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The title says it all really, but I think that since the battery is disconnected there is now an 'open circuit'. I know that charge can only flow if the circuit is complete (closed).

But the part that puzzles me here is, what is there to stop the electrons from the negatively charged capacitor plate 'flowing back' to where the negative terminal of the battery was located before?

Update:

Answers so far seem to indicate that I am asking 'where' the charge would go if the circuit was closed again (after the battery was disconnected) - by means of completing the circuit with a piece of metal or even yourself. This is not what I am asking; I am asking where the charge would go if the circuit was left open (still with no battery connected).

I have added a schematic below to clarify what I am actually asking. But put simply, would the charge stay on the capacitor plate(s) or would there be some 'leakage' or 'backflow' of charge away from the charged plates?

Capacitor before and after charging

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    $\begingroup$ To be sure, what do you mean by "charge"? If a capacitor is charged with a battery, the capacitor is still electrically neutral. The battery has given up some of its stored energy to the capacitor (and some to heat). There is no electrical charge stored in the capacitor, only electrical energy via the separation of charge. Are you asking where the energy stored on the capacitor goes when the battery is disconnected? $\endgroup$
    – Alfred Centauri
    Aug 22, 2021 at 1:59
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    $\begingroup$ Charge can exist without it "flowing." $\endgroup$
    – Solomon Slow
    Aug 22, 2021 at 14:01
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    $\begingroup$ Re the edit: you can't really disconnect the capacitor from the rest of reality. Over a long enough time, the charge will leak out into the air (or whatever dielectric) between the plates and neutralize the charge difference between the plates, or it will find a path to ground (such as through you, if you touch it). $\endgroup$
    – Andrew
    Aug 22, 2021 at 14:50
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    $\begingroup$ I can give a practical example: I was playing and experimenting with capacitors a few weeks ago and I left a 470 μF capacitor charged to about 100 mV. I checked the voltage now and it's still 100 mV. Leakage was therefore negligible. $\endgroup$
    – Some Student
    Aug 22, 2021 at 15:46
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    $\begingroup$ Just my two cents here - all the comments here on dielectric leakage are an unneeded complication. OP's question (as I understand it) isn't answered or illuminated by dielectric leakage. $\endgroup$
    – Alfred Centauri
    Aug 22, 2021 at 17:27

9 Answers 9

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The charge won't go anywhere and the capacitor will remain charged until you short the plates of the capacitor. Where there was once a battery terminal there is now an insulator and that stops the electrons.

Also, the terminal will be made of metal that has a negligible capacitance so can't store significant amounts of charge.

And there is no net charge taken from the battery. The battery will push electrons from one of the capacitor's plate to the other.

Regarding your update: A theoretical perfect capacitor will never lose any volts. A real capacitor will always lose volts because air has some conductance and so does whatever dielectric is used to separate the plates. Even though a practical capacitor will lose volts, the loss may be small enough that it won't discharge in years.

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  • $\begingroup$ Hi, thanks for the answer, what do you mean by "short the plates"? $\endgroup$
    – N. Gin labs
    Aug 22, 2021 at 14:43
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    $\begingroup$ It's electric jargon for "connect plus terminal with minus terminal". $\endgroup$
    – Some Student
    Aug 22, 2021 at 15:34
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    $\begingroup$ This. In a real electronic circuit, there will be a current path to intentionally discharge the cap when the power supply is disconnected. Otherwise, the stored charge on a large-value high-voltage capacitor can kill you just as effectively as sticking your fingers into a mains power socket, even hours after the circuit was powered off. $\endgroup$
    – alephzero
    Aug 22, 2021 at 20:37
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    $\begingroup$ @N.Ginlabs A short circuit (i.e. "short[ing] the plates") is, put simply, when the positive and negative terminals are connected with nothing in between them. A battery in the circuit would be in between the terminals, so this configuration is not a short circuit $\endgroup$
    – JolonB
    Aug 22, 2021 at 23:21
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    $\begingroup$ @N.Ginlabs: "Short" means connect directly with a wire. (Either a real wire with tiny but non-zero resistance and inductance, so you get huge but finite currents for a very short time, or in theory with an ideal wire.) $\endgroup$
    – Peter Cordes
    Aug 23, 2021 at 4:27
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Nothing will stop them, and they do: the now-disconnected capacitor wires are themselves charged to exactly the same voltage as the capacitor plates. This means if you were to grab those wires each in your hands, you would get a full-on electrical shock from the stored charge in the plates.

