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What I'm asking about is if it's possible to discharge one side of a capacitor. I've found a question about that, but I also found the answers to be a little unclear or questionable. I imagine that if you were to do so, you would need to put energy into it, because you would be removing charge from one side of the capacitor and creating electrical potential energy. Could this be done if you connected the positive and negative ends of a battery to a negative and a positive side of two capacitors? Could doing that cause capacitors to break down?

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    \$\begingroup\$ I think you are asking "What if I started with a capacitor where there was \$10^{20}\$ excess electrons on the negative plate that were moved there from the positive plate (which now has \$10^{20}\$ too few electrons) and then decided to remove the excess \$10^{20}\$ electrons [magically] so that the negative plate was neutral while the positive side still was missing those \$10^{20}\$ electrons that were earlier removed?" Is that about it? \$\endgroup\$
    – jonk
    Nov 6, 2017 at 0:05
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    \$\begingroup\$ The charge "on" a capacitor is half the charge difference between the two plates. "Discharging one side" sounds like a naive way to just say "discharging it halfway". \$\endgroup\$
    – user253751
    Nov 6, 2017 at 0:05
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    \$\begingroup\$ What is the sound of one hand clapping? \$\endgroup\$
    – Dave Tweed
    Nov 6, 2017 at 0:20
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    \$\begingroup\$ sound of one hand clapping youtube.com/watch?v=b6YSfEKMeC8 \$\endgroup\$
    – jsotola
    Nov 6, 2017 at 5:10
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    \$\begingroup\$ Oof. This is like the question, "Where do I get a magnet with only one pole?" and makes me worry the asker is looking into free energy... \$\endgroup\$
    – Adam Davis
    Nov 6, 2017 at 17:03

5 Answers 5

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It is physically possible for there to be more electrons on one side of a capacitor without there being a corresponding number of holes (absences of electrons) on the other side. In fact, your proposed configuration of two capacitors and a battery would do that — but by a very, very small amount — about the same amount as if you cut a single capacitor in half and spread the plates apart to the same locations, then connected the battery.

This effect, which applies to any conductor, not just capacitor plates, is called self-capacitance, as opposed to mutual capacitance. It is defined in the same way as capacitance,

$$C = \frac{q}{V}$$

— but it is immensely smaller for a given physical size. The amount of charge 1.5 volts — or 9 volts or 240 volts — can push into such a capacitor is so small that it has a negligible effect in typical circuits — we do not bother to think about it.

(It is also true that there is some amount of (mutual) capacitance between the unconnected ends of the two capacitors. Every pair of conductors is a capacitor, but they're usually bad ones with small area and large plate separation! Both self-capacitance and mutual capacitance contribute to how much charge you can stuff into a conductor for a given voltage.)

In electrostatic systems, working with kilovolts and up, the effects of self-capacitance can become significant. If you walk across a carpet and touch a CMOS IC, destroying it, what was the immediate source of the energy at the discharge? It was your body having a net positive or negative charge. The opposing charge was left behind on the carpet. The self-capacitance is the ratio between that amount of carried charge and the voltage between you and the carpet. (Where did the large voltage come from? Separating the “plates”. Where did the initial charge transfer come from? The triboelectric effect.)

A physical example of essentially an “only one side charged” capacitor is a Van de Graaff electrostatic generator. The sphere on top is one plate; the entire surroundings including the Earth (assuming the generator is grounded, as it usually would be) is the other, but the Earth is so much bigger that the charge imbalance is insignificant for it but very significant for the sphere.

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  • \$\begingroup\$ That does make sense. If you did want to maximize the charge on one side, would you want to use a capacitor with the highest voltage possible? \$\endgroup\$
    – Tom
    Nov 6, 2017 at 2:50
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    \$\begingroup\$ @Tom Yes, if the maximum voltage is reached the dielectric fails and the other side of the capacitor would also become charged. \$\endgroup\$
    – Kevin Reid
    Nov 6, 2017 at 3:34
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No.

The charge on a capacitor is defined by the voltage difference between the two plates, the geometry of the plates, and the chemical properties of the dielectric.

