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Veritasium "How Electricity Actually Works"
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aetherist:

--- Quote from: electrodacus on May 07, 2022, 05:20:18 am ---
--- Quote from: aetherist on May 07, 2022, 03:04:47 am ---Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.

(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.
--- End quote ---
I need to ask what is your qualification.
I will not bother to read the rest of what you wrote.

There is absolutely zero evidence that energy in this particular case can travel outside the wire.
I guess my only way to explain is some sort of mechanical analogies but you did not mentioned anything about my last analogy with compressed gas in a cylinder with two chambers separated by an elastic membrane.
If there are more molecules in one chamber then if they have a path they will like to move in the other chamber in order to get to the lowest energy state so equal amount of molecules in both chambers.
While as many molecules will enter the low pressure chamber as they will exit form the high pressure one there will be energy delivered by them.
Why will you think there will be any difference for electrons stored in a capacitor ?

Also why when transferring energy between two identical capacitors half of the energy is lost and using a thermal camera you can see where all that energy went and it is in conductors making it quite obvious that energy traveled trough wires and due to resistance half of transferred energy was lost.
--- End quote ---
I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.

The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.

Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.

[C]  If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.

In both cases no energy is transferred via the air.

But the amount of energy transfer will usually not be the same in [G] & [C].
SandyCox:
Let's return to Veritasium's original topic.

Consider the steady-state (DC) case where all the voltages and currents are constant. Also assume that the wires are perfect conductors. The electronics only carry a small amount of kinetic energy as they travel through the wires. Once they reach the resistor (lightbulb) they are accelerated by the electric field associated with the voltage across the resistor. In the process, electric potential energy of the electrons is converted to kinetic energy. This kinetic energy is, in turn, converted to thermal energy through collisions with atoms in the resistor.

Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?
aetherist:

--- Quote from: SandyCox on May 07, 2022, 10:23:09 am ---Let's return to Veritasium's original topic.

Consider the steady-state (DC) case where all the voltages and currents are constant. Also assume that the wires are perfect conductors. The electronics only carry a small amount of kinetic energy as they travel through the wires. Once they reach the resistor (lightbulb) they are accelerated by the electric field associated with the voltage across the resistor. In the process, electric potential energy of the electrons is converted to kinetic energy. This kinetic energy is, in turn, converted to thermal energy through collisions with atoms in the resistor.
--- End quote ---

Why should electrons drift faster when they reach a resistor?
Why can't electrons go slower when they reach a resistor, but have more collisions or resistance than when going faster in the wire?
If electrons have zero resistance when in a perfect conductor then they must drift at almost c/1.  If so then how could they accelerate when they get to the resistor?

--- Quote from: SandyCox on May 07, 2022, 10:23:09 am ---Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?
--- End quote ---
The gravitational potential energy is not carried by anything.
The existence of the spaceship has no effect on any gravitational field.
The spaceship's gravitational potential energy is available, & it has a certain value, but it is due to the spaceship's relative position to Earth [caveat] in the surrounding gravitational field including Earth's gravitational field.
In any case, if Earth's escape velocity is  11.8 km/s, & if the Sun's escape velocity at Earth's radius is  42 km/s, then the spaceship's direction of approach to Earth, & the spaceship's velocity would affect whether the Earth or the Sun were the major factor.
SandyCox:
I will reply to your post after you received the Noble prize.
TimFox:
Remember that the drift of electrons (or holes in a semiconductor) can be measured using the Hall effect.
In solid-state physics, the solid structure imposes a mobility factor (in m2V s) on the charge carriers.
The conductivity of the medium is proportional to the product of mobility and carrier density.
In the article cited below, quite a few properties of the material contribute to the mobility:  drifting electrons encounter an obstacle course as they try to follow the E field.
Note that the Hall effect can discern the polarity of the charge carriers:  if both polarities be present, then there is a net effect.
For a summary, see  https://en.wikipedia.org/wiki/Electron_mobility
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