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Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??

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electrodacus:

--- Quote from: hamster_nz on July 22, 2022, 06:36:19 am ---You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.

--- End quote ---

You and Derek claim "energy doesn't flow in wires". There is zero evidence for that.
Both sides have the same metal surface area on the outside so what makes the higher resistance side receive 90% of the energy ?

 

hamster_nz:

--- Quote from: electrodacus on July 22, 2022, 06:58:00 am ---
--- Quote from: hamster_nz on July 22, 2022, 06:36:19 am ---You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.

--- End quote ---

You and Derek claim "energy doesn't flow in wires". There is zero evidence for that.
Both sides have the same metal surface area on the outside so what makes the higher resistance side receive 90% of the energy ?

 

--- End quote ---
Zero evidence? (glances as a wireless charger) yeah, if you say so.

And of course if you choose to use a stupid model so you can get 'counter intuitive' results... for example, where in the hardware store can I find this copper pipe you talk about which has 9 ohm resistance over 21 meters  :-//? Must have a cross-sectional area of a fraction of a mm.

One side has an 18V field over its 21m length, the other has a 2V field over the same length. Of course the charges in the stronger field will be doing more work (assuming both are made out of the same material, and the difference in resistance is due to cross-sectional area).

If you didn't want charges to do work in your wires, go get better wires, where the field strength in the wire will be lower for the same current, so less energy is doing work pushing changes around in those wire.

Naej:

--- Quote from: Nominal Animal on July 22, 2022, 01:17:56 am ---Let's assume you do that by using say Coulomb gauge, which says that the divergence of the magnetic vector potential is always everywhere zero, ∇·A = 0.  This is not an arbitrary choice, as it leads to specific properties, which are not a result of Maxwell's laws, but the result of this specific gauge applied to the potential approach to Maxwell's model.

--- End quote ---
It's the gauge chosen by Maxwell if my memory is correct.

--- Quote from: Nominal Animal on July 22, 2022, 01:17:56 am ---Finally, the term electric potential energy describes the potential energy associated with the electric field related to the charges present, those related to the Coulomb forces I mentioned; and those only.  It does not mean that all potential energy in a system that includes moving charges is in the electric field!

Can you see what I am trying to describe?  That I am not saying that anyone is wrong per se, I am saying that they are using such approximations that allow their viewpoint to be true, but that also change the system being considered so that the arguers are not actually talking about the same system anymore!

--- End quote ---
What's the approximation here? That it is not quantised?

Nominal Animal:

--- Quote from: Naej on July 22, 2022, 09:24:48 am ---What's the approximation here?
--- End quote ---
Those arguing different places for the majority of energy flow are making different approximations.

One uses the steady state model, ignoring everything that happens when the circuit is formed, and then argues with someone who only looks at the initial nonequilibrium state or an AC system saying they are "missing the entire point".

It is useless to try and unravel the argument, until they come to a mutual agreement of an exact experimental setup.

They won't, because each one wants to keep their own approximations that let them define the originally vague and fuzzy system in terms that make their chosen answer stick.  It's very, very silly.

electrodacus:

--- Quote from: hamster_nz on July 22, 2022, 08:54:59 am ---
--- Quote from: electrodacus on July 22, 2022, 06:58:00 am ---
--- Quote from: hamster_nz on July 22, 2022, 06:36:19 am ---You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.

--- End quote ---

You and Derek claim "energy doesn't flow in wires". There is zero evidence for that.
Both sides have the same metal surface area on the outside so what makes the higher resistance side receive 90% of the energy ?

 

--- End quote ---
Zero evidence? (glances as a wireless charger) yeah, if you say so.

And of course if you choose to use a stupid model so you can get 'counter intuitive' results... for example, where in the hardware store can I find this copper pipe you talk about which has 9 ohm resistance over 21 meters  :-//? Must have a cross-sectional area of a fraction of a mm.

One side has an 18V field over its 21m length, the other has a 2V field over the same length. Of course the charges in the stronger field will be doing more work (assuming both are made out of the same material, and the difference in resistance is due to cross-sectional area).

If you didn't want charges to do work in your wires, go get better wires, where the field strength in the wire will be lower for the same current, so less energy is doing work pushing changes around in those wire.

--- End quote ---

Wireless chargers that you refer to are based on magnetic coupling and that is not DC.  In this circuit with 20V battery you have a constant magnetic field same as from a permanent magnet.
Just have you wireless charging enabled phone on a permanent magnet and see if you get any charge.
Both the magnetic and electric fields are constant over those 5 seconds in this circuit supplied by a 20Vdc battery.

The resistance values are for ease of calculation and yes it is possible to have 9Ohm for a 21m long pipe (I have no specified the diameter or thickness of the pipe wall) I also did not specified the metal that pipe is made out off.

The test is specifically setup to show that current travels through wires (the entire section) not coming from outside and traveling only at the surface.
There is nothing else exiting the battery other than charged particles (electrons) and those travel through wires.
The magnetic field you measure around the wire is due to the motion of those electrons and the electric field is due to charge imbalance.
So you get the electron first, then due to electron you get the electric and magnetic field not the other way around.

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