Veritasium "How Electricity Actually Works"

[SiliconWizard]

While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.

[electrodacus]

So then how does the energy pass across the empty space between the switch and the lamp in the initial transient few nanoseconds?

How does it get from the bottom two wires to the top two wires?

"Capacitive coupling"  doesn't equate to the energy only flows in wires.

Replace the battery (source) with a charged capacitor so just imagine a simple two plate capacitor that has excess of electrons on one plate and a deficit of electrons on the other.
Then have two other much smaller capacity capacitors that are empty (no stored energy) in series with a light bulb or resistor (same thing).
That group of two capacitors and a resistor all in series can be equivalent with a single capacitor so when you connect (using wires) that to the charged capacitor is like modifying the existing charged capacitor by increasing his capacity thus charges will be redistributed.
If you can imagine all that you will notice that there was no flow outside wires (this includes the capacitor electrodes).

You can also imagine just a charged capacitor on witch you increase the plate area. While you do so the energy stored will remain the same for ideal case while in real case part of it is lost due to plates having a non zero resistance so when electrons move to rearrange you will have some IR losses.
All energy travels trough the plates in this case so trough conductors/wires.   

[electrodacus]

While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.

There is no chicken-and-egg problem as there you ask witch one was first but here you start with either the chicken or the egg so that problem of what was first is non existent.

[rfeecs]

So then how does the energy pass across the empty space between the switch and the lamp in the initial transient few nanoseconds?

How does it get from the bottom two wires to the top two wires?

"Capacitive coupling"  doesn't equate to the energy only flows in wires.

... All energy travels trough the plates in this case so trough conductors/wires.

Apparently you mean current, not energy.

[electrodacus]

Apparently you mean current, not energy.

If there is no current there is no energy.
There can be current and no energy flow if current flow is constant and it travels trough a superconductor loop.

I think that the charged capacitor where you change the size of the plates is the most simplified example I can think of that represents the initial few moments after the switch is closed.

[rfeecs]

Apparently you mean current, not energy.

If there is no current there is no energy.

So there is no energy in electromagnetic waves?

There's no energy in a laser beam?

No energy in sunlight?

Then how does solar energy work?  You seem to be in that business.

[rfeecs]

While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.

Something like do moving charges cause fields or do fields cause charges to move?

Perhaps he addressed that in saying electrons don't push each other.

He comes down on the side of fields cause charges to move.

[rfeecs]

I don't think he adequately debunks the idea that charges gain potential energy (V) and then that potential energy is converted to kinetic energy.  The P=IV model.

He doesn't mention potential energy, just kinetic energy.  He argues that the electrons drift too slowly to carry the energy.

But the (retarded) potential moves at the speed of light.

I imagine he didn't want to mention potential because it introduces another concept that might be complicated to explain.

[electrodacus]

So there is no energy in electromagnetic waves?

There's no energy in a laser beam?

No energy in sunlight?

Then how does solar energy work?  You seem to be in that business.

The resistor (lamp) is not powered by lasers or sunlight.  Such a lamp powered by laser or sunlight will be a a mirror.

A capacitor and an inductor are energy storage devices.
Storing energy is not the same with using energy to do work (convert one type of energy in to another).

If the ends of that loop in Derek's video where open then whose long parallel wires will only be a capacitor so when switch will have been closed that capacitor will have charged the exact same way it happened with the closed loop tested.
The difference will have been that after those few ns of charging this capacitor there will be no current flow thus no voltage drop on that 1.1kOhm resistor and so no power available for the resistor/lamp thus no energy.
The capacitor will have remained charged so if you look at the balance of energy you will see this

Energy delivered from battery =  IR loss on the internal battery ESR + IR loss in wires + IR loss in the 1.1kOhm resistor + energy stored in the capacitor made by the parallel wires.

[rfeecs]

Energy delivered from battery =  IR loss on the internal battery ESR + IR loss in wires + IR loss in the 1.1kOhm resistor + energy stored in the capacitor made by the parallel wires.

You are using a lumped model that is not adequate.

This circuit is three dimensional.  There is a one meter gap between the switch and the load.  You haven't explained how energy gets across that gap.