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    $\begingroup$ Which is why large caps should always have a shorting strap when not attached in a circuit. $\endgroup$
    – Jon Custer
    Aug 22, 2021 at 2:28
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    $\begingroup$ @JonCuster, yes yes. otherwise one runs the risk of dying exponentially as the cap discharges. $\endgroup$
    – niels nielsen
    Aug 22, 2021 at 4:41
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    $\begingroup$ For a big enough cap, there is a real risk of dying in a fairly sudden manner (more Heaviside function, not exponential…). $\endgroup$
    – Jon Custer
    Aug 22, 2021 at 13:31
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    $\begingroup$ @N.Ginlabs: Yes, the circuit is open, which stops a persistent current from flowing. The shock you get as the capacitor plates equalise through you is not persistent; it stops once the charge imbalance has disappeared. $\endgroup$
    – Vikki
    Aug 22, 2021 at 17:16
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    $\begingroup$ As kids we used to take the charged high voltage flash bulb capacitors out of disposable cameras (anyone remember those?) and, for a gag, would drop them into people's shirt from behind - really dumb way to give someone a 300V surprise. $\endgroup$
    – J...
    Aug 22, 2021 at 18:21
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The dielectric medium between the plates is not perfect, a DC current will allow the charge to leak between the plates reducing the stored voltage difference to zero over time. The leak (or "insulation resistance") can be an important parameter when selecting capacitors in an electrical application.

Also see for example this.

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  • $\begingroup$ Thanks for the interesting answer, although I never said anything about putting dielectric between the plates; there is only a vacuum between the plates. $\endgroup$
    – N. Gin labs
    Aug 22, 2021 at 16:09
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    $\begingroup$ While there may be some value in pointing out things like this in a comment, I honestly don't see how this answer goes to the heart of OP's question as I understand it. $\endgroup$
    – Alfred Centauri
    Aug 22, 2021 at 16:59
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    $\begingroup$ @AlfredCentauri Well the OP writes "capacitor" in the title; and a capacitor is not two plates in a vacuum, a capacitor (as used in an electrical appliance or electronics) consists of two plates/electrodes with a dielectric medium. At no place in the question does the OP write that what s/he really means is what happens to two bare metal plates in a vacuum.. but sure that is another question that requires another answer if so :) $\endgroup$
    – BjornW
    Aug 22, 2021 at 19:53
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    $\begingroup$ bjornW. it is often the case, particularly in a question like this, that "capacitor" means ideal capacitor or at least a good approximation thereof. If it were the case that OP's question can only be answered in the context of a non-ideal, physical capacitor then I would accept your point as taken. But that isn't the case is it? $\endgroup$
    – Alfred Centauri
    Aug 22, 2021 at 21:57
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    $\begingroup$ If we're in the non-ideal world, vacuum is not a perfect insulator. $\endgroup$
    – zwol
    Aug 23, 2021 at 1:37
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What is there to stop the electrons from the negatively charged capacitor plate 'flowing back' to where the negative terminal of the battery was located before?

It will "flow back", and therefore the entire wire (+ the capacitor plate) is charged. That's why you will get an electric shock by touching any part of the wire or plate. Another way of thinking about it is that the wire & plate are at the same electric potential.

You can get a sense of this from the gold-leaf electroscope. The entire leaf rises when it's charged.

