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Homework Help: Capacitance of a parallel plate capacitor with 2 dielectrics

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the capacitance of the parallel plate capacitor with 2 dielectrics below. Given that the parallel plate capacitor has area A = WL and the separation between plates is d.

    vfedyr.png


    3. The attempt at a solution

    My method is to use integration. First solve for two small capacitors dC1, dC2 and then sum them up in parallel.

    [tex] y = \frac{d}{L}x [/tex]

    [tex] dC_1 = \frac{\kappa_1 \epsilon_0 WL dx}{d(L-x)} [/tex]

    [tex]dC_2 = \frac{\kappa_2 \epsilon_0 WL}{d} \frac{1}{x} dx [/tex]

    [tex] dC = \frac{dC_1 dC_2}{dC_1 + dC_2} = \frac{\kappa_1 \kappa_2 \epsilon_0 WL}{d} \cdot \frac{1}{\kappa_2 L + x(\kappa_1 - \kappa_2)} [/tex]

    [tex]C = \frac{\kappa_1\kappa_2 \epsilon_0 WL}{d(\kappa_1 - \kappa_2)} \left[ ln (\kappa_2 L + x(\kappa_1 - \kappa_2) \right]_0^L[/tex]

    The expression is definitely wrong. If we let the two dielectrics have equal value k, we should expect C to follow that of a single dielectric, C' = kC.
     
  2. jcsd
  3. Feb 2, 2010 #2

    ehild

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    Homework Helper

    You have one more L in the expressions of dC. The area of an elementary capacitor is Wdx.

    ehild
     
  4. Feb 2, 2010 #3
    Ok, but L is a constant and having amended that, it doesn't make the final integral any more correct. The ln form of the last equation is still clearly wrong.
     
  5. Feb 2, 2010 #4

    ehild

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    Homework Helper

    Are you sure that the ln is wrong? You can not simply substitute K1=K2, it will lead to 0/0. You have to take the limit K2/K1-->1.



    ehild
     
    Last edited: Feb 2, 2010
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