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.
""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.
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.
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.
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 Vbb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.
I put in the effort to read what each one of you has to say.
QuoteI 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.
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.
[How will energy travel outside wire if you shield the wire from electric fields?
[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 are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.
What is your answer to my original question? How long will it take?
What is your answer to my original question? How long will it take?
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.
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?
So how does energy transfer from side of a vacuum tube diode to the other?
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.
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?
Nonsense!
Do you understand how shielding works? Have you heard of a Faraday cage?