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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: hamster_nz on May 02, 2022, 02:37:28 am ---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.

--- End quote ---

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:

--- Quote from: hamster_nz on May 02, 2022, 03:10:07 am ---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.

--- End quote ---

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: ---    +------------+
    |            |
  ---- 0V      ---- 0V
  ----         ----
    |            |
Gnd +----/ ------+

--- End code ---
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: ---    +------------+
    |            |
  ---- 10V     ---- 0V
  ----         ----
    |            |
GND +----/ ------+
          10V

--- End code ---

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: ---    +------------+
    |            |
  ---- 5V      ---- 5V
  ----         ----
    |            |
GND +------------+
           0V

--- End code ---
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

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