When a parallel-plate capacitor has two different dielectrics as shown below, it can be considered equivalent to two capacitors in series, one taking the value of one of the dielectrics and the other of the other dielectric.
Similarly, if the dielectrics´boundary is perpendicular to the parallel plates of the capacitor, it can be considered to be two capcitors in parallel:
Here, it could be considered that k1 would make up a capacitor and k2 and k3 two capacitors in series.
What is an intuitive way to explain why we can consider a single capacitor with mixed dielectrics t be equivalent to capacitors in series and parallel, taking each case separately? Maybe showing it with diagrams and the polarisation of molecules would make it clearer why they can be considered single capacitors in series and parallel, because even though I know how to manipulate equations, I can't see where this assumption came from that they can be considered capacitors in series or parallel.
Best Answer
In Figure V.16 the equipotential surfaces between the two plates are all parallel to the plates.
So the line (surfaces) labelled $V$ is an equipotential.
Putting in an infinitesimally thin sheet of conductor along that equipotential surface makes no difference to the electrical properties of the capacitor.
Now increasing the thickness of the conductor again makes no difference to the electrical properties of the capacitor.
Cutting the conductor in half parallel to the plates and connecting a wire between between the central conductors makes no difference to the electrical properties of the capacitor.
So you now have two capacitors in series.
In the second example any "edge" effects around the centre of the capacitor can be assumed to have a negligible effect if the separation of the plates is very, very small compared with a linear dimension of a plate.
In this case cut the capacitor in half and connect the plates together with a wire so now you have two capacitors in parallel.
Split the right hand capacitor into two capacitors in series as explained above.