Material between capacitor plates - viewed as two capacitors in series

[Mircro]

Homework Statement:

Between the plates of the capacitor which have area $S=60cm^2$ there is paraffine plate thick $d_1=4cm$ and a layer of air thick $d_2=2cm$.
Determine the capacity of the capacitor if relative permittivity of paraffine is $\varepsilon _{r1}=2,3$.

Relevant Equations:

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$
$C=\varepsilon \cdot \frac{S}{d}$

This is a second grade high school problem, translated from my native language.

I don't have a problem with calculating, but with understanding the concept. There is an instruction with the assignment that says: The capacitor can be viewed as a combination of two capacitors in series with relative permittivities $\varepsilon _{r1}=2,3$ and $\varepsilon_{r2}=1$.

I would like to develop a deeper understanding of this problem so I could recognise similar problems in the future, hopefully even without the given instruction.

So why this can be viewed as two capacitors in series? Can we say that the charge accumulates on the first plate, then on the other side of the of the paraffine plate, so we have first plate, paraffine plate and the second plate, so it is like two capacitors with one plate (paraffine plate) in common? I was thinking: because of the way that this system (hope we can call it that way) is put together in space, there will be the same amount of charge in every of the plates, which is a characteristics of capacitors connected in series.

Thanks in advance!

 

Across a capacitor, there will be a voltage drop which relates to the integral of the electric field between the plates. If you have two different dielectrics in series, then you have two different electric field strengths and the voltage drop is a sum of two integrals.

If you replace the single compound capacitor with two capacitors in series each containing only one type of dielectric, the two outermost plates are still of charges $Q$ and $-Q$, and the electric fields in each respective dielectric are the same as before, so we still get the same voltage drop when we add the two.

The electric fields in each of the dielectrics are the same as before since the electric field next to an (infinite) charged plane in a dielectric only depends on the surface charge density and the permittivity in that dielectric, and not the separation of the plates!

So if we imagine the stacked capacitor as two capacitors in series, our component has the same voltage drop and same terminal charge, so is indistinguishable.

 

[Cutter Ketch]

The two plates create an electric field. Paraffin is an insulator. The charge is not free to flow, but it CAN polarize. All of the electrons will move just a little toward the positive plate. (And the positive atom cores will squeeze toward the negative plate to the degree that is possible, but not much)

Picture a block of electrons the shape and size of the paraffin. To start it is exactly overlapped with a block of atom cores the exact same size and shape and total charge. To begin with the whole thing is neutral. Now picture moving the whole block of electrons a little one direction. Where the blocks overlap the charge is still neutral. You get a sliver, a thin plate of excess negative charge on one side and a sliver of excess positive charge on the other. The charge in the paraffin polarizing in the applied field comes out looking like two thin plates of charge, very much like a charged capacitor with the opposite direction to the plates.

 

[Mircro]

I have a problem grasping some parts of your answers, especially the part with integrals and similar, since my knowledge is still limited to a high school level. But both answers gave me additional insight to the intuitive understanding of the process, which was the most important for me at this moment.

So thank you very much!