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how How does electric energy travel at speed of light when electron drift speed is so slow it cannot dissipate the voltage difference in that time?

Edit: Please let me clarify I read answers on this topic and I know the analogy of sound wherein although the air molecules from source don't reach you as sound, the wave pushes on neighboring molecules and your ear only receives the molecules near you. But the problem with this analogy is that: considering drift speed of electrons, do electrons equally distribute themselves along the wire immeediatelyimmediately as soon as you connect the circuit..i.e does the voltage drop immediately? If so, how?

Sorry if my question is ambiguosambiguous, I will expand it here. I know the following things are true (please tell if any is wrong).

Electrical potential is due to unequal number of charges at the two ends of a wire. This can be used to do work just like water at high altitude rolling down can do work. The electrons of the negative side pass through the metal wire to the positive end because of repulsion freomfrom other electrons. The electromagnetic energy travels at the speed of light (?probably using electrons as a medium or wave guidewaveguide) through the metallic wire. So my question is, if the energy gained in this scenario is due to the potential difference dissipating ) i.e the electrons moving to otehrother end and equalising the difference, how can the energy travel so fast? Also, how does AC current transfer energy then?

Thanks.

how does electric energy travel at speed of light when electron drift speed is so slow it cannot dissipate the voltage difference in that time?

Edit: Please let me clarify I read answers on this topic and I know the analogy of sound wherein although the air molecules from source don't reach you as sound, the wave pushes on neighboring molecules and your ear only receives the molecules near you. But the problem with this analogy is that: considering drift speed of electrons, do electrons equally distribute themselves along the wire immeediately as soon as you connect the circuit..i.e does the voltage drop immediately? If so, how?

Sorry if my question is ambiguos, I will expand it here. I know the following things are true (please tell if any is wrong).

Electrical potential is due to unequal number of charges at the two ends of a wire. This can be used to do work just like water at high altitude rolling down can do work. The electrons of the negative side pass through the metal wire to the positive end because of repulsion freom other electrons. The electromagnetic energy travels at the speed of light (?probably using electrons as a medium or wave guide) through the metallic wire. So my question is, if the energy gained in this scenario is due to the potential difference dissipating ) i.e the electrons moving to otehr end and equalising the difference, how can the energy travel so fast? Also, how does AC current transfer energy then?

Thanks.

How does electric energy travel at speed of light when electron drift speed is so slow it cannot dissipate the voltage difference in that time?

Edit: Please let me clarify I read answers on this topic and I know the analogy of sound wherein although the air molecules from source don't reach you as sound, the wave pushes on neighboring molecules and your ear only receives the molecules near you. But the problem with this analogy is that: considering drift speed of electrons, do electrons equally distribute themselves along the wire immediately as soon as you connect the circuit..i.e does the voltage drop immediately? If so, how?

Sorry if my question is ambiguous, I will expand it here. I know the following things are true (please tell if any is wrong).

Electrical potential is due to unequal number of charges at the two ends of a wire. This can be used to do work just like water at high altitude rolling down can do work. The electrons of the negative side pass through the metal wire to the positive end because of repulsion from other electrons. The electromagnetic energy travels at the speed of light (?probably using electrons as a medium or waveguide) through the metallic wire. So my question is, if the energy gained in this scenario is due to the potential difference dissipating ) i.e the electrons moving to other end and equalising the difference, how can the energy travel so fast? Also, how does AC current transfer energy then?

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how does electric energy travel at speed of light when electron drift speed is so slow it cannot dissipate the voltage difference in that time?

Edit: Please let me clarify I read answers on this topic and I know the analogy of sound wherein although the air molecules from source don't reach you as sound, the wave pushes on neighboring molecules and your ear only receives the molecules near you. But the problem with this analogy is that: considering drift speed of electrons, do electrons equally distribute themselves along the wire immeediately as soon as you connect the circuit..i.e does the voltage drop immediately? If so, how?

Sorry if my question is ambiguos, I will expand it here. I know the following things are true (please tell if any is wrong).

Electrical potential is due to unequal number of charges at the two ends of a wire. This can be used to do work just like water at high altitude rolling down can do work. The electrons of the negative side pass through the metal wire to the positive end because of repulsion freom other electrons. The electromagnetic energy travels at the speed of light (?probably using electrons as a medium or wave guide) through the metallic wire. So my question is, if the energy gained in this scenario is due to the potential difference dissipating ) i.e the electrons moving to otehr end and equalising the difference, how can the energy travel so fast? Also, how does AC current transfer energy then?

Thanks.