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Capacitors in series have identical charges. We can explain how the capacitors end up with identical charge by following a chain reaction of events, in which the charging of each capacitor causes the charging of the next capacitor. We start with capacitor 3 and work upward to capacitor 1. When the battery is first connected to the series of capacitors, it produces charge -q on the bottom plate of capacitor 3. That charge then repels negative charge from the top plate of capacitor 3 (leaving it with charge +q). The repelled negative charge moves to the bottom plate of capacitor 2 (giving it charge -q). That charge on the bottom plate of capacitor 2 then repels negative charge from the top plate of capacitor 2 (leaving it with charge +q) to the bottom plate of capacitor 1 (giving it charge -q). Finally, the charge on the bottom plate of capacitor 1 helps move negative charge from the top plate of capacitor 1 to the battery, leaving that top plate with charge +q.

Why would inducing a charge of +q on one plate cause the other plate to acquire a charge of -q? I get that it would attract electrons from the other side, but the plates aren't the same distance from the electrons, so wouldn't the charge be less than q?

For example, consider the following capacitor in series (the other capacitors are on its left). If you induce a charge on the right plate, electrons on the left side will build up on the left plate. But since the left plate is closer to the electrons on the left wire than the right plate, the electrons will stop moving when the field on the left plate is greater than -q.

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  • $\begingroup$ I'm asking about a capacitor in series, not a single capacitor $\endgroup$ – dfg Feb 4 '14 at 3:40
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Since the capacitor is initially electrically neutral, the excess charge of $q$ needs to be compensated by something. We know that there must be a charge of $-q$ somewhere in the system, and in conductors, charge is only ever found at the surface.

You are confusing the charge with the electric field. The field itself cannot be "greater than $-q$" as you say, because it has different units!

EDIT: More explanation Okay, so your question is why the heck does the charge turn out to be the same? Your argument is that somehow the distance of the plates plays a role.

Well, here's my argument why the charges on equal sites of a capacitor must be the same: At least for plate capacitors it's easy to see using Gauss law, and the fact that INSIDE a conductor, the electric field must be zero. But then, imagine a box with one side parallel going through the left plate and the other side parallel going through the right plate of the capacitor, and then connect those surfaces far away from the capacitor (ignoring stray fields).

Since the field in the conductor is zero, the Gaussian integral gives 0, and this tells us that the enclosed charge must be zero. Since we have charge $q$ on one plate, we know that there must be charge $-q$ on the other plate.

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  • $\begingroup$ Sorry I don't understand. The excess charge on one plate is compensated by having a deficit of charge on the plate of the capacitor connected to the other end of the battery. So why does there need to be more movement of charge? $\endgroup$ – dfg Feb 4 '14 at 2:11
  • $\begingroup$ Because the charges on one plate of a capacitor will electrically attract opposite charges on the other plate of the capacitor. $\endgroup$ – Lagerbaer Feb 5 '14 at 2:35
  • $\begingroup$ Right, I get that, but the attraction will stop when the charge on the second plate is less than q because the electrons are closer to the second plate. For example, if these are the plates: -----||----- and the plate on the left is charged and the plate on the right isn't. Electrons on the right side will start being attracted to the right plate but will stop being attracted before the charge on the right plate is q since the right plate is closer to the electrons. Do you see what I mean? $\endgroup$ – dfg Feb 5 '14 at 3:09
  • $\begingroup$ I guess your question has less to do with capacitors per se and more with induction and the field of conducting plates. I have edited my answer. $\endgroup$ – Lagerbaer Feb 6 '14 at 3:59
  • $\begingroup$ Great explanation! Thanks to your answer I now get why the plates have to have equal charge, but why was my original idea (that I expressed in my previous comment) invalid? $\endgroup$ – dfg Feb 6 '14 at 5:42

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