What represents the charge on a capacitor?

[LarryS]

TL;DR

The electrons involved in the current in a copper wire have a relatively low drift velocity. Yet capacitors reach full charge relatively fast. Trying to reconcile these two facts.

Consider a simple DC circuit containing a 9V battery, a switch, a 10 kΩ resister and a 100 μF capacitor all in series. When the switch is thrown, it will take basically 5 seconds for the capacitor to reach full charge.

Based on what I have read online, the charge on the above capacitor is represented by an abundance of electrons on one of the plates. The total charge being the number of electrons times the elementary charge e.

I have also read that electrons have a very low, around 1 mm per second, drift velocity in copper wires. That most of the energy in the current is actually from EM waves in the wire. So, based on that, I does not seem that enough electrons would reach the capacitor plate during the 5 seconds of charging. So, what represents the charge on the capacitor?
Thanks in advance.

 

[Drakkith]

There are somewhere around 10¹⁹ atoms in a cubic millimeter of copper, each of which has at least one free valence electron which means there are a lot of electrons in even a small section of wire that can move about under an applied voltage. So even a small drift velocity can result in a huge electric current.

That most of the energy in the current is actually from EM waves in the wire.

It is true that the energy delivered by a circuit is not dependent solely on the current. A load requiring 100 watts of power can often be designed to get that wattage from 10 volts and 10 amps, 5 volts and 20 amps, 20 volts and 5 amps, or whatever combination of V times A that you desire as long as it equals 100.

Not only that, but you can step up/down both the voltage and the amperage at will. A load that requires 10 amps can connect to a circuit that delivers only 5 amps by using a transformer or other means of changing the voltage/current.

So, based on that, I does not seem that enough electrons would reach the capacitor plate during the 5 seconds of charging. So, what represents the charge on the capacitor?

The charge on the capacitor is the number of excess electric charges on each plate (or equivalent if you don't have a capacitor with 2 plates). This number depends on the voltage and capacitance of said capacitor. A capacitor with a very high capacitance requires more charge to reach the same voltage as a capacitor of less capacitance, and increasing the voltage of a capacitor requires increasing the charge stored on that capacitor (at least in simple capacitors).

So, based on that, I does not seem that enough electrons would reach the capacitor plate during the 5 seconds of charging.

Your example circuit has a maximum current of less than 1 milliamp, and your capacitor holds only 0.9 millicoulombs of charge. However, a drift velocity of 1 mm/s through a copper wire with a cross sectional area of 1 mm² represents roughly 13 amps, enough to charge your capacitor in milliseconds.

Don't underestimate the power of 10^xx numbers of electric charges moving at slow velocities.

Finally, remember that drift velocity is not the 'true' velocity that electrons move at. They are like air molecules in a calm breeze. The breeze may be slow but the individual molecules are bouncing around at half the speed of sound. Similarly, electrons in a metal are moving about the metal lattice at enormous velocities. At absolute zero the velocity of an electron of copper in the highest occupied state has a velocity (called the fermi velocity) of about 1,570,000 m/s. So the electric current you see in a circuit is simply the result of a very slight reduction in the velocity of electrons going one way through the circuit and the slight increase in velocity of electrons going the other way.

 

[sophiecentaur]

Don't underestimate the power of 10xx numbers of electric charges moving at slow velocities.

It's the same sort of thing that makes multi-lane highways so good at passing vast amounts of traffic. You have to do the sums with the right figures before you can overcome 'wrong' intuitive conclusions.

If you think in terms of Coulombs and Volts and then how many valence electrons are present in a piece of copper, you can get an idea of the actual numbers involved. It's instructive to calculate the Energy (Work) and Forces involved in bringing two charges of +1C together from Infinity to a distance of 1m from each other. (The formulae are readily available all over the place.)