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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: hamster_nz on May 23, 2022, 08:26:21 pm ---
Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.

The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.

If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.

Pipes are essential to the system working, even though they carry no usable energy.

The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.

--- End quote ---

I guess I need to be more explicit as you did not understood any of my compressed air analogies.

Air molecules are the electrons and compressed air cylinder with with two chambers represents the capacitor.
A discharged capacitor is like an air cylinder that has the same pressure in both chambers say atmospheric pressure but it can be any value as long as the same number of air particles are found in both chambers and inside the hose's.

Now you can charge the capacitor using say mechanical energy to pump air from one chamber to the other one. You can use the same pump in reverse to do work with the stored energy.
Since now there are more air molecules (electrons) in one chamber (capacitor plate) than the other one you have stored energy.
If you connect the two chambers through a pipe air molecules (electrons) will flow (electric current) from the higher pressure chamber to the other one until pressure (voltage) is equalized. 

IanB:

--- Quote from: electrodacus on May 23, 2022, 08:08:21 pm ---
--- Quote from: IanB on May 23, 2022, 07:50:51 pm ---I see. So if you have three capacitors in series, the energy to charge the middle capacitor flows through the outer capacitors. If energy does not flow through the outer capacitors the middle capacitor will not be charged. If energy does flow through the outer capacitors the middle capacitor will be charged.

--- End quote ---

At no point in time any energy flows through the capacitor.
--- End quote ---

What I wrote was not a question. It follows directly from what you wrote. If you contradict it, you contradict yourself.

Sredni:
Now, I am well aware that after the Faraday vs. Kirchhoff threads I am the last person who should point this out but...

This short video summarize perfectly this thread:



As for the ongoing discussion about the role of the wires, I would like to point out the role of surface charges.
To simplify: we need surface charge of different polarities at the resistor ends. The surface charge distribution will cause conduction electrons inside the resistor to be accelerated and then collide with the ion lattice, releasing heat (classical ED). Now, why don't the electrons that arrive on the other side of the resistor cancel the positive charge on the surface on that side? Because we have the wire to scoop them away (and to replenish on the other side before they enter).

The role of the wires is that to provide the boundary conditions that will keep the surface charge distribution at the resistor. And what makes the charge distribute on the surface of the conductors and resistor? The fields all around them. The electric field inside is the result of the surface charge distribution.

hamster_nz:

--- Quote from: electrodacus on May 23, 2022, 08:46:06 pm ---
--- Quote from: hamster_nz on May 23, 2022, 08:26:21 pm ---
Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.

The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.

If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.

Pipes are essential to the system working, even though they carry no usable energy.

The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.

--- End quote ---

I guess I need to be more explicit as you did not understood any of my compressed air analogies.

Air molecules are the electrons and compressed air cylinder with with two chambers represents the capacitor.
A discharged capacitor is like an air cylinder that has the same pressure in both chambers say atmospheric pressure but it can be any value as long as the same number of air particles are found in both chambers and inside the hose's.

Now you can charge the capacitor using say mechanical energy to pump air from one chamber to the other one. You can use the same pump in reverse to do work with the stored energy.
Since now there are more air molecules (electrons) in one chamber (capacitor plate) than the other one you have stored energy.
If you connect the two chambers through a pipe air molecules (electrons) will flow (electric current) from the higher pressure chamber to the other one until pressure (voltage) is equalized.

--- End quote ---

What? This was in reply to your post:


--- Quote ---It is like saying that energy from a compressed air tank is delivered to a compressed air tool outside the hose.

--- End quote ---

I was pointing out a system where a vacuum tank and pipes can deliver energy to an air-powered tool, using the air outside of the hose, by literally delivering nothing in the pipes.

You just don't like it because the open air molecules have no 'special' added energy, and they also have no 'special' added energy when in the vacuum pipes on the way back to the tank & pump, pipes are required, those pipes are empty, and yet somehow still deliver energy to the tool.

electrodacus:

--- Quote from: IanB on May 23, 2022, 08:57:49 pm ---
What I wrote was not a question. It follows directly from what you wrote. If you contradict it, you contradict yourself.

--- End quote ---

It can not follow from what I wrote as I never wrote that energy flows through a capacitor but energy flows in or out of a capacitor.
Is like you are unable to see a certain color (that color will be called energy storage).

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