How can a capacitor store charge whilst also passing current?

Question

It's frequently said that capacitors store charge. Just reading through Wikipedia, I find:

Daniel Gralath was the first to combine several jars in parallel into a "battery" to increase the charge storage capacity. Benjamin Franklin investigated the Leyden jar and came to the conclusion that the charge was stored on the glass, not in the water as others had assumed.

Because the conductors (or plates) are close together, the opposite charges on the conductors attract one another due to their electric fields, allowing the capacitor to store more charge for a given voltage than if the conductors were separated, giving the capacitor a large capacitance.

Here Q is the charge stored in the capacitor

Charge is measured in coulombs, and I know from the definition of capacitance that if a 1F capacitor has a voltage of 1V, then 1C of charge is stored in it. If a coulomb is 6.241×10¹⁸ electrons, then there should be 6.241×10¹⁸ electrons in this capacitor somewhere.

But now consider this. If I use a capacitor as a load to some AC voltage source, some current will flow (the precise amount depending on the voltage, frequency, and capacitance):

^{simulate this circuit – Schematic created using CircuitLab}

I know that current is flowing all the way around this circuit, because if I put a lightbulb on either side of the capacitor, it will light. But if the current is flowing around this circuit, how does the capacitor "store charge"? In other words, how can I ever put electrons into the capacitor if the current is flowing around the circuit, which means for all the electrons I put in the capacitor, the same number come out the other side? If I can't put electrons in without taking some out, then how can the capacitor be storing them?

Answer 1

It's easy. A capacitor doesn't store charge, it stores energy. The net charge in a complete capacitor (rather than considering a single plate or the insulator) never changes. An increase of negative charge on one plate is exactly balanced by a decrease in negative charge on the other plate. Therefore, as current enters one terminal an equal current must leave the other terminal.

Answer 2

This is kind of the cartoon version, but it works in my head.

There's an insulative gap in the capacitor, so individual electrons can't travel from one terminal to the other. So the electrons going in aren't the same ones coming out the other side! Instead, the incoming electrons "stop" on one plate. But that electron's electric field repels an electron from the other side, which proceeds out of the other plate, eventually reaching the source. We have a complete circuit, but electrons are building up on one plate, and holes are building up on the other!

Now, there's a limit to how many electrons can build up on the plate. Electrons repel each other, so the more there are, the harder it is for another one to stick. We need something that forces them to stay on the plate together. That's voltage. Conversely, the fact that the electrons are trying to repel each other is also a voltage, a force trying to move electrons around a circuit.

Now, when an incoming electron knocks one loose from the other plate, the outgoing electron has less energy than the incoming one, which accounts for the voltage drop across the charged capacitor.

Of course, electrons aren't holding still, even if there's nowhere for them to go on a macro scale. They're all repelling each other, "bouncing" off each other's electric field. If those fields get too intense (the voltage gets too high), the interactions can cause an electron to penetrate the dielectric barrier between plates. When the voltage across the plates gets too high, the leakage current of the cap goes up. And if that goes on too long, the dielectric gets damaged, and you don't have a very good cap any more.

Answer 3

Charge can mean a lot of things. We can talk about charging a capacitor with energy, like we charge bombs, or prepaid credit cards. We can also take about electric charge, which is measured in coulombs.

About 6.241×10¹⁸ electrons does indeed make 1C of charge. However, when people talk about charge in a capacitor, they aren't talking about electrons in a capacitor, like one would talk about cookies in a cookie jar. They are talking about something else. It's confusing, but it's what they do anyway.

What they are actually talking about is the integral of current. That is, the mean current that's been flowing, times how long it's been flowing. If current is measured in amperes, and time in seconds, then when you take current and multiply it by time, you get a thing measured in ampere-seconds. And, if you recall, an ampere means one coulomb per second. Thus:

$$ A = \frac{1C}{s} \\ 1A\cdot s = \frac{1C}{s} \cdot s = 1C$$

That is, an ampere-second is a coulomb. The integral of current is charge. So when someone says a capacitor is "storing 1C of charge", they don't mean there is 1C of electrons in the capacitor, they mean 1C of charge has passed through the capacitor. The capacitor is "storing" that much charge in the sense that it now contains enough energy to push 1C of charge back the other way.

Answer 4

It's better to think of a capacitor as an energy storage device than as a charge storage device. When current flows into a capacitor, a voltage accumulates at the terminals. This voltage is separated by the distance between the plates and thus creates an electric field. This field is where the energy is stored. Inductors, on the other hand, store energy with magnetic fields.

As the current flows, opposite charges accumulate on each opposite plate of the capacitor. The electrons are trying to go around the circuit, but they get stopped at the plate of the capacitor, leaving a negative charge on one side and a positive charge on the other. The magnitude of each charge can be described by the equation:

C = Q/V

The current will keep flowing and charge will keep accumulating until the circuit with the capacitor is stable. For example, if the circuit was simply a battery, a resistor, and a capacitor in series, current would continue to flow until the capacitor voltage was equal to the battery voltage. Thus, in a steady-state DC circuit, where no currents are changing, a capacitor appears as an open circuit with the accumulated charge proportional to the voltage across the terminals and the capacitance.

However, for any circuit that is not DC, a better way to describe the behavior of capacitors is:

I = C*(dV/dt)

Therefore, if you have a sine wave voltage source, the current flowing "through" the capacitor is constantly changing and the accumulated charge is never steady. Imagine tipping a half full water bottle back and forth. The water is not flowing continuously like current in a DC circuit, but it is still doing work. If you had some bizarre turbine device in the water bottle, it would be constantly spinning, stopping only to change direction when the bottle is tipped the other way.

Finally, in a DC circuit, equal and opposite charges are stored on each side plate of the capacitor. The capacitor does not store electrons at all. It stores a charge. Electrons from one side travel all the way around the circuit to the other side as provoked by an external voltage difference. The result is an concentration of electrons on one side and an absence on the other, a charge. In an AC circuit, this same phenomenon happens, but is consistently changing. As soon the supply voltage changes, the electrons are not attracted to the plates the same way and begin to mobilize. If these electrons happen to pass through a load, like a lightbulb, on the way, they will do work and the lightbulb will turn on. Thus, the current is not actually flowing around the circuit. It is simply sloshing back and forth like water in a bottle. However, all it takes to light the bulb is moving electrons. The bulb doesn't care which way they are moving, and your eyes can't perceive the change in direction so long as the switching speed is fast enough.

I would also like to note that we are talking about ideal capacitors. In practice, at high enough frequencies, capacitors will look like inductors (V = L*(di/dt)).

Edit:

To answer the specific question: Where is the charge stored in a capacitor?

Within in a complete capacitor, no net charge is stored. However, using the parallel plate model, equal and opposite charges of magnitude Q are located on each of the plates. When an external voltage is applied to a capacitor, the electrons flee from the plate with a higher potential and are attracted to the plate with a lower potential. These accumulated electrons form a negative charge on that plate and the absence of electrons from the other plate form a positive charge. The actual magnitude of each, total charge Q is determined by the voltage V and the capacitance C.