1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]
    [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


    User Avatar
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook