2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.
If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.
Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.
If the switch is closer to charged capacitor / source then the wire after the switch forms another capacitor with the wire of the load thus effectively the small current Derek's is seeing is the one needed to charge that extra capacitor (the transmission line).
With the switch far away you also have a capacitor made by the transmission wires (that is fully charged before starting the test) and what the switch is doing is shorting that capacitor but from the far end of the capacitor thus electron wave that move the energy trough wires needs time to travel that distance.
Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).
The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light.
This distance is say 10 m (or praps
10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say
34 ns.Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity.
If the wire is not insulated then this too propagates at the speed of light, & the delay for say
10.5 m is say 35 ns, nearly the same as for the early (air) signal.
If the wire is insulated then this 35 ns would be 50% greater, ie
52.5 ns. But the 34 ns for the early signal would remain at
34 ns.
Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.