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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: bsfeechannel on May 06, 2022, 04:10:52 pm ---Another problem is that some people who do not accept that the energy flows through space, are prepared to understand that the Poynting vector, perpendicular to the surface of the load, really conveys the energy that it will dissipate. But they don't understand that the wires are also resistors, and that the Poynting vector, perpendicular to whatever surface of the wire, be it radial or axial, will cause that energy to be dissipated by the wire.

In other words, you cannot hand energy to a wire in one side and expect that this energy will reappear at the other side. The wire will dissipate it entirely.

You can prove that by connecting any piece of wire to the poles of a battery. The energy will not be transferred from the negative pole to the positive pole. It'll simply be dissipated. Completely.

Sounds ridiculous, but people do not connect the dots.

--- End quote ---

How will you charge a discharged capacitor from a charged one if you can not transfer energy trough a wire ?
Do you just keep the capacitors close and wait for energy to be transferred or do you connect them with wires ?
And yes part of the energy (not all) will end up as heat due to wire resistance. In the case of two identical capacitors one charged and one discharged exactly half of the energy will end up as heat with the other half of the energy remaining in the two capacitors in equal quantity so a quarter of the initial energy in each capacitor.

bsfeechannel:

--- Quote from: electrodacus on May 06, 2022, 04:26:25 pm ---How will you charge a discharged capacitor from a charged one if you can not transfer energy trough a wire ?
--- End quote ---
If you are charging or discharging, a wire will come in handy for the transfer of the charges.

--- Quote ---Do you just keep the capacitors close and wait for energy to be transferred or do you connect them with wires ?
--- End quote ---
Have in mind that wires, capacitors, inductors, antennas, etc. are all conductors and that nature does not distinguish between them.

If you manage to put the plates of both capacitors in contact without the use of wires, the charges will redistribute and you'll have a loss of energy due to the current through the plates. And part of the rest, that was in the air between the plates of the first capacitor, will be transferred in between the plates of the second capacitor through the air.

Nice try, though.

aetherist:

--- Quote from: electrodacus on May 06, 2022, 12:02:52 am ---
--- Quote from: aetherist on May 05, 2022, 11:49:46 pm ---Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's  3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.

However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.

U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.

In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
--- End quote ---
All energy transfer from source to load/lamp is done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.
--- End quote ---
It seems to me that u have defined electricity as being energy carried by wires. And then u say that electric energy can only be carried by wires.
It seems to me that u are ignoring wireless energy transfer.

hamster_nz:
The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.

1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

The motion of electrons is not transferring the energy - it does not change with energy being transferred.

Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).
 

aetherist:

--- Quote from: hamster_nz on May 06, 2022, 11:01:53 pm ---The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.

1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.

The motion of electrons is not transferring the energy - it does not change with energy being transferred.

Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).

--- End quote ---
Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.

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