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I encountered the following problem: A parallel plate capacitor has plate area $A$ and plate separation $d$.The space between the plates is filled up to a thickness $x<d$ with a dielectric of dielectric constant $K$. Calculate the capacitance of the system.

Here's how the solution was given in the textbook:

The given system is equivalent to the series combination of two capacitors, one between between $a$ and $c$ and the other between $c$ and $b$. This is because the potential at the upper surface of the dielectric is constant and we can imagine a thin metal plate being placed there. [What followed next was series calculations of the capacitors.]

My doubt is that how can we assume that between $a$ and $c$, and between $c$ and $b$ to be different capacitors? If we place a charge $+Q$ on the plate $a$, then since $(a,b)$ is a capacitor $b$ must have a charge of $-Q$ (by the definition of capacitor), but then $c$ will have a charge of $Q(1-\frac{1}{K})$, so $(b,c)$ cannot be a capacitor as they have different magnitude of charges. Can someone explain why $(a,c)$ and $(c,b)$ can be taken as individual capacitors? Thanks image->

enter image description here

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I think that you have a misconception as to the role of the dielectric.

Imagine a parallel plate capacitor plate separation $d$, area $A$ with a vacuum between the plates and not connected to the surroundings so $q$ stays constant.

Charge $\pm q$ on the plates produces an electric field $E = \frac {q}{\epsilon_0 A} = \frac v d$ between the plates where $v$ is the potential difference between the plates and the capacitance of such an arrangement is $C = \frac qv = \frac{\epsilon_0 A}{d}$.

Now introduce into the space between the plates a dielectric (constant $k$) whose thickness is very slightly less than the separation of the plates as shown in the diagram below.

enter image description here

Charges $\pm q'$ are induced on the dielectric with $q'=q(1-\frac 1k)$ and a new potential difference between the plates of $v'$.
Note that the charge $q$ is unchanged as the conducting plates are electrically isolated from the surroundings.

The electric field between the plates is the sum of the electric field due to $\pm q$ and the electric field due to $\pm q'$.

$E = \frac {q}{\epsilon_0 A} - \frac {q'}{\epsilon_0 A} = \frac {q}{\epsilon_0 A} - \left (\frac {q}{\epsilon_0 A}-\frac {q'}{k\epsilon_0 A}\right) = \frac {q}{k\epsilon_0 A}=\frac {v'}{d}\Rightarrow C'=\frac {q}{v'} =\frac{k\epsilon_0 A}{d}$

In the derivation for the capacitance of a "mixed" dielectric capacitor, just above the interface between the top of the dielectric and the vacuum is considered to be an equipotential as is the bottom plate of the capacitor which is just below the bottom of the dielectric.

enter image description here

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