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Relative dielectric constant - More than one dielectric in a capacitor

  1. Oct 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A parallel plane capacitor with conducting plates area [tex]\Sigma[/tex]= 400cm2, distant d=5mm, is half filled of mica (k1=5) and half filled of paraffin (k2=2). Calculate:

    a) C (Capacitance of the capacitor)

    ...others


    2. Relevant equations

    None

    3. The attempt at a solution

    I haven't got too many problems solving this exercise but i just don't know how to get to the correct result of k (relative dielectric constant) when there's more than one dielectric in a parallel plate capacitor.

    Esn436pag104.jpg
     
  2. jcsd
  3. Oct 6, 2010 #2

    vela

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    Calculate the voltage drop across the two dielectrics separately. Their sum will equal the voltage drop across the capacitor.
     
  4. Oct 7, 2010 #3
    Ok, then:

    V=(E*d)/k ; V1=(E*(d/2))/k1; V2=(E*(d/2))/k2;

    V=V1+V2;

    (E*d)/k=((E*d)*(k1+k2))/(2k1k2)

    k=(2k1k2)/(k1+k2)

    this should be correct thanks.

    -------------------------------------------------------

    Another question: if i would have had a capacitor like this:

    Esn435pag104.jpg

    i can't find k like i did before 'cause k would be like (k1*k2)/k1+k2 while it should be like
    (k1+k2)/2
     
    Last edited: Oct 7, 2010
  5. Oct 7, 2010 #4

    vela

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    You can think of it as two capacitors in parallel.
     
  6. Oct 7, 2010 #5
    ok, thanks i got it now, the sum of the capacitances of the two capacitors (cause they are in parallel) gives me the total capacitance.

    So, in the first exercise i've asked you, the capacitors can be considered as connected in series?
     
  7. Oct 7, 2010 #6

    vela

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    Yes, effectively it's two capacitors in series.
     
  8. Oct 7, 2010 #7
    ok, thank you very much.
     
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