How to Determine Capacitance with Variable Dielectric Dimensions?

[MetalManuel]

Homework Statement

Homework Equations

\int\vec{E}\cdot\vec{dA}=\frac{Q_{enc}}{\epsilon_{0}}
\int\vec{E}\cdot\vec{dA}=\frac{Q_{enc}}{k\epsilon_{0}}
V=\int\vec{E}\cdot\vec{dS}
C=\frac{q}{V}
C_{eq}=C_{1}+C_{2}+...
\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+...

The Attempt at a Solution

I didn't want to do the calculations in itex because it would take forever, but this is what i have so far. I just did the integrations quickly and went straight to the point and put in the equations.

With the answer that I get I can't seem to get the answer. I have spent a lot of hours on this problem and it took me a while to figure out the proper areas and dimensions, but even after all that I am stuck. What is my problem? I have a feeling it has to do with the 4h^2 (it's not the right dimensions for the dielectric area).

Thanks

 

[Spinnor]

The volume of the dielectric is L*L*L/2 and is a constant? Where do you use this fact? I apologize if the answer is obvious.

 

[MetalManuel]

The volume of the dielectric is L*L*L/2 and is a constant? Where do you use this fact? I apologize if the answer is obvious.

The volume of \frac{L^{3}}{2} is the total volume of the dielectric, but you can change the dimensions of the dielectric as long as long as l=w. For example, you can say that the dielectric is now \frac{L}{8}, 2L and 2L because the volume is still the same. You have to account for all the possible shapes that this dielectric can take.

edit:Imagine there's a piece of clay in between the capacitors, this will be the dielectric. You can shape the clay however you want and it will still have constant volume, just different dimensions. In this case though the clay would have to be a cubic shape, the two dimensions equal to each other.

 

[MetalManuel]

I think I have figured it out, my mind is racing. Will update if I figure it out.

 

[MetalManuel]

So i finally solved it, this is how I am feeling right now.
[PLAIN]http://r3dsky.com/wp-content/uploads/2011/04/1302819681876.jpg

This problem took me so long to figure out, I'm glad the torture is finally over.It says:
We know that the area is constant for the dielectric(for the square side, not the thickness, l=w). My initial thoughts of the area being 4h^2 was incorrect because h could be << than l and w, therefore I thought that the area could be described with a function: