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Suppose I have two capacitors in series, but with an initial trapped charge $Q_0$ between both capacitors.

What will be the total electrostatic energy stored on the system if we apply a voltage $V_g$ accros this capacitor?

I first thought this would be a simple electrostatic problem, but I guess that $Q = C \cdot V$ for a capacitor doesn't really apply here since in this case the charge doesn't have to be the same on both plates of a capacitor. Has anyone an idea how to solve this?

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  • $\begingroup$ It's not clear what you are asking. What do you mean by "trapped charge there"? $\endgroup$
    – Bob D
    Oct 19, 2023 at 10:51
  • $\begingroup$ There is a net charge on the wire between C1 and C2. You can google "trapped charge on series capacitors" for more information. $\endgroup$ Oct 19, 2023 at 11:22
  • $\begingroup$ And just how does that happen? $\endgroup$
    – Bob D
    Oct 19, 2023 at 11:24
  • $\begingroup$ You still haven't told me exactly how excess charge is only on the wire between C1 and C2 before connecting the voltage source. I doubt anyone will connect your question with "Quantum coherence with a single Cooper pair". $\endgroup$
    – Bob D
    Oct 19, 2023 at 11:47

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Consider the individual charges of the capacitors $Q_1,Q_2$. The charge of the wire is: $$ Q_0 = Q_2-Q_1 $$ The sign depends on the convention chosen, I chose the one where the current flows in the clockwise direction. Note that due to charge neutrality, you'll need the other portion of the wire to be charged as $-Q_0$. If you don't want to respect charge neutrality, check out the last paragraph.

If you initially have $V_g=0$, then: $$ U_1 + U_2 = 0 $$ so: $$ \frac{Q_1}{C_1}+\frac{Q_2}{C_2} = 0 $$ Using your prescription, you can solve for the charges, the voltages and the energies to get: $$ \begin{align} Q_2 &= \frac{C_2}{C_1+C_2}Q_0 & Q_1 &= -\frac{C_1}{C_1+C_2}Q_0 \\ U_2 &= \frac{1}{C_1+C_2}Q_0 & U_1 &= -\frac{1}{C_1+C_2}Q_0 \\ E_2 &= \frac{C_2}{2(C_1+C_2)^2}Q_0^2 & E_1 &= \frac{C_1}{2(C_1+C_2)^2}Q_0^2 \end{align} $$ with the total energy: \begin{align} E &= E_1+E_2 \\ &= \frac{Q_0^2}{2(C_1+C_2)} \end{align} Thus, the trapped charge acts like a capacitor of capacitance $C_0 = C_1+C_2$. In particular, trapping a charge requires energy as expected.

Now you can also add an external voltage. The second equation is changed to: $$ U_1 + U_2 + V = 0 $$ You now get: $$ \begin{align} Q_2 &= \frac{C_2}{C_1+C_2}Q_0-\frac{C_1C_2}{C_1+C_2}V & Q_1 &= -\frac{C_1}{C_1+C_2}Q_0-\frac{C_1C_2}{C_1+C_2}V\\ U_2 &= \frac{1}{C_1+C_2}Q_0-\frac{C_1}{C_1+C_2}V & U_1 &= -\frac{1}{C_1+C_2}Q_0 -\frac{C_2}{C_1+C_1}V\\ E_2 &= \frac{C_2}{2(C_1+C_2)^2}(Q_0-C_1V)^2 & E_1 &= \frac{C_1}{2(C_1+C_2)^2}(Q_0+C_2V)^2 \end{align} $$ with the total energy: \begin{align} E &= E_1+E_2 \\ &= \frac{Q_0^2}{2(C_1+C_2)}+\frac{C_1C_2V^2}{2(C_1+C_2)} \end{align} Your circuit can therefore be seen as two effective capacitors. The first one corresponds to the trapped charge $Q_0$, giving it its charge and has capacitance $C_0$. The second one corresponds to the imposed voltage $V$, giving it its voltage and has capacitance $C = \frac{C_1C_2}{C_1+C_2}$, the usual formula for capacitors in series.

If you really want your circuit to have a net charge, then you'll need to add additional capacitance coefficients. They will relate the net charge of the circuit to the net potential of the circuit with respect to infinity. The same analysis applies.

Hope this helps.

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    $\begingroup$ "Consider the individual charges of the capacitors $Q_1,Q_2$." The problem is the OP is not saying (yet) that there are charges on the two capacitors, just on the wire between them, which I have asked for an explanation (yet to be given). $\endgroup$
    – Bob D
    Oct 19, 2023 at 12:26
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    $\begingroup$ Yes, for the quantitative discussion, I assumed charge neutrality (I suspect that was the spirit of the OP). The last paragraph is for the case that there is only the charge $Q_0$ and no charge neutrality. $\endgroup$
    – LPZ
    Oct 19, 2023 at 13:16

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