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?

The following diagram might clarify what I mean. Its an explanation of why I think the charges on both plates aren't equal:

enter image description here

Please note I know that you can't use Coloumb's Law with plates. But Coloumb's Law allows superposition, so consider the above diagram to be the sum of all the individual point charges on the plates.

  • \$\begingroup\$ If there's an imbalance in the charges then how are they supposed to remain on the plate? \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 4 '14 at 4:00
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    \$\begingroup\$ Can you clarify what you are asking? Are you asking why charges on a capacitor's plates must be equal and opposite? \$\endgroup\$ – SomeEE Feb 9 '14 at 17:27
  • \$\begingroup\$ @MathEE Kinda. The book says the charges on both plates would be equal and opposite because one plate would attract electrons until it has enough electrons for the field to be zero. The book says the field will be zero when both plates have equal charge since their fields will cancel out. I'm asking why they would be equal since the plates aren't the same distance from the electrons. \$\endgroup\$ – dfg Feb 9 '14 at 20:06
  • \$\begingroup\$ @MathEE I added in a diagram to clarify what I mean. Does it help? \$\endgroup\$ – dfg Feb 9 '14 at 20:14
  • \$\begingroup\$ You'd get a better answer and more attention on physics.stackexchange.com \$\endgroup\$ – Scott Seidman Feb 9 '14 at 20:37

Your gap is too wide. Make it very, very very narrow. And rolled into a cylinder. Like a real capacitor.

Yes, +q could be less than -q, but only if the attraction/repulsion effects of electrons in the connecting wires were nearly as large as the attraction/repulsion down between the capacitor plates. (In that case the plates wouldn't be a near-perfect electrical shield for the fields produced by the wires.) But with real-world capacitors, this doesn't happen, and instead the field between the plate is totally enormous compared to the tiny fields produced by electrons in the wires. If +q only differs from -q by a millionth of a percent, we ignore it. See Engineer's capacitor vs. Physicist's capacitor, a split metal ball, versus two separate balls.

For capacitors used in circuitry, if we dump some charge on one capacitor terminal, exactly half of it will seemingly migrate to the other terminal. Weird. But "physicist-style capacitors" with small, wide-spaced plates are different, and an extra electron on the wire will make +q not equal to -q.

In detail: if the capacitance across the plates is 10,000pF, and the capacitance to Earth of each wire and plate is 0.01 pF, then the opposite plate's charges will ignore any small +q and/or -q on the connecting wires. The attraction/repulsion of electrons in the wires doesn't significantly alter the enormous +q and -q on the inner side of the capacitor plates.

Engineers use real-world components: wide capacitor plates with very narrow gaps; gaps the thickness of insulating film. But if you were a physicist, your capacitors might be metal spheres with large gaps between, or metal disks where the space between the plates was large when compared to their diameter. (Or you'd draw a capacitor symbol where the gap between plates was enormous and easy to see.) In this case the attraction/repulsion of electrons on the connecting wires would have an effect on the balance of +q -q between capacitor plates.


Another weird concept: make a solid stack of thousands of disc capacitors: foil disk, dielectric disk, foil disk, etc. Use half-inch wide disks, and stack them up into a narrow foot-long rod. Now connect one end to 1,000 volts. The same kilovolt will appear on the other end! The rod is acting like a conductor. Yet its DC resistance is just about infinite. Series capacitors! Each little capacitor induces charge on the next and the next, all the way to the end.

  • \$\begingroup\$ Sorry, I don't really understand. My fault, my question wasn't very clear. I added in a diagram to clarify what I mean. \$\endgroup\$ – dfg Feb 9 '14 at 20:14
  • \$\begingroup\$ @dfg added more. \$\endgroup\$ – wbeaty Feb 9 '14 at 21:47

That is a pretty tortured explanation. It makes it sound like the electron's influence flows more like water than a force communicated near the speed of light. Note that the pairs of plates in the center of the series, like the + of C1 and the - of C2, ---||---these---||--- are not connected to anything else. Their total charge must be conserved, so if one side is +q, the other must be -q, and so on through the string of caps. To answer about plates with gaps, I would consider the electric field. What is the voltage across each cap in a series, and why?


The force on the electron in your diagram is NOT $$\frac{k\cdot q\cdot e}{d^2}$$ That formula is only valid for point charges (more on this later).

On capacitors, in the ideal model, which is the one you should concern yourself with for this question, an electrical field is generated, constant over all the space, with no change in regards to distance. The Positive charged plate will create fields pointing outward, and the negative charged plate will create a field oriented inward, represented by the arrows below:

enter image description here

And the force on any \$q\$ charged particle is equal to this constant field \$E\$ times the charge on the particle.

As you can see, if we have equal charges on the capacitors, the field in the external regions is zero.

The field on the internal regions is constant, oriented towards the negatively charged capacitor, and (assuming the plates are equal) has magnitude \$2\cdot E\$, as represented below:

enter image description here

Your comment on "superposition" is also wrong, or at least grossly oversimplified. Considering now a \$real\$ model. If you actually want to superpose small charges on the plate with the charge on the wire, you would have to:

  1. Find a generic point on the plane in a referenced coordinate system.
  2. Find the distance between this point and the charge.
  3. Apply coulombs law over this small charge.
  4. Integrate over the entire plane.

This is for one capacitor. What you would find, after using a computer program to do this on the entire system, is that the field close to the finite plane would be similar to the ideal model. The fields around the borders would be a bit different, and the external fields would die out (note: they are dying out both because the fields from the positive and negative plates are cancelling each other out, but also because the field dies down due to the infinite plate model not being valid over bigger distances).

This result can be visualized in the picture below. The force on each electron particle is proportional to the sum of all the "arrows" of where you place it.

enter image description here

You can see that on the point where your electron in your drawing is, there are no field components from the capacitor, so the capacitor has virtually no effect on it.

That's about as much detail as I'm going to get on the "real" model of the capacitor.

To sum things up, do yourself a favor and think about the ideal infinite-plane model, and you will see that it makes perfect sense for the charges to be equal. Your idea of superposition is wrong, and the correct way to use it would just end up giving a result identical to the original model, for all intents and purposes.

  • \$\begingroup\$ I think OP is talking about adding extra charges to just one capacitor terminal. Models with balanced +q and -q don't answer his question. \$\endgroup\$ – wbeaty Feb 9 '14 at 21:59
  • \$\begingroup\$ His question is pretty clear: --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?-- He thinks that for the electron in his drawing to be in equilibrium, one plate would have to have more charge than the other. So I'm answering no, that is not the case, and that his superposition argument is wrong \$\endgroup\$ – triplebig Feb 9 '14 at 22:03

Why you would induce a charge of -q on the other plate if the first plate is given a charge of +q?

Its due to conservation of charge. Lets assume from the start that in a single capacitor everything is neutral, and that nothing is polarized. Electrons and protons are evenly distributed on both sides of a capacitor. Then you apply a voltage across the capacitor, which separates X amounts of electrons (aka the -q) from the protons (aka the +q). You have now given a +q to one side, the other side automatically has -q.

In a series capacitor, consider the following:

enter image description here

When you apply a voltage V and stuff in a +q charge on the positive side of C1 (that you have taken from the negative side of C2), it induces a -q on the negative side of C2. This inturn polarizes a +q on the positive side of C2 and -q on the negative side of C1. How can this be you say? Well here is the key....realize that the negative plate of C1 and the positive plate of C2 are physically connected and isolated from the rest of the circuit, therefore the total charge on them must remain constant! The charges must all balance.


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