Calculating Capacitance for Non-Uniform Dielectric: Finding the Right Formula

[godtripp]

This isn't a homework problem, just something to explore my understanding so correct me if any of my assumptions are wrong.

I have given myself the situation where I have a capacitor of area "A" and separation "d"

Between them is a non-uniform but continuous dielectric, so that the dielectric on the right plate is significantly more dense than that of the left.

I have also given the dielectric the function: $$\epsilon = \epsilon_{0}+\alpha x$$
where x is the distance from the left plate.


Now I can find my answer by taking an infinite sum of infinitesimal capacitors in series.

I'm just confused now on how to properly build the formula. Should I take the derivative of
my capacitance formula with respect to d, plug in for epsilon as a function of distance and integrate? Or how?

I've listed a few of my ideas on formulas below. If anyone can point out the right one and why.

$$C = \frac{\epsilon_{ave} A}{d}$$
Where
$$\epsilon_{ave}=\frac{1}{d}\int \epsilon_{0}+\alpha x dx$$

$$C_{eq} = \int \frac{-(\epsilon_{0}+\alpha x)A dx}{x^2}$$

$$\frac{1}{C_{eq}}= \int \frac{dx}{(\epsilon_{0}+\alpha x)A}$$


$$\frac{1}{C_{eq}}= \int \frac{x dx}{(\epsilon_{0}+\alpha x)A}$$

 

[tiny-tim]

Hi godtripp!

(have an epsilon: ε and an alpha: α and an integral: ∫ )

You have to integrate 1/C to get 1/C_eq.

For a slice of thickness dx, d(1/C) = dx/εA,

and so 1/C_eq = ∫ d(1/C) = ∫ dx/εA

(and, just to check … when ε is constant, that's x/εA ! )

 

[godtripp]

Thank you tim!