You can look at the simulation in the video.  When the switch is closed a spherical wave propagates out in all directions.  It should be very clear that some of that energy is going out away from the bulb.  Some of the energy is lost.  Your model doesn't account for the energy radiated away.

Consequently your equation is wrong.

[SiliconWizard]

While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.

Something like do moving charges cause fields or do fields cause charges to move?

That's one of them, yes.

Perhaps he addressed that in saying electrons don't push each other.
He comes down on the side of fields cause charges to move.

So, if he said so, that must be settled?

To extend the matter a bit, what about gravity?

[electrodacus]

You are using a lumped model that is not adequate.

This circuit is three dimensional.  There is a one meter gap between the switch and the load.  You haven't explained how energy gets across that gap.

You can look at the simulation in the video.  When the switch is closed a spherical wave propagates out in all directions.  It should be very clear that some of that energy is going out away from the bulb.  Some of the energy is lost.  Your model doesn't account for the energy radiated away.

Consequently your equation is wrong.

There is no energy radiated away. This is a copy paste of a replay I just made to someone else but is applies to your question also.

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.

[rfeecs]

There is no energy radiated away.

I disagree.   

I wonder how you think radios work.  By energy storage I suppose.

[electrodacus]

I disagree.   

I wonder how you think radios work.  By energy storage I suppose.

Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.

The circuit simplifies to charging an empty capacitor with a charged capacitor.  If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.

[rfeecs]

I disagree.   

I wonder how you think radios work.  By energy storage I suppose.

Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.

The circuit simplifies to charging an empty capacitor with a charged capacitor.  If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.

I have already done that.  I have said there is radiation.  You say there isn't.  Enough said.

The simulation shows what happens.

If I wanted to model it, I would start where the battery/switch is connected to two wires.  These wires form an antenna, or an odd looking transmission line.  They can be thought of as a skinny bi-cone antenna.  The infinite bi-cone looks like a transmission line and has constant impedance.  It has spherical symmetry and the wave propagates out spherically.  In this case, the wires are not conical but straight, but the propagation is approximately spherical and the impedance will change along the line but levels off to a slow increase in impedance.

When the switch closes, there is a transient voltage change, as in a Heaviside step function.  This transient is what starts the energy propagating out in all directions, roughly spherically.

After a period of time, the wave front hits the top pair of wires.  Now we have another antenna / transmission line.  The electric field across the load will cause a current and voltage wave to propagate along this line in a similar manner as the source antenna.  Clearly the signal is much smaller because the field has spread out spherically. 

The two antennae are clearly coupled and form another set of transmission lines, the twin line that has been often mentioned.  So to properly model this I would consider these to be coupled lines.  The odd mode impedance of the twin line is well known.  The even mode impedance will be formed by the bi-cone type lines.

Yes, this model is ridiculously more complicated that your capacitor model.  But it can model the fact that the lower pair of wires initially have a higher current than the upper pair of wires and give the correct current for both wires.

Your model cannot do this.

[hamster_nz]

I disagree.   

I wonder how you think radios work.  By energy storage I suppose.

Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.

The circuit simplifies to charging an empty capacitor with a charged capacitor.  If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.

Is what you are saying "If we pick a simplified model, it simplifies to this simple model, that doesn't explain the details of the physical results, but the model agrees with the model, so the model is good".

I'm wait for the "I'll add in ESR resistors and parasitic inductance to take my first-order approximation to a second order approximation, so we get virtual coils and resistors along with the virtual capacitors I've added because my model demands it, not because there are actual coils and capacitors in the experimental apparatus."

[Naej]

I don't think he adequately debunks the idea that charges gain potential energy (V) and then that potential energy is converted to kinetic energy.  The P=IV model.

He doesn't mention potential energy, just kinetic energy.  He argues that the electrons drift too slowly to carry the energy.

But the (retarded) potential moves at the speed of light.

I imagine he didn't want to mention potential because it introduces another concept that might be complicated to explain.

Electron's potential energy is converted into heat in a resistor.
And he didn't debunk it because it is true.

[rfeecs]

I don't think he adequately debunks the idea that charges gain potential energy (V) and then that potential energy is converted to kinetic energy.  The P=IV model.

