Potential between swapped plates between two plate capacitors Suppose we have two plate capacitors $C_1$ and $C_2$. Each one is charged up to the same voltage $V$ using two different circuits with no common connections between them. Each capacitor is then removed from their circuit, reading a voltage of $+V $ between the $+$ and $-$ of each capacitor. Now their negative plates are swapped, i.e., $C_1$- plate replaces $C_2$- and vice versa. What voltage would be read between the $+$ and $-$ plate of the new "Frankenstein" capacitors?
 A: If the capacitors $C_1$ and $C_2$ are identical, their plates hold the same charge after being charged to the same voltage $V$. After swapping the plates, the situation is indistinguishable from before (there are just two identical capacitors with the same charge), so the voltage of both new capacitors will be $V$ as well.
If the capacitors are different, meaning that the capacitances $C_1$ and $C_2$ differ, the initial charge on each of the plates can be calculated as
$$
Q_i = C_i V~,
$$
where $i \in \{1,2\}$. Thus, the total charge on each plate after the plate-swapping is known. If the capacitance of the new capacitors is known as well, the above formula can be used to calculate the new voltages. Otherwise, you will have to solve the Poisson equation,
$$
\Delta \phi(\vec r) = - \frac{\rho(\vec r)}{\varepsilon_0}~,
$$
where $\phi$ is the electrical potential, $\rho$ is the charge density, $\vec r$ is the position, $\Delta$ is the Laplacian and $\varepsilon_0$ is the dielectric constant of the vacuum. The total charge on each of the plates serves as boundary condition for the solution.
Remark 1: The above holds true only for ideal insulation, meaning that the capacitors do not lose any charge after being disconected from the power source.
Remark 2: When using the new capacitances to calculate the voltages, be sure to use a reference potential for which one plate of the new capacitor holds the charge $+q'$ and the other one $-q'$. This means, the charge to use for the calculation of the voltage is actually $(Q_1 + Q_2)/2$.
A: There is not clear if both capacitors are identical? Are "C1" and "C2" are just names of these identical capacitors or capacitors are different in "C1" and "C2" are capacitances of these capacitors?
If capacitors are identical looks like the answer is obvious: the voltage reading on both "Frankenstein" capacitors would be V. The electric charge on each plate was +Q, -Q, +Q, -Q, and nothing changed when we switched two identical plates with same charges.
There is an inaccuracy here. If we add some charge to both plates of a capacitor the voltage on this capacitor will not change. So may be initial charges on our plates were (Q+q), (-Q+q), +Q, -Q. Then reading on "Frankenstein" capacitors will be different and depend on the value q.
If capacitors are different it's also not possible to calculate capacitors readings after plates swap. There were some charges capacitors plates before the switch: +Q1, -Q1, +Q2, -Q2. After the switch we will have some configuration of plates with same charges, but we have very little information about this configuration to calculate the voltage. What does it even mean to "switch plates" if the distance between plates of first capacitor is different from the distance between plates of the second capacitor?
