Veritasium "How Electricity Actually Works"

[snarkysparky]

""The two or three capacitors in series are just one capacitor.""

Factually they are not!

I ask to consider the actual case where there are three distinct capacitors.  And the middle one receives energy from somewhere. 

If you don't think capacitors can transmit energy you should be able to explain this.

[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.

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.

As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um).  Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes.  In our initial calculations, we had forgotten that the material available (for free) was double-sided.  We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.

[electrodacus]

""The two or three capacitors in series are just one capacitor.""

Factually they are not!

I ask to consider the actual case where there are three distinct capacitors.  And the middle one receives energy from somewhere. 

If you don't think capacitors can transmit energy you should be able to explain this.

What about two or three capacitors in parallel?  They are factually two or three in parallel but is there any difference in functionality compared to a single larger capacity capacitor ?
Same thing applies for two or three in series is just that leakage is a bit of a problem as it will not be equal so they may charge a bit inequal and if used close to the voltage limit you will usually add some parallel resistors if used this way in a circuit.

When connecting identical capacitors in parallel you increase the plate area with distance remaining the same.
When connecting identical capacitors in series you just increase the distance between plates of the equivalent capacitor that is why capacity drops.

Example c) and example d) are identical with the exception that you can disconnect the two middle plates leaving you with 3 capacitors after you do that.

e) -|  |_| |_|  |-

Say you have e) so 3 capacitors in series but the one in middle has plates closer say 2x the capacity of those at the ends.
When you disconnect those 3 capacitors after you charged them
Do you think:

1) all have the same amount of energy stored equally divided.
2) the one in the middle contains twice as much energy as each of the other ones.
3) the one in the middle contains half of the energy each of the other two contain.

[electrodacus]

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.

I put in the effort to read what each one of you has to say.  If you understand how capacitors and transmission lines work you will not need to even do the experiment to see that what I say holds true.

Likely you never seen any sort of shield that prevents a load getting energy from a source that is connected to load through wires. So you know for sure even without testing that lamp/resistor will receive energy from battery no matter what shield I create around.
The question is only if you think the shield that I proposed or any similar shield can stop energy getting to lamp/resistor in those first 65ns required for the energy to be delivered through the long wires.

Question is if this if proven true to you it is enough or you can still invent something to allow for energy transfer outside wire to still be true?

[electrodacus]

As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um).  Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes.  In our initial calculations, we had forgotten that the material available (for free) was double-sided.  We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.

That was a large sheet (not something I had in mind) and the double metalization will not have been helpful quite the opposite as mylar as outside layer will have helped reduce the risk of a discharge when sheet got to close to the grid. You also needs some sort of transformer to add audio signal.
Yes in my case b) the heat loss due to conduction is fairly small.

 

[TimFox]

As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um).  Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes.  In our initial calculations, we had forgotten that the material available (for free) was double-sided.  We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.

That was a large sheet (not something I had in mind) and the double metalization will not have been helpful quite the opposite as mylar as outside layer will have helped reduce the risk of a discharge when sheet got to close to the grid. You also needs some sort of transformer to add audio signal.
Yes in my case b) the heat loss due to conduction is fairly small.

Huge sheets:  these were large high-energy physics contraptions.  Full-size electrostatics (Quad and KLH) were common then, 6DQ5s were still in production, and lots of military-surplus high-voltage transformers were available.
A transformer is not needed if you can couple directly to the plates of the push-pull driver, using the V_bb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.

[electrodacus]

Huge sheets:  these were large high-energy physics contraptions.  Full-size electrostatics (Quad and KLH) were common then, 6DQ5s were still in production, and lots of military-surplus high-voltage transformers were available.
A transformer is not needed if you can couple directly to the plates of the push-pull driver, using the V_bb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.

Yes it can work with some high voltage push-pull mosfet or IGBT and optically isolated gate driver but is more complex than just a transformer with center tap.
I did Electrical Engineering so high voltage / high energy stuff and we had laboratories for lightning simulation tho I was more passionate about low voltage automation electronics.
As a hobby I never did projects exceeding 60Vdc open circuit the safe low voltage standard limit.
Case b) and c) where done to show that the internal plates that are fully isolated from supply have no net charge. You need to go to something like d) and then disconnect the two surfaces in order to isolate the sides with different charges.

For example the setup in Derek's test was more like
f) -| |_[resistor]_| |-
So capacitors on each side and resistor/lamp in between (that is the transient part).
So while current flows through resistor it is electrons from one plate moving to the other plate as the capacitors charge from the source.
If source is then removed and you short the capacitors all stored energy will flow back through resistor to have those plates neutral again.
So amount of energy through the resistor is directly proportional with the amount of charge being stored in the capacitors and then when they are discharged the same amount will flow back.
So electrons are the ones that travel through wire and transport the energy.
If we shield one quarter of Derek's transmission line and connect that shied on the opposite side of the resistor this capacitive energy storage will be done around the resistor so no more energy before those 65ns or so but then despite this shielding energy arrives through wires at the resistor/lamp.
And it can be DC or AC it will still get to that lamp resistor through wires despite the shield that removes the capacitive effect of the line.         

