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Suppose I energize one capacitor by connecting it across a battery, allowing it to achieve some potential difference V0 across its plates, then disconnect it and allow both of its leads to float in air. Next I take a second identical capacitor, whose leads have recently been shorted together, and connect one of its leads to one of the leads of the first capacitor. If the charges on the plates of the first capacitor were +Q and -Q before connecting the two capacitors (and of course 0 on both plates of the second capacitor), how do the charges distribute themselves among the four plates of the two connected capacitors?

I have attempted to answer this for myself using the method of contributor Manisheath at the post How does instant charging of one plate affect the potential of the other plate of a floating capacitor?. What I get, with the charged capacitor on the left and the charge of +Q on the leftmost plate of the left capacitor, is

0|+Q -Q|0 ----------------------------------------- 0|0 0|0

where the "--------" line represents the wire connecting the two capacitors.

In other words, no charge will flow. Is this correct? To me, this seems counter-intuitive, since on energetic grounds, I would expect stored energy to 'want' to spread out to a condition of lowest energy density per unit volume, mass, moles, whatever (without violating conservation of energy, of course). Recall, the energy stored in one capacitor is

E = 0.5*C*V^2 = 0.5*Q*V, since Q = C*V.

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Sure, energy WANTS to equalize. But without a path to equalize it can't. This is why skyscrapers can stand a thousand feet tall next to a flat ocean beach. And why pure hydrogen can exist happily in a bottle with that bottle surrounded by free oxygen. Etc. – mwengler Jun 15 '12 at 16:03
If your capacitors are real, extended physical objects, then currents will flow, depending on the relative potential of each plate to the others. For n conductors, there are n*(n+1)/2 coefficients of potential, which generalize the term "capacitance". In this case you mention two capacitances, but there are really 4*5/2 = 10 of these coefficients to this problem and they are all changing, as you are moving the capacitors around. So not only will currents flow when two electrodes are touching, but displacement currents will flow even while you are moving the capacitors without touching. – CuriousOne Aug 11 '14 at 23:12

Your guess is correct -- no current flows. If you want to be more accurate, you could say "negligible current flows". There will be a tiny bit of current thanks to stray capacitance.

Stray capacitance means that there is an imaginary capacitor between any two objects in the same room. So you can imagine a capacitor connecting the allegedly-open leads. If you draw the circuit now, you'll see that charge can flow from one capacitor to the other after all. But stray capacitance tends to be very small, so you can calculate that only a tiny tiny amount of charge will transfer between two capacitors. This depends on the situation I guess...if the "open" leads are both very very close to each other or to a common ground plane, the stray capacitance might not be that small, so maybe a measureable amount of charge will transfer to the uncharged capacitor. If the open leads are sitting in the air a few meters apart, the stray capacitance will definitely be very small and a totally negligible amount of charge will transfer.

You have a healthy intuition about matter rearranging itself to lower the energy density. Unfortunately the charge has a restricted range of can't pass through the capacitor dielectrics, nor through the air. Therefore it is impossible to for the charge to rearrange itself in order to half-charge each capacitor.

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Something is not clear to me. If one capacitor is charged with $Q$ and the other is not then I connect them in series, what prevents the charge to be equally distributed between the inner plates that became now connected? That new charge came on that plate, which was zero originally, what prevents it from inducing a charge on the other neutral plate so the 2nd capacitor becomes charged? – Revo Nov 7 '12 at 19:18
@Revo, that would require either a large charge on the open leads (which is not realistic unless they've got nontrivial capacitance, stray or otherwise) or for a net charge of $Q/2$ on two capacitor plates (the non-connected one on the old one and the connected one on the new one) which will incur a massive Coulomb potential energy. – Emilio Pisanty Nov 7 '12 at 20:16
@EmilioPisanty There is something I am still missing. Let me call the 4 plates of the capacitors 1, 2, 3, 4 as shown ----1| |2--------3| |4----, the dotted lines are conductors, wires say – Revo Nov 8 '12 at 0:05
@EmilioPisanty Now assume 12 was the charged capacitor with $Q$ and 34 was the uncharged one. Now tell me if my following understanding is right or wrong and if wrong why. After connecting the capacitors in series $Q$ on plate 2 will flow to plate 3 and will be distributed equally between plates 2 and 3. Now since 3 is charged, it will induce an opposite charge on plate 4 (not sure if it will be equal in magnitude or not). Similarly not sure what will be the charge on plate 1 – Revo Nov 8 '12 at 0:11
No, that is wrong. – Emilio Pisanty Nov 8 '12 at 12:35

I'm not quite sure I understand your methology, but I would write the capacitors down like this: q+ | q- ----- 0 | 0 with --- representing the wire, | representing the gap or dielectricum between the plates, q or 0 beeing where the plates are.

So, you have a difference of potential between one plate of the empty capacitor and the connected on of the full C, current will flow. I think your intuition is basically sound though I would look at charges and potential. In a symetrically charged capacitor, there's a potential to ground (or anything not charg) from either plate.

I think the problem is you notation - what do the zeros in 0|+Q -Q|0 represent, what the |?

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