That is.. the charge is between the plates, across the dielectric, not on the plates.

You need to understand it is the presence, or absence of electrons on one plate that drive away or attract electrons on the other plate. You can't change one without changing the other.

As such, the concept of removing charge from one plate is incorrect.

If you remove electrons from the negatively side of the capacitor, the voltage across the plates would drop, as would the charge in the entire capacitor, not just that side of the capacitor.

In fact, the only way to remove the electrons is to change the applied voltage across the capacitor. So we just went round in a nice circle. This is of course what we do all the time when we discharge a capacitor, we apply zero volts across it.

EDIT

There is one way you could achieve what you suggest and that is to use actual plates in a capacitor configuration. Charge them up then disconnect them from the source and then separate the plates. Both plates would still be "charged". You could then discharge one of them to ground and then put them back together. You would then have an unbalanced capacitor. Of course, as soon as you hooked it up to anything, it would immediately try to rebalance itself.

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  • \$\begingroup\$ I see. Still, if you removed electrons from the negatively charged side, wouldn’t you effectively end up with a neutral plate/surface on that side and a positively charged plate on the other side? (For a parallel plate capacitor). \$\endgroup\$
    – Tom
    Nov 6, 2017 at 0:00
  • \$\begingroup\$ @TOM. No, you need to understand it is the presence, or absence of electrons on one plate that drive away or attract electrons on the other plate. You can't change one without changing the other. \$\endgroup\$
    – Trevor_G
    Nov 6, 2017 at 0:02
  • \$\begingroup\$ "if you removed electrons from the negatively charged side..." How exactly do you propose to do that? The capacitor overall must remain electrically neutral. And if you did accomplish that, what you would be doing is positively charging the capacitor as a whole with regard to the rest of the universe (or wherever you chose to put those electrons you took away). \$\endgroup\$
    – Dave Tweed
    Nov 6, 2017 at 0:25
  • \$\begingroup\$ @DaveTweed that's the follow on paragraph.. \$\endgroup\$
    – Trevor_G
    Nov 6, 2017 at 0:29
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    \$\begingroup\$ Trevor, your answer is a little too hand-wavy for me. Once you start moving the plates relative to each other, you're doing work on the system, which just muddies the waters. \$\endgroup\$
    – Dave Tweed
    Nov 6, 2017 at 0:40
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Capacitor discharge is a process of reducing the stored charge in the capacitor. That is going to be a relative operation of the component itself and not something that you do to just one lead of the capacitor.

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If you charge the capacitor and separate out the capacitor plates, more charge will dissipate to ground when you ground side A of the capacitor and leave the side B floating, compared to the case where the plates are close together. Side A will now have less charge than side B. If you then unground side A and ground side B, side B will now have less charge than side A, this sequence will continue until the charge on both plates approaches zero.

When you do this with a capacitor with a microscopic distance between the plates, it would take an extremely long time until the charge reaches zero using this method because the induced charge on plate B is almost the same as the charge on plate A.

This is basic electrostatic induction theory. Connecting one of the plates to earth basically increases the size of the plate and the charges arrange accordingly due to the applied field of the other plate. If plate A is charged and B is uncharged, and the plates are extremely close together, the field applied to the larger conductor formed by plate B and earth causes the opposite charges to plate A to pool in plate B because it is more exposed to the field than the immediately surrounding surface of earth and much more exposed than the rest of earth. The conductor consisting of the connected plate B and earth is now polarised. The potential on plate B now is the negative of the potential experienced at B due to the charge at A according to coulombs law. When plate B is ungrounded, the result is that this charge configuration is now locked in place even when the external field from plate A is no longer applied.

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You just need to think outside of the box. Of course you can discharge one side of a capacitor.

If you charge a capacitor, one side has electrons and the other is equally electron deficient. Now create a pulse with a nuke EMP. No one will tell you that you just didn't discharged the one plate only (the other plate already had few electrons to move). And yes it probably doesn't take a nuke to do it, but this is not in some book.

If approached from a closed circuit perspective, the answer will be no. But open circuit may be different, what if you pulsed the positive side with a large negative pulse, the negative plate being attached to Earth ground with a diode?

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