Capacitor charging: Where do extra electrons come from?

[LarryS]

TL;DR Summary

Do the electrons come from that part of the conducting wires that is very CLOSE to the capacitor?

Given a DC circuit consisting of a battery, a switch and a parallel plate capacitor and no resistor. The wires are relatively thick copper. Also, the copper wire leads into the capacitor are each, say, about one meter long. Now when the capacitor is fully charged, the negative plate will contain extra electrons. I’m trying to visualize WHERE those extra electrons actually come from. I’m thinking that they must come from that part of the copper leads that is VERY CLOSE to the capacitor. The charging time of the capacitor is very fast because of the low resistance. Therefore they cannot come from the battery or the leads that are close to the battery because the drift velocity of the electrons is very slow and the length of the copper leads is long (one meter).

Am I right? Is the capacitor supplied electrons from that part of the copper leads that are very close to it?

 

[Baluncore]

Free electron drift velocity is very low, but the changes in drift velocity propagate at close to the speed of light. Individual electrons do not travel far, they all move a very short distance in the same direction.

The negative plate will have excess free electrons that are pushed further out of the metal surface, those will come from the conductive capacitor plates, to be replaced by free electrons from the cloud, that moves along the conductive wire.

On the positive plate, the electrons will retract slightly into the metal lattice, moving along the conductor, through the battery to reach the negative plate.

 

[tech99]

When the switch is operated, a pulse of opposite polarity is sent at the speed of light along each wire to the capacitor. The pulses are a compression wave in the electron cloud. The two pulses charge the capacitor. It then discharges into the circuit, whereupon an inductive back EMF is produced, and an unending oscillation, where energy is exchanged between L and C, commences. If there is resistance in the circuit then the oscillation gradually fades away so that the capacitor has battery potential across it and zero current. Energy equal to 0.5 CV^2 will have been expended in the resistance.

 

[sophiecentaur]

TL;DR Summary: Do the electrons come from that part of the conducting wires that is very CLOSE to the capacitor?

Given a DC circuit

This condition makes assumptions which sort of beg the question. Just call it a "Circuit" with initial conditions and then find out what happens. In many situations it's not DC.

When the switch is operated, a pulse of opposite polarity is sent at the speed of light along each wire to the capacitor. The pulses are a compression wave in the electron cloud. The two pulses charge the capacitor. It then discharges into the circuit, whereupon an inductive back EMF is produced, and an unending oscillation, where energy is exchanged between L and C, commences. If there is resistance in the circuit then the oscillation gradually fades away so that the capacitor has battery potential across it and zero current. Energy equal to 0.5 CV^2 will have been expended in the resistance.

Whist this is totally correct and of interest once people have got over the basics, it does add a touch of confusion when someone is fumbling. Fact is that we initially assume some significant resistance in the charging path (Battery+ ... R...Wire ...Capacitor ....wire...R.... Battery- ) This will limit the rate of charge flow because there will usually be damping. The current flowing will be unidirectional ('DC') and decay exponentially until the PD across the C is equal to the Battery volts.

Electrons will build up on the - plate and a - charge will build up on the + plate. In between the plates, the dielectric will Polarise and the electrons around each molecule will be displaced a small amount in the direction of the + plate (but no flow)

Once this makes sense, we move on to the situation where the R is low and the Magnetic Field due to high current flow will store Energy, temporarily. Energy 'sloshes' from supply to capacitor and back again and an oscillation decays to zero when the C is fully charged. Some of this Energy will be radiated as an EM wave.