[PlainName]

I put in the effort to read what each one of you has to say.

That may be, but what you write in response to a post is usually irrelevant. It's just waffle as a diversion. For instance, just above this snarkysparky asked about three capacitors in series and clearly stated that it was the middle one that was interesting. Your response is 16 lines rabbiting on about parallel capacitors.

Either you didn't understand their short and to the point post, so just latched onto 'multiple capacitors' and made a guess at what it was about, or you deliberately obfuscated by diverting to a parallel setup (which is completely irrelevant).

Then, just to rub it in you then asked three questions designed to steer the discussion away from the series setup, and you will act miffed when those aren't answered.

[electrodacus]

I put in the effort to read what each one of you has to say.

That may be, but what you write in response to a post is usually irrelevant. It's just waffle as a diversion. For instance, just above this snarkysparky asked about three capacitors in series and clearly stated that it was the middle one that was interesting. Your response is 16 lines rabbiting on about parallel capacitors.

Either you didn't understand their short and to the point post, so just latched onto 'multiple capacitors' and made a guess at what it was about, or you deliberately obfuscated by diverting to a parallel setup (which is completely irrelevant).

Then, just to rub it in you then asked three questions designed to steer the discussion away from the series setup, and you will act miffed when those aren't answered.

What is the difference between this two ?

1) -| | | |-
2) -| |_| |_| |-

If you can not disconnect those links at 2) it is exactly the same as 1)
At 2) the electrons flow through that wire link to another plate but the net charge is zero as no electrons from battery get there.
As soon as you disconnect the links you have 3 separate and isolated capacitors and if all equal each contains 1/3 of the energy that battery delivered.
But you needed to take the action to cut the connections after the capacitor was charged.
No energy ever flowed from battery to the middle capacitor.
If you get all the middle plates out either on 1) or on 2) (without disconnecting the connections at 2) then the remaining two outer plates will contain the initial amount of energy. We will assume that the dielectric thickness remains about the same so plates have negligible thickness.

And yes I post long replays to explain how things actually work.
You still did not answered the main question for the last few posts.    How will energy travel outside wire if you shield the wire from electric fields?
And also avoided to answer if proving that to you will be sufficient evidence or will you find some other invented theory to deny that electrical energy travels through wires.

[PlainName]

And also avoided to answer if proving that to you will be sufficient evidence or will you find some other invented theory to deny that electrical energy travels through wires.

If you can prove it one way or the other I would be happy to accept the definitive explanation.

[SandyCox]

[How will energy travel outside wire if you shield the wire from electric fields?

Let's think of the following setup:
1. Place both the battery and bulb in separate sealed boxes made of a perfect conductor.
2. Replace the two cables with coaxial cable made of perfect conductors and a lossless dielectric.
3. Connect the shields of the two coaxial cables to the two sealed boxes. (No pigtails allowed)
4. Connect the inner conductors to the bulb, switch, and battery, like in Vertasium's original video.

How long will it now take for the bulb to turn on?

[electrodacus]

[How will energy travel outside wire if you shield the wire from electric fields?

Let's think of the following setup:
1. Place both the battery and bulb in separate sealed boxes made of a perfect conductor.
2. Replace the two cables with coaxial cable made of perfect conductors and a lossless dielectric.
3. Connect the shields of the two coaxial cables to the two sealed boxes. (No pigtails allowed)
4. Connect the inner conductors to the bulb, switch, and battery, like in Vertasium's original video.

How long will it now take for the bulb to turn on?

You did not mentioned connecting the shield to anything.
Just connect the shield to say one side of the lamp.
But you do not even need all that complication to shield everything.
All you need is to shield a quarter of the wire so 10m of shield in Derek's experiment and connect that to the opposite side of the lamp/resistor.

 ______________________[resistor]_________________________
| ______________________________|                                               |
|                                                                                                             |
|                                                                                                             |
|                                                                                                             |
|___________________/ ____-  +____________________________|

Hope you can see the diagram.  The shield is only electrically connected to the right side of the resistor/lamp
What will happen is that the shield will now be the second plate of the capacitor formed with the lower wire and current for charging the capacitor will travel through that shield and connected to the opposite side of the resistor.
So there will still be current through the shield on left side and the line on the right side as the line capacitance charge but no more current through the resistor thus no more power and no energy until after the 65ns or so the electron wave needs to get to resistor through wire.

So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.

