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Sphericial Capacitance w/Dielectrics

  1. Oct 31, 2007 #1
    A spherical capacitor is made of two insulating spherical shells with different dielectric strengths, k1 and k2, situated between two spherical metallic shells and separated by a vacuum gap. Calculate the capacitance of this system.


    Total C = Q(free) / [tex]\Delta[/tex]V where [tex]\Lambda[/tex]V is the varied potential between the two metal shells.

    [tex]\Lambda[/tex]V = [tex]\int^{a}_{d}[/tex]E(r)dr = - [tex]\int^{a}_{b}[/tex][(1/4*pi*Eps) * (1/k1)*(Qfree / r^2) * dr] - [tex]\int^{b}_{c}[/tex][(1/4*pi*Eps) * (Qfree / r^2) * dr] - [tex]\int^{c}_{d}[/tex][(1/4*pi*Eps) * (1/k2) * (Qfree / r^2) * dr]

    Q free will be negated in the final equation, since C = Q/V and V includes Q in its numerator (hope that makes sense). So after integrating V through each of the mediums, the two dielectrics and the vacuum, I get the sum :

    Ctotal = = Ke[C(dielectric 1) + C(vacuum) + C(dielectric 2) = Ke[(1/k1)*(1/a - 1/b) + (1/b - 1/c) + (1/k2)*(1/c - 1/d)]

    Is this correct?
  2. jcsd
  3. Oct 31, 2007 #2


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    Homework Helper

    above looks good.

    how did you get this part? stick to what you were doing above... get delta V... then take Q/delta V... don't take any shortcuts...
  4. Oct 31, 2007 #3
    I did this next...

    delV = - (Qfree/(4*pi*Eps)) [ [tex]\int^{a}_{b}[/tex] (dr/(k1*r^2)) + [tex]\int^{b}_{c}[/tex] (dr/r^2) + [tex]\int^{c}_{d}[/tex] (dr/(k2*r^2))

    The general form of the solution for int(dr/r^2) is -(1/r). This lets the pre-existing negative cancel out, and I just have to substitute to get the expression I have above... I've got it written down, just didn't want to type it all out. :)

    Also, the solution I came up with should be equal to 1/Ctot, not just Ctot. I think that's right...
  5. Oct 31, 2007 #4


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    Yes, exactly that was 1/Ctot. everything looks good.
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