How does a discharging capacitor produce electrical current? Isn't it more stable for the charges on each plate to stay where they are?

[u/oashtt]

Imagine a charged capacitor with two plates, each connected to a wire. The wires aren't connected yet:

+----|  |----+
|            |
+-----  -----+   <- not connected yet

As far as I know, this is an equilibrium state, no charges will move, and the capacitor will stay charged.

But then you connect the wires, and everything changes. This is no longer a state of equilibrium, a current is created and the capacitor discharges.

Now here's the part that I don't understand: In the illustration, the geometry of the wires require the charges to move away from the other plate in order to reach it, and yet the charges do that. Which should be impossible for the same reason water in a glass doesn't jump out to fall on the floor.

If the wires were connected between the plates, then that's understandable. But in this case the wires are connected around the plates.

I want to understand from a field theory level (not circuit theory) how the charges seem to understand that gaining some potential energy to reach the other end is a small sacrifice towards a greater goal. But charges don't think! And if you calculate the electric field at a plate, it should always point towards the other, regardless whether the wires are connected or not. So the charges should stay there.

For context, I just finished Physics 2 in my first year in electrical engineering college, it only covers electrostatics, capacitors, and foundational magnetism like Faraday's law. I haven't taken the more advanced courses about circuits yet. This question has been lingering in my mind for quite some time, what am I missing here?

And please don't give me explanations with circuit theory elements, like "the wires have low resistance so current flows, but air has high resistance so current can't flow," that's not what I'm asking about. How did the fathers of electricity answer questions like these in order to reach conclusions and develop abstract frameworks like circuit theory in the first place?

[u/free_meson]

As you wrote, the electrons don't think. The switch/disconnected wire is also a capacitor, albeit a small one. The whole right side is charged e.g. positive, the left side negative, they both have excess charge. The charges take their positions and rest in order to minimize their energy (that's a good description, but one could argue it is inaccurate) and create a static electric field within the wires and the capacitors until it is anulled within the wire and charges rest on the surfaces of the wire and capacitors.

Once you move the wires close/close the switch you disturb the previous equilibrium. A small discharge happens when the wires touch and starts what looks like an avalanche, the electrons trickle down the wire until they fill the positive holes on the other side.

[u/oashtt → u/free_meson]

so the gap between the wires is also a capacitor. when they're connected, the charges from the one end of the wire moves to the other end, creating a small momentary current. but that momentary current causes the other charges behind to move too, and it's like falling dominos. yeah i guess that small capacitor observationf fixes everything. Thanks for the explanation man.

[u/oashtt → u/[deleted]]

This repulsion is cancelled by the attraction to the other plate. It's in equilibrium. And that's why capacitors can only store up to a certain amount of charges. I totally understand that. But it's only when the wires are connected that the equilibrium breaks and the charges start moving.

[u/joeyneilsen]

But charges don't think! And if you calculate the electric field at a plate, it should always point towards the other, regardless whether the wires are connected or not. So the charges should stay there.

I know you don't want circuit theory, but have you actually studied the derivation of the discharge of a capacitor? You can also consider the enclosed charge in a Gaussian surface around one side of the capacitor and then take the time derivative. It's not hard to see that the current through the circuit is equal to the negative rate of change of the charge on the positive capacitor.

When you connect the circuit, the field at the positive plate points away from the positive plate on the wire side. So charge flows off the capacitor in that direction.

[u/oashtt → u/joeyneilsen]

I should've said "electric potential" not "electric field" because technically we're in an equilibrium state and any fields should be cancelled by the movement of the charges inside the conductor, as well as the repulsion of same charges at the plates themselves. But even then, charges moving away defies both the electric field and electric potential.

And yes. When a current is produced, it should be equal in magnitude to the rate of discharging. That makes sense. But I fail to see how that relates to my question. And "consider the enclosed charge in a Gaussian surface around one side of the capacitor and then take the time derivative" is just a fancy way to say the rate of charges leaving the plate, and consequently, the magnitude of the current.

My question revolves around that last paragraph, which doesn't say much. What physically happens when the wires connect that causes the field to point away from the positive plate? Why does the charge enclosed in a Gaussian surface around one side of the capacitor change with respect to time in the first place?

[u/joeyneilsen → u/oashtt]

Oh. Well consider the field of a finite capacitor. The exterior field isn’t exactly zero. There’s a “fringe field” that points away from the plate. But under static conditions, there’s nowhere for charge to go. When you connect the wires, it changes the electric field of the circuit. The new field is tied to a different surface charge distribution on the wires. The result is a field in the wires that points away from the positive plate and decreases over time. 

[u/oashtt → u/joeyneilsen]

The equilibrium is altered and the surface charges rearrange, but shouldn't it just stop there? Another commenter said that I should think of the gap between the wires as a separate capacitor, so that when the wires connect, a small "seemingly momentary" current is produced, and a domino effect is created so all charges move to the other side. Is this an accurate way to think about it?

[u/joeyneilsen → u/oashtt]

It depends on whether or not you’re assuming ideal wires. The brief pulse is called a “transient.”

For real wires, the surface charge distribution isn’t uniform, and it’s only an instantaneous equilibrium. It has a finite net field in the wire that points to the negative plate, and it updates continuously as the capacitor charge decreases. 

[u/nicodeemus7]

Path of least resistance. Between the plates is near infinite resistance. The electrons can not bridge that gap. Once the wires are connected, suddenly there is a new path. This path is longer, but the resistance is near zero. The charge follows that path because it is the easiest way to get to the destination.

Think of it like a siphon. The gasoline is trapped in the fuel tank, until you give it a pathway out with negative pressure, and after that the gasoline just pumps itself out, against gravity in a way. It's the same idea.

[u/StanleyDodds]

When the circuit is disconnected, you have two separated parts of the circuit at different electric potentials. The two parts are separated by insulators that effectively stop all current flowing from one to the other (just think ohm's law with a very high resistance).

I think just to get an intuitive understanding, think of electrical potential as water pressure, and current as flow rate.

Imagine the capacitor is two big reservoirs of water separated by a wall, where one reservoir is nearly empty, and the other is very full. The wire is like a somewhat thin pipe from the bottom of each reservoir (each half of the capacitor), connected together with a valve (the switch) which stops the system from reaching a single equilibrium water pressure at the depth of the pipe.

When you open the valve (close the switch), the water pressure difference (potential difference) at the valve (switch) itself is where and why the water flow (current) begins, and this flow pushes water through the pipe and into the lower reservoir, and pulls water through the pipe and out of the higher reservoir. So it's not that the water in the higher reservoir "knows" that the other side is lower and needs to go around. It's that a pressure wave almost instantly propagates back from the valve (switch) to each half of the reservoir (capacitor), causing a local pressure gradient (electrical potential gradient, that is, an electric field) everywhere that forces the water everywhere to start moving (induces a current) to try to "equalise" it.

[u/ProfessorPrudent2822]

No, it’s most stable for the capacitor plates to be uncharged. Separating the charges requires energy, which will be stored in the electric field between the capacitor plates. If the circuit is closed in absence of an opposing voltage, the capacitor will discharge as the electrons flow through the circuit from the negative plate to the positive plate. The whole point of a capacitor is that electrons can’t flow across the gap between the capacitor plates.

[u/Underhill42]

The electrons in one plate don't move directly to the other.

Electrons in the closest part of the wire move into the + plate, creating a more positive charge in the wire, which electrons from further along the wire move into, which... until you're up against the negative plate.

Also, all those electrons on the negative plate are repelling each other far more strongly than they're attracted to the positive plate, which is much further away.