Consider a parallel plate capacitor of area $A$ with a distance between the plates $d$, disconnected from the battery. I wonder, how would the capacitance of the system change if we placed a conductor or an insulator (which also have some width $l$ < $d$) of smaller area $A'$ between the plates?
Apparently, we can't neglect edge effects in such scenario, so it is not clear how to calculate voltage between the plates. This system is equivalent to two capacitors in series (1-3 and 4-2), however, both of these capacitors seem to have different charges on each plate. In case of conductor, since the field inside is zero, charge on each side must be $Qconductor=Qplates*A'/A$, and in case of insulator, since the field inside must be decreased by $\epsilon$ - $Qinsulator=Qplates*A'*(1-1/(\epsilon))/A$. So we get two capacitors of different surface areas and with different charges. How can I calculate the capacitance of any of them? Is there another way to deal with the initial problem?