step function), the load impedance, what is the current at the load and the source (vs time)?

Clearly you can calculate this with classical field theory.  You can calculate the impedances from the dimensions.  You can do an accurate simulation.

She says for this problem, using classical theory is even harder.  You have to do all sorts of computer calculations.  But for QED its just probability.  Easy.

The fact is with classical theory you can actually get a numerical answer that you can use.  Can you do that with QED?  Or does the complexity immediately get out of hand for even a much simpler system, like a single molecule?  Here's a video discussing the difficulty of that:



There are lots of models.  Each has limitations.  Use the right tool for the job.

[T3sl4co1l]

So let me double post this from the other thread, because I have yet to find a satisfying answer:

Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):

I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat.  Let's say I 'generated' 1 joule of energy.

Has this energy traveled along the 1000 meters path?

What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?

Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?

Potential energy is just, what it is.  It's a property of the mass's altitude (in this case, or velocity too if you want to count that), not some internal state that comes along for the ride.  It's not like it's got a battery in it.   However, if you include yourself in the system, you're a battery (of sorts), that's transporting energy.

Or if you have a definition to apply it to, like: "transporting energy" sounds like integral path length times potential energy along that path, in which case, you'd be transporting potential energy with respect to that 10m drop any time you're above its floor, and negative potential while below.

Tim

[electrodacus]

I posted this comment on Derek's video, slighly expanded:

Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?

The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.

Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.

In both cases DC and AC/transients all energy transfer from source to load is trough wires not outside the wires.
His test setup can be reduced to one charged capacitor (as the source instead of the battery) and one discharged capacitor (simulating the transmission line with the ends open).
Due to the fact that capacitors have an ESR half of the energy will be lost during the energy transfer from one capacitor to another and that loss will be as heat in the wires and capacitor plates that are also wires.
There is no field in the empty capacitor and the field will start to form as electrons are transferred and an imbalance of electrons is formed between the two plates.

[HuronKing]

Can we lock the other thread or something? This discussion is super confusing to read when we've got entire lines of thought being double-posted in both threads. It's super confusing to follow who is responding to what. 

"Let's see - is this the BIG thread talking about how Veritasium is right but actually wrong or the LITTLE thread talking about how Veritasium is wrong but actually right?" 

[rfeecs]

I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat.  Let's say I 'generated' 1 joule of energy.

Has this energy traveled along the 1000 meters path?

If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it.  Energy was 'dissipated'  in the volume of space around where you dropped it.  So we can say energy flowed through the space between.  But we can't say exactly the path that the energy took.

Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock.  How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.

I hate analogies.  Water analogy, rock analogy, whatever. 

[electrodacus]

I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat.  Let's say I 'generated' 1 joule of energy.

Has this energy traveled along the 1000 meters path?

If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it.  Energy was 'dissipated'  in the volume of space around where you dropped it.  So we can say energy flowed through the space between.  But we can't say exactly the path that the energy took.

Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock.  How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.

I hate analogies.  Water analogy, rock analogy, whatever. 

Yes this sort of analogies are not very useful.
Still I will try to give an answer
Before you let that mass drop there is that 1J of stored potential energy.
When you let that drop the potential energy is converted gradually in to kinetic energy (not all potential energy will end up as kinetic energy since part will be lost as heat due to friction with air).
Say 0.9J ended up as kinetic energy and if you have a perfectly non elastic collision with ground (not possible in real world) then all that kinetic energy will end up as heat (it just reminded me of a funny you tube video where someone cooked some meat using a robot hand to slap it).
If collision is elastic then not all kinetic energy will end up as heat at initial contact just a part will end up as heat the other part will be stored in compressing the mass (assuming the mass is elastic only) and then that stored energy will be converted back to kinetic energy as the mass will be accelerated back up.
In perfect elastic collision and no friction with air the mass will get back up to 1000m and since there is no loss it will be a perpetuum mobile (not possible in real life).

Yes the energy has traveled along the path. You can see with a thermal camera how jut the air that got in contact with the mass heated up and there is no heat below the mass where the mas did not get to and also no heat on the ground until the mass gets there to deliver energy same as electron needs to get there to deliver energy and electron travels trough wires thus energy is delivered trough wire.

[SiliconWizard]

If "energy doesn't flow in wires" why he even needs the wires ?

Well, as I understand it, a medium is needed to hold charges, and if there are no charges, there's no field?
The question then is more about fields making charges move rather than charges moving creating fields. Or something.