Apparent paradox of current through two parallel plate capacitors in series in an open circuit This apparent paradox has been wrecking my brain. Suppose we charge up two parallel plate capacitors (with distance $d$ between each pair of plates) independently with different batteries to voltage $V$ and then take them out of their respective circuits. We then connect them in series with no battery and without closing the circuit. When we connect them in series without closing the circuit, will there be transient current flowing? I think not - the circuit isn't closed.
But here is what's confusing. Suppose that before connecting the two capacitors in series, we increased the distance $d$ between each capacitor's pair of plates. As far as I understand, all that will do is increase each capacitor's voltage, reducing capacitance but preserving charge. When we connect the two capacitors in series again, there again shouldn't be any current flowing between them as long as we aren't closing the circuit.
But here is where I am confused. If we move the distance $d$ between each capacitor's pair of plates to infinity, connecting them in series is now equivalent to shorting two charged capacitor plates, i.e. the + plate of the first capacitor and the - plate of the other capacitor. This supposedly should generate current. Where am I going wrong in thinking about this?
Thank you
 A: First, when charging the capacitors, the 2 circuits should have the same ground/reference voltage for us to deduce that they have the same voltage across the plates.

will there be transient current flowing?

let's assume that the plates connected are the ones that have a positive charge, it is rather intuitive that the charges won't move. now if the plates with opposite charges are connected, the charges will start flowing until there is no difference, much like how in thermodynamics the exchange in heat is taking place until there is no difference in temperature. The 2 plates connected to each other will become neutral and the plates at the other ends will preserve charge, so if you only look at the outer plates, it's like you have a "new capacitor"!

This supposedly should generate current

Again, if you have less electrons on one plate compared to the other, meaning one plate is negative and the other positive, it is only intuitive to see how electrons will flow from the side with more to the other with less, like water falling down in a waterfall.
I don't really understand what you mean when talking about the distance, but yes, if there are opposite in charge there will be movement of electrons, so a transient current. you don't necessarily need for the circuit to be closed to have transient bursts of current. Another example is a rod, moving through a magnetic field on it's own. once it starts moving, negative charges and positive charges accumulate on opposing ends and this happens as soon as the rod starts moving, but after the charges are at the opposite ends, nothing more will happen.
PS: a drawing of what you mean could make your question much clearer. As it goes, "A picture is worth a thousand words".
EDIT: Calculation of this current, assuming resistance $R$ between the 2 connected plates with potential difference $V$ across the caps:
Let's take the easy scenario, where two infinitely big plates are connected to each other (the scenario in which the distance between cap plates tends to infinity, making the question just the connection of 2 plates), one having charge $-Q$ and one $+Q$. Let's call the charge on one $Q_1$ and the other $Q_2$, which are functions of time. One assumption that needs to be made here is that the distance between them, i.e. the length of the wire is much smaller than the dimension of the plates, and that the plates are similar. There are a few things that we can be certain  of, one is that the rate of exchange of charge between the plates is proportional to the difference in charge, could be written as
$$\frac{dq}{dt}\;\alpha \;|Q_2-Q_1|$$
It is also known that $Q_2 + Q_1 = 0$. replacing this in the equation above, you will get:
$$\frac{dq}{dt}\;\alpha\; 2Q_2$$
One can endeavor to figure out the sign conventions intuitively. If plate two has more positive charge, the positive charges will "deplete" (i.e. electrons moving in the other dir'n), so the rate of change of $Q_2$ is negative meaning it is decreasing, this can be written as
$$-\frac{dQ_2}{dt} = 2Q_2k'$$
In this equation,$\frac{dQ_2}{dt}$ is the current, and you can see how it decreases with decreasing charge. $k'$ is a constant to do with the geometry and the scenario specific settings of the question. Solving the equation will yield:
$$Q_2 = Ae^{-2k't} \; = \; Ae^{kt}$$
$k$ is just $-2k'$. we know that at $t=0$, this charge is $+Q$, so $A=Q$. Current will then be:
$$\frac{dQ_2}{dt} \;=\; kQe^{kt}$$
This is just the very basic, and I hope you can now appreciate how you can have different levels of abstraction when solving these types of questions. Avoiding these sort of complications is one of the reasons engineers use "lumped models" when dealing with circuits.
If we take into account the effect of the 2 outer plates, it will become a bit more complicated. Then the charges will exchange until equilibrium, but equilibrium is no longer when the 2 connected plated are neutral, because now the attraction force from the outer plates have to be taken into account, meaning equilibrium is when the force from the outer plate is equal to the tendency of the charges to go to plate they are connected to.