He doesn't mention potential energy, just kinetic energy.  He argues that the electrons drift too slowly to carry the energy.

But the (retarded) potential moves at the speed of light.

I imagine he didn't want to mention potential because it introduces another concept that might be complicated to explain.

Electron's potential energy is converted into heat in a resistor.
And he didn't debunk it because it is true.

Yes, of course it's true.  But his "misconception number 1" is "electrons carry energy from battery to bulb".  He tries to prove this without mentioning potential energy which seems to be part of the most common explanation.

[electrodacus]

I have already done that.  I have said there is radiation.  You say there isn't.  Enough said.

The simulation shows what happens.

If I wanted to model it, I would start where the battery/switch is connected to two wires.  These wires form an antenna, or an odd looking transmission line.  They can be thought of as a skinny bi-cone antenna.  The infinite bi-cone looks like a transmission line and has constant impedance.  It has spherical symmetry and the wave propagates out spherically.  In this case, the wires are not conical but straight, but the propagation is approximately spherical and the impedance will change along the line but levels off to a slow increase in impedance.

When the switch closes, there is a transient voltage change, as in a Heaviside step function.  This transient is what starts the energy propagating out in all directions, roughly spherically.

After a period of time, the wave front hits the top pair of wires.  Now we have another antenna / transmission line.  The electric field across the load will cause a current and voltage wave to propagate along this line in a similar manner as the source antenna.  Clearly the signal is much smaller because the field has spread out spherically. 

The two antennae are clearly coupled and form another set of transmission lines, the twin line that has been often mentioned.  So to properly model this I would consider these to be coupled lines.  The odd mode impedance of the twin line is well known.  The even mode impedance will be formed by the bi-cone type lines.

Yes, this model is ridiculously more complicated that your capacitor model.  But it can model the fact that the lower pair of wires initially have a lower current than the upper pair of wires and give the correct current for both wires.

Your model cannot do this.

We are discussing about a field and you mention radiation. The infrared radiation for example from a lamp or resistor or from battery is not what transfers the energy from battery to lamp.


Do you agree that battery (or charged capacitor) is the only energy source in that example ?
If you do then just move the switch far from the battery and lamp/load then see what is the time required for the lamp to see any current.
You can not claim that energy comes from the switch if you agreed that battery/charged capacitor is the only source.

When the switch closes it just connects one of the plate of the charged capacitor to a discharged capacitor (transmission line).
There is no electric field before closing the switch and first electrons will move in to the wire or out of the wire (depending witch plate you connected) and just because of that you will start to have an electric field.

Here is the super simplified example

  ________________ common plate
  +++++++

  _______/  ________
  -----------


This simplified example of Derek's experiment.
Those are real capacitors so the plates have both resistance and inductance. Capacitor on the left is the charged one while the one on the right is discharged.
What do you think happens when you close the switch ?

[rfeecs]

What do you think happens when you close the switch ?

 

[hamster_nz]

Here is the super simplified example

  ________________ common plate
  +++++++

  _______/  ________
  -----------


This simplified example of Derek's experiment.
Those are real capacitors so the plates have both resistance and inductance. Capacitor on the left is the charged one while the one on the right is discharged.
What do you think happens when you close the switch ?

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.

[electrodacus]

What do you think happens when you close the switch ?

 

Here is the exact example if you do not like the simplification

  ________________________________________________
  +++++++++++++++++++++              ___________________
                                                1KOhm___________________
  _____________________/   _________________________
  -----------------------------------



Those are real capacitors so the plates have both resistance and inductance. Capacitor on the left is the charged the equivalent of the battery and the two capacitors on the right are representing the transmission line and they are in series with 1KOhm resistor. The ends are not closed as is not relevant for the first few ns.
What do you think happens when you close the switch ?

[hamster_nz]

Here is the exact example if you do not like the simplification

Can you please upgrade to at least crayons and paper? Because this is what the rest of us are looking at:

[electrodacus]

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.

[electrodacus]

Here is the exact example if you do not like the simplification

Can you please upgrade to at least crayons and paper? Because this is what the rest of us are looking at:

Sorry I did not expect the forum will scale with monitor resolution (you likely watch this on your phone).
I made is much shorter hopefully it will fit.