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Non-Uniform Dielectric

  • Thread starter godtripp
  • Start date
  • #1
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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: [tex]\epsilon = \epsilon_{0}+\alpha x[/tex]
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.

[tex] C = \frac{\epsilon_{ave} A}{d} [/tex]
Where
[tex]\epsilon_{ave}=\frac{1}{d}\int \epsilon_{0}+\alpha x dx[/tex]

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

[tex] \frac{1}{C_{eq}}= \int \frac{dx}{(\epsilon_{0}+\alpha x)A} [/tex]


[tex] \frac{1}{C_{eq}}= \int \frac{x dx}{(\epsilon_{0}+\alpha x)A} [/tex]
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi godtripp! :smile:

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

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

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

and so 1/Ceq = ∫ d(1/C) = ∫ dx/εA :smile:

(and, just to check … when ε is constant, that's x/εA ! :wink:)
 
  • #3
54
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Thank you tim!
 

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