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Non-Uniform Dielectric

  1. Apr 1, 2010 #1
    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]
     
  2. jcsd
  3. Apr 1, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    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:)
     
  4. Apr 1, 2010 #3
    Thank you tim!
     
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