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    $\begingroup$ Why do you say that the charge will flow back so that the wire becomes charged? My understanding is that the wire connected to the negative plate is negatively charged before the battery is disconnected. Is your understanding different from mine? $\endgroup$
    – Alfred Centauri
    Aug 22, 2021 at 17:09
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    $\begingroup$ @AlfredCentauri Im with you on that. Unless the wires move closer together upon disconnecting, the charge in them won’t change. $\endgroup$
    – Al Brown
    Aug 23, 2021 at 3:56
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I know that charge can only flow if the circuit is complete (closed).

This is a bit of an oversimplification that you may find in sloppy physics books. In order for there to be a steady current, you need to have a closed circuit. But you can have a transient (non-steady) current flow in other situations. For example, charge in a conductor will rearrange itself so the electric field in the conductor is zero -- during the time when the conductor is reaching equilibrium, current is flowing, even though there is not a closed circuit.

In this case,

(a) the dielectric between the capacitor plates is never a perfect insulator in reality, so over time charge will flow from one plate to the other and neutralize the charge imbalance between the plates.

and

(b) random motion of electrons, and the repulsive force from other charges on the capacitor plate, will tend to mean electrons drift away from the plate and redistribute themselves along any conductors connected to the capacitor plate, ultimately working their way to ground.

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    $\begingroup$ I don't uite understand what you have in mind in your section (b). In particular, the "ultimately working their way to ground". Would you elaborate on that? It seems to me that if electrons from the negative plate worked their way to ground, the capacitor would be left with net positive charge. $\endgroup$
    – Alfred Centauri
    Aug 22, 2021 at 17:15
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    $\begingroup$ @AlfredCentauri I might not have expressed it well but I don't think I'm saying anything deep. If there's there's some stray path to ground, charge can flow from the capacitor to ground. $\endgroup$
    – Andrew
    Aug 22, 2021 at 17:22
  • $\begingroup$ @Andrew Thanks for the answer, but, "(a) the dielectric between the capacitor plates" - At no point in my question did I mention a dielectric being put between the plates. $\endgroup$
    – N. Gin labs
    Aug 22, 2021 at 17:35
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    $\begingroup$ @N.Ginlabs I assumed when you asked your question you were interested in what would happen to real capacitors. $\endgroup$
    – Andrew
    Aug 22, 2021 at 17:43
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    $\begingroup$ Elections can also make their way from ground. $\endgroup$
    – Lamar Latrell
    Aug 22, 2021 at 18:35
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what is there to stop the electrons from the negatively charged capacitor plate 'flowing back' to where the negative terminal of the battery was located before?

The electric field that exists between the plates of the capacitor due to the separation of charge. The electrons on the negatively charged plate are attracted to the positively charged plate. It's actually a minimum energy configuration.

Also, while the battery is connected to the charged capacitor, the negative plate, connecting wire, and negative terminal of the battery are all negatively charged. Similarly for the positive plate, connecting wire, and positive terminal. Indeed, there is an electric field between the connecting wires that gives rise to the voltage across them.

When the battery is disconnected, an ideal voltmeter connected across the connecting wires will continue to read the battery voltage since that is the voltage the capacitor and connecting wires charged to. Electrons didn't need to leave the negative plate to establish this voltage - the charge distribution giving rise to this voltage was already there.

Update:

With the edit to the question which added diagrams, I can now more easily make a point that I only touched on in my initial answer. Note: I'll be leaving out what I feel to be unnecessary complications in order to make the essential point.

The two leads that connect to the capacitor plates are conductors and, as such, form a capacitor in parallel with the primary capacitor formed by the plates. This stray or parasitic capacitance is almost always negligible compared to the primary capacitance but, particularly in RF circuits, it must sometimes be taken into account. This 'inter-lead' capacitance is of course dependent on the geometry, e.g., lead spacing, length etc.

Regardless, this capacitor is also charged by the battery to the battery voltage and this means that the wires connected to the plates have a small amount of charge density to establish the electric field between the leads and thus the voltage across.