[SandyCox]

You are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.

[electrodacus]

You are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.

That will not be called a shield if it is not connected. Earth has nothing to do with this isolated system so connecting the shield to earth will do nothing unless you want to shield against external influence.

The shield I proposed is the simplest that will shield against electric field and so it shows that even with blocking the electric field from getting to Lamp/resistor the energy travels through wires and gets to resistor.
To "shield" against energy traveling through wire all you need to do is cut a gap of say 1mm in the wire  

[SandyCox]

What is your answer to my original question? How long will it take?

[electrodacus]

What is your answer to my original question? How long will it take?

If as you say the shield is not connected to any point to the circuit then is not actually a shield so you still get energy through that lamp/resistor well before 65ns.

In simplest form what you say is

-| | |-    a capacitor with a plate in the middle so like two capacitors in series. You can not call the middle plate a shield unless is connected to one of the sides.

[PlainName]

What is your answer to my original question? How long will it take?

I wish you luck and great patience 

[hamster_nz]

So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.

um, what "electric field generated by the lower wire"?

You were arguing that the energy is all in the wires. So why the need to shield from what you consider a non-existent electric field?

[electrodacus]

So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.

um, what "electric field generated by the lower wire"?

You were arguing that the energy is all in the wires. So why the need to shield from what you consider a non-existent electric field?

The lower line and top line form a capacitor that will be charged. There is one capacitor made by the lines on each side.
If you add that shield then the shield will be the other plate of that capacitor.
If you connect that to the other side of the resistor the current to charge the line capacitance will no longer flow through the lamp/resistor.

Since the only so called "proof" Derek had was that energy arriving earlier than 65ns. Adding this shield will get rid of that energy and still energy will get to resistor and travel through wires.

Derek's claim "energy doesn't travel through wires" and I can say just make a cut in the wire and see if any energy gets to lamp/resistor.

[Alex Eisenhut]

So how does energy transfer from side of a vacuum tube diode to the other?

[electrodacus]

So how does energy transfer from side of a vacuum tube diode to the other?

I'm not so old to have ever used a vacuum tube diode but a quick google revealed how that worked.
This is a direct quote that should provide you the answer:
"When the cathode of a vacuum tube is heated, electrons are emitted. If the anode has an electric potential higher than the cathode, the emitted electrons are attracted by it and an electric current start to flow "




So electrons do travel through vacuum in that particular case.
There are no electrons traveling from one wire to the other 1m apart wire in Derek's circuit thus energy travels through wire.
But the vacuum tube diode is a good example of energy traveling through vacuum so outside of the wire. It just that the vacuum tube diode has nothing to do with Derek's circuit where all energy delivered to lamp travels through wire.

[SandyCox]

What is your answer to my original question? How long will it take?

If as you say the shield is not connected to any point to the circuit then is not actually a shield so you still get energy through that lamp/resistor well before 65ns.

In simplest form what you say is

-| | |-    a capacitor with a plate in the middle so like two capacitors in series. You can not call the middle plate a shield unless is connected to one of the sides.

A transmission line and a capacitor is not the same thing. Maybe you should think again.

What happens inside a perfectly conducting sealed enclosure stays inside the enclosure, and what happens outside stays outside. Magnetic and electric fields cannot exit or enter the enclosure.

So by shielding the electric and magnetic fields, you are forcing the "electricity" to go all the way round the loop.

[electrodacus]

A transmission line and a capacitor is not the same thing. Maybe you should think again.

What happens inside a perfectly conducting sealed enclosure stays inside the enclosure, and what happens outside stays outside. Magnetic and electric fields cannot exit or enter the enclosure.

So by shielding the electric and magnetic fields, you are forcing the "electricity" to go all the way round the loop.

They are basically the same thing. Both the capacitor and the transmission line have resistance, inductance and capacitance just in different proportions.
If the shield as you say is not connected to any part of your circuit it will not do anything for this particular case.
The plate between the two capacitors plates unconnected to anything is useless.
-| | |-
While the simpler smaller shield I proposed will eliminate the energy being delivered to the lamp/resistor in the first 65ns as now the line capacitance is between that shield so current to charge that capacitor will flow through shield and not resistor. 

[SandyCox]

If the shield as you say is not connected to any part of your circuit it will not do anything for this particular case.
The plate between the two capacitors plates unconnected to anything is useless.

Nonsense!

Do you understand how shielding works? Have you heard of a Faraday cage?

[electrodacus]

Nonsense!

Do you understand how shielding works? Have you heard of a Faraday cage?

What you have proposed is a special type of cage, is a rectangular donut


But leaving all this aside. Is there a way you can intrerup the energy flow from source to lamp/resistor without making a cut in the wire ?
Because if you can not that is another proof that energy travels through wire.