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An uncharged capacitor has an equal amount of positive and negative charges in both plates, meaning there are ions in both plates which altogether have a neutral charge.

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When you connect an uncharged capacitor to a battery, the electrons from the battery’s anode move to the capacitor's plate connected to the anode (plate A on the illustration below), whereas electrons from the other plate of the capacitor (B) move to the battery’s cathode.

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Does this mean that the battery is not moving electrons from one plate to the other, rather it’s only moving charges?

Does this mean that all electrons are removed from the ions on the plate (B), and all that’s left on the plate (B) are cations, whereas the plate (A) now has free-floating electrons from the battery, as well as the previously present ions?

If that’s the case, then the illustration above isn’t correct, right? Otherwise, how can there be an equal amount of neutral charges inside both plates once the capacitor is charged?

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2 Answers 2

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We generally don't think about electrons and ions in this kind of circuit. We just use charge. We normally assume that the capacitor as a whole remains neutral (the way to say this is that we ignore "stray capacitance" between the capacitor and the environment). If the capacitor is "charged", that means that the plates have equal and opposite charge.

Now if the plates are metal, the negative charge can be understood as extra electrons in the conduction band. The positive charge can be understood as a deficit of electrons in the conduction band. But the electrons in the conduction band are not localized: they don't belong to any particular nucleus, so there are no identifiable ions.

In an "electrolytic" capacitor, one of the plates is an electrolyte, so its charge may be understood as an ionic imbalance. In a semiconductor capacitor, the charge carriers may be holes. But if you just consider charge, you don't have to worry about what it represents at the atomic level.

Note that, in practice, the amount of charge stored is tiny compared to the total charge of electrons in the capacitor. In a semiconductor, you can deplete the free electrons or holes, but the bound electrons greatly outnumber them.

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  • $\begingroup$ "But the electrons in the conduction band are not localized: they don't belong to any particular nucleus, so there are no identifiable ions." I assume this is after charging? Do they belong to a nucleus in neutral state, and then get pulled away from the nucleus when the battery is attached, or are they not localized to begin with? And if the latter is the case, is that just the natural state of the metal that it contains such unlocalized electrons? $\endgroup$
    – user1329190
    Feb 19 at 18:08
  • $\begingroup$ @user1329190 A metal is characterized by a dense collection of unlocalized free electrons. For copper, the density of free electrons is $\approx8.4 \times10^{28} m^{-3}$. We would not generally model all of the electrons as delocalized (for example, the tightly bound K shell electrons are effectively localized). $\endgroup$
    – John Doty
    Feb 19 at 18:17
  • $\begingroup$ Oh, reading up on bands... Is it just that there's so many atoms, and there's so much overlapping orbitals, that it's kinda nonsensical to delve into whether electrons belong to any particular nucleus. They're there, in the overlapping orbitals, and those that are "loosest" get away. It only makes sense to observe it in totality, through formulas. $\endgroup$
    – user1329190
    Feb 19 at 18:17
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...meaning there are ions in both plates which altogether have a neutral charge.

Ions are defined as "atoms or molecules with a net electric charge due to the loss or gain of one or more electrons". There are no ions on the plates of the capacitor. There are fixed protons and free electrons constantly moving (thermally) about the positive charge in a kind of "electron cloud}".

Diagram of "uncharged capacitor"

Your diagram of the "uncharged" capacitor is misleading. You show the positive and negative charge lined up. In reality they are randomly oriented. See Fig 1 below.

When you connect an uncharged capacitor to a battery, the electrons from the battery’s anode move to the capacitor's plate connected to the anode (plate A on the illustration below), whereas electrons from the other plate of the capacitor (B) move to the battery’s cathode.

Better to say that the cathode pulls free electrons off of one plate, making it positively charged, and the anode deposits an equal amount of electrons on the other plate, making it negatively charged. The battery does not supply electrons and protons. It supplies the energy to move electrons from one plate to the other. See FIG 2 below.

Your diagram of the charged capacitor is incorrect. It shows a total of 4 more electrons and 4 more protons than were on the uncharged capacitor. It appears you are under the incorrect impression that the battery supplies electrons and protons to the capacitor. It does not. It only performs the necessary work to move electrons already existing in the capacitor from one plate to the other.

Does this mean that the battery is not moving electrons from one plate to the other, rather it’s only moving charges?

The only charges that can be taken off one plate and deposited on the other are electrons. The protons are fixed in place and cannot move.

Does this mean that all electrons are removed from the ions on the plate (B), and all that’s left on the plate (B) are cations, whereas the plate (A) now has free-floating electrons from the battery, as well as the previously present ions?

I'm not sure I understand what you are asking. The amount of net charge $Q$ on the capacitor will depend on the capacitance and battery voltage according to

$$Q=CV$$

Hope this helps.

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  • $\begingroup$ It does help, but I still don't get how the electrons are being moved from one plate to another through the battery. One wire is connecting one plate to the anode, and another is connecting cathode to the other plate. If electrons were being moved from one plate to the other, it would mean they're passing through the battery, but the battery would have to have some sort of barrier to prevent exactly that from happening, otherwise, the postive and negative charges already in the battery would neutralize, as there'd be nothing keeping them apart. From what I've read... $\endgroup$
    – user1329190
    Feb 20 at 11:21
  • $\begingroup$ the chemical reaction inside the battery is releasing electrons from the electrolyte, which are building on the one side (anode), them being ready to be released once the battery is connected to a circuit, whereas the chemical reaction on the other side makes the cathode more "open" to accept incoming electrons. Which would mean that one and same electron is not traveling from (A) to (B), but one electron is traveling from (A) to cathode, after which a different electron is released from anode to (B). (Or an electron is first released from anode, after which another is drawn to the cathode?) $\endgroup$
    – user1329190
    Feb 20 at 11:23
  • $\begingroup$ What goes on inside a battery is a whole separate question. I suggest you do some research and then, if you have specific questions submit them in a new post $\endgroup$
    – Bob D
    Feb 20 at 11:36
  • $\begingroup$ You might want to start here: science.org.au/curious/technology-future/batteries $\endgroup$
    – Bob D
    Feb 20 at 11:53
  • $\begingroup$ Yeah, that's where I read it. "There are a couple of chemical reactions going on that we need to understand. At the anode, the electrode reacts with the electrolyte in a reaction that produces electrons. These electrons accumulate at the anode. Meanwhile, at the cathode, another chemical reaction occurs simultaneously that enables that electrode to accept electrons. " Then: "What we end up with is electrons being attracted to the cathode from the anode (and the anode not trying to fight very much), and when provided with an easy pathway to get there—a conducting wire—we can harness... $\endgroup$
    – user1329190
    Feb 20 at 14:25

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