When the battery is disconnected (in some 'ideal' way that doesn't disturb the leads etc.), this stray capacitance remains charged (again, ignoring all the physical ways charge might 'leak' from one lead to the other).

That is to say, it is not my understanding that electrons from the negative plate charge this stray capacitance after the battery is disconnected, it was already charged.

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As others have said, the charge stays in the capacitor until it either gets somewhere to go or leaks away over time. Two real-world examples:

Project in physics class--charge up a 8,000uf, 5?V capacitor, then use that power to power something transporting the capacitor as far as possible. (I made the minimum possible vehicle--the capacitor itself, the drive motor was on the launcher. Unfortunately, I didn't have the metalworking equipment to get it to launch straight enough. Had I gotten a clean launch I could have bounced it off the far end of the hall, but in practice it always spun out before it got that far.)

And, even more dramatic, a You-tube video where he stacked up some ultracapacitors, spent quite a bit of time charging them up then used the power in them basically as wanna-be arc-welder. It could actually melt metal, but didn't really have the power to weld.

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  • $\begingroup$ Wow interesting. $\endgroup$
    – Al Brown
    Sep 4, 2021 at 1:08
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    $\begingroup$ @AlBrown I found the video astounding--I had no idea capacitors that big existed. $\endgroup$
    – Loren Pechtel
    Sep 5, 2021 at 1:51
  • $\begingroup$ Send the link if convenient pls $\endgroup$
    – Al Brown
    Sep 5, 2021 at 4:47
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    $\begingroup$ @AlBrown I think this is it: youtube.com/watch?v=EoWMF3VkI6U $\endgroup$
    – Loren Pechtel
    Sep 6, 2021 at 1:13
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May I point out that no matter how many times you state there is no dielectric between your capacitor plates, there is, even in a perfect vacuum. dielectric is simply jargon for "the stuff between the plates that inhibits current flow" which a perfect vacuum is.

You should be picturing electron flow as water flow, the battery provides pressure which gets forced into the capacitor, once disconnected the capacitor provides pressure back to the origin. Nothing can leave the wire, except in the form of heat, which is a function of current. A momentary reversal of current is all you'd see if you use exceptionally long and non-ideal wires. But once the system is at equilibrium absolutely nothing else will happen, no more losses since the current is no longer flowing.

You have to understand that you're asking about non-ideal (real world) situations. An ideal wire has no resistance, no capacitance, no inductance, none of all those really fun things that complicate a system. An ideal wire is generally ignored since the non-ideal wire is more akin to a hose with holes in it but a hose with no holes in it that capacitor can supply as much pressure as it wants, but the water is going no where.

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Question clear from title. Surprised at six answers.

Simplest Conception:

Under voltage $V$, was charge $+q$ on plate 1 and $-q$ on plate 2, such that $q=cV$.

Then we disconnect, and charge cannot escape. The half-circuit connected to plate 1 has $+q$ and other has $-q$.

Since no voltage supplied, charges move to the lowest voltage configuration, and the voltage will not necessarily be $V$ anymore. To minimize voltage, oppositely charged ions (the electrons and the “holes”) will get close as possible to each other. Answers saying do or don’t flow are guesses (cept the Alfred one assuming no location changes)

Depends on where the isolated components are. How did we disconnect? If any part of those wires or routings are close, a reduction in voltage will be achieved by a change in charge distribution. There will still be some voltage, and could be same.

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    $\begingroup$ Hi Al, isn't it $Q=CV$ rather than $Q=C/V$? Also, isn't $Q$ the magnitude of the charge on either plate? $\endgroup$
    – Alfred Centauri
    Aug 23, 2021 at 22:18
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    $\begingroup$ @AlfredCentauri yes to both and much thanks. Great place to interact and learn. 🙏🏻👍🏻 $\endgroup$
    – Al Brown
    Aug 24, 2021 at 0:18

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