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Energy of a Capacitor System with partially inserted Dielectric

  1. May 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Picture2-3.jpg

    The above is a capacitor system. A dielectric is inserted into capacitor C3 and I need to find the total energy of the system. The dielectric is not fully in, only partially (lets say a length x).


    3. The attempt at a solution

    I first found the capacitance of C3 with the dielectric partially inserted. I know that dielectric makes a system of 2 parallel capacitors.

    Hence

    [tex]C3 = \frac{L\epsilon_{0}}{d} (x(\epsilon_{r}-1) + L)[/tex]

    where L is the length and width of the parallel capacitor.

    Total capacitance of the system = [tex]\frac{(C3 + C2)C1}{C3 + C2 + C1}[/tex]

    To find the total energy, I use the formula [tex]1/2 \times \frac{Q^{2}}{C}[/tex]

    which becomes:

    [tex]1/2 Q^{2} \times \frac{C3 + C2 + C1}{(C3 + C2)C1}[/tex]

    Does this seem correct? The answer give to me by the professor is:

    [tex]U =1/2 \frac{Q^{2}}{C1} + \frac{Q^{2}}{C2 + C3}[/tex]

    I'm just kinda whether I'm wrong or the prof is...

    Please advise.
     
  2. jcsd
  3. May 10, 2009 #2

    tiny-tim

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    Hi saad87! :wink:
    erm :redface: … they're the same! (assuming the professor's "1/2" is for both bits) :smile:
     
  4. May 10, 2009 #3
    Really? Could please show me how? I really can't understand how they are equal!

    Do we use Partial fractions to expand the fractions or am I way off?
     
  5. May 10, 2009 #4

    tiny-tim

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    1/C1 = (C2 + C3)/(C2 + C3)C1

    1/(C2 + C3) = C1/(C2 + C3)C1 :wink:

    (generally, 1/a + 1/b = (a + b)/ab )
     
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