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I am trying to find equivalent circuit for a circuit composed of interconnected capacitors when isolated pieces of conductor may carry nonzero electric charge, and I have been struggling with this problem for some time now.

I want to understand what is the equivalent circuit for the situation depicted below. The only difference from the usual series capacitor circuit is the nonzero charge $q_m$ trapped on the piece of conductor marked with a dashed red square.

enter image description here

Such situation can be created by performing the following steps:
(1) A capacitor (capacitance $C$) is initially charged to hold some electric charge $q_m$;
(2) The capacitor is then connected in series with an identical, uncharged capacitor. The charge $q_m$ becomes trapped between the two capacitors;
(3) Finally, some voltage $U$ is applied across the circuit.

This is what I did to find equivalent circuit for the depicted situation:

Let's assume that electric potential of the leftmost electrode is $\varphi_L$, and electric potential on the rightmost electrode is $\varphi_R$, such that $U = \varphi_R - \varphi_L$. Electric potential of the charged piece of conductor in the middle is $\varphi_m$. Electric charges stored on the leftmost and rightmost plates are:

$q_L = C (\varphi_L-\varphi_m)\\ q_R = C (\varphi_R-\varphi_m)$

The total charge of the middle conductor is $q_m$, and therefore:

$q_m = C (\varphi_m - \varphi_L + \varphi_m - \varphi_R)$.

The potential $\varphi_m$ can now be excluded from the expressions for $q_L,q_R$, which yields:

$q_L + \frac{q_m}{2} = \frac{C}{2} (\varphi_L-\varphi_R)\\ q_R + \frac{q_m}{2} = \frac{C}{2} (\varphi_R-\varphi_L)$

Now, these are equations for charge stored on a capacitor with capacitance $C/2$, when the voltage across the capacitor is $\varphi_R-\varphi_L \equiv U$. It would seem that the seach for the alternative circuit is over now. However, electrostatic energies of the two circuits differ: for the original circuit it is:

$W_{orig} = \frac{C}{4}(\varphi_R-\varphi_L)^{2} + \frac{q_m^2}{4C}$,

while energy of the "equivalent" circuit is:

$W_{eq} = \frac{C}{4}(\varphi_R-\varphi_L)^{2}$.

Therefore the single capacitor $C/2$ can not be equivalent to the original series circuit!

Instead it seems to me that the original circuit may be equivalent to a set of two independent circuits, where one of them is the $C/2$ capacitor connected to a voltage source, and the other one is a piece of isolated conductor with capacitance $2C$ and charge $q_m$.

So far I have not found any books confirming or contradicting my solution or the physical interpretation. Do you have any ideas, where I should look for explanation? Or maybe you can confirm my line of thought or point out what is wrong with it? Any help will be much appreciated!

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The problem here - and it is an easily missed one - is in assuming that energy is conserved in your system. Consider just a single charged capacitor with a charge q and and a capacitance C. Connect it across another capacitor which is uncharged but has the same capacitance C. The energy of the system before connecting the two capacitors together is $Q^2/(2C)$ and the energy after connecting the two capacitors together is $Q^2/(4C)$. So there is a "missing" energy of $Q^2/(4C)$. That energy is necessarily lost (to heat, flash and heat of a spark, etc.) because it is impossible to conserve both charge and the energy of the system by connecting the capacitors in this way.

Your setup of capacitors and a voltage supply is a bit more complicated than this simple example, but the basic problem is the same: Charge is conserved, but you can't assume that the energy of the system is conserved.

See also my answer in Physics Stack Exchange: Energy Loss In Capacitors

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  • $\begingroup$ Thank you for your input! Energy loss due to dissipation would indeed be the case if I was comparing energies of the initially charged capacitor $C$ and the circuit with two capacitors in series. In this case, however the whole thing with connecting charged capacitors merely provides means for putting nonzero charge $q_m$ on the marked piece of conductor. The aim is to find equivalent circuit for the situation depicted in my question. I will edit the question to communicate my problem more clearly. $\endgroup$
    – Arturs C.
    Commented Mar 11, 2016 at 21:16
  • $\begingroup$ @ArtursC. - If you simply want an equivalent circuit to a capacitor C initially charged to some voltage V in series with a capacitor C with no initial charge, then that's just a capacitor with capacitance C/2 charged to a voltage V. If you regard the two capacitors in the first case as a black box, there's no difference between the two cases. If you discharge the combined system, you will end up with an internal charge on the inner plates of the two capacitors and, therefore, opposing voltages on them, but that is unobservable. The only observable is the total voltage, which would be zero. $\endgroup$
    – user93237
    Commented Mar 11, 2016 at 21:47
  • $\begingroup$ can you point out which step in my calculation would be equivalent to stating that I will treat my circuit as a black box? $\endgroup$
    – Arturs C.
    Commented Mar 15, 2016 at 13:02
  • $\begingroup$ @ArtursC. - Seems like you were asking for what the equivalent circuit was for a system of two capacitors as described above. If so, then the question involves regarding the pair as a "black box" unit and asking for an electrical system which gives the same electrical response without regard for the inner workings of the "black box". As I mentioned, this "black box" does contain a "hidden variable" (i.e., the charge on the inner plates of the two capacitors), but that hidden variable cannot be measured by any standard electrical measurements I can think of. $\endgroup$
    – user93237
    Commented Mar 15, 2016 at 17:03
  • $\begingroup$ Thank you, this seems reasonable. I was just wondering, which part of the presented calculations makes that transfer from the original system to a "black box" - I did come up with a correct equivalent circuit after all! Now that I have looked at my problem from this perspective it seems that the "black box" comes into play when $q_L+q_m/2$ is substituted by charge on a plate of the equivalent single capacitor circuit. The initial problem is defined by two variables: potential difference $U$ and charge $q_m$, while the equivalent circuit has only one dimension $U$. $\endgroup$
    – Arturs C.
    Commented Mar 15, 2016 at 22:17

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