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Parallel-plate capacitor: Two dielectric materials

  1. May 27, 2008 #1
    [SOLVED] Parallel-plate capacitor: Two dielectric materials

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

    A parallel-plate capacitor with area A = 5.56 cm^2 and separation d = 5.56 mm has the left half of the gap filled with material of dielectric constant K1 = 7.00 and the right half filled with material of dielectric constant K2 = 12.0. What is the capacitance?


    2. Relevant equations

    [tex]C = \frac{\kappa \epsilon_{0} A}{d} [/tex]

    [tex] \frac {1}{C_{tot}} = \frac {1}{C_{left}} + \frac {1}{C_{right}}[/tex]

    3. The attempt at a solution

    Since it's one capacitor (two plates) and since the dielectric materials are next to each other (as opposed to stacked) I approached the problem as if the two halves of the capacitor were separate parallel capacitors and calculated:

    [tex]C_{left} = 3.10 pF [/tex]
    [tex]C_{right} = 5.31 pF [/tex]

    [tex]C_{tot} = 1.96 pF [/tex]

    But this is not amongst the given possible answers. Where did I go wrong?

    Cheers

    phyz
     
  2. jcsd
  3. May 27, 2008 #2
    This is a serial connection of capacitors so you must add right and left halves capacity:
    [tex]C=C_{right}+C_{left}[/tex]
     
    Last edited: May 27, 2008
  4. May 27, 2008 #3
    Thanks for your reply...however, could you please explain why this setup acts as capacitors in series?
     
  5. May 27, 2008 #4
    I apologize to bad equation [tex]C=C_{1}+C_{2}[/tex]
    It's wrong... You have a good solution, because
    [tex]V=\frac{q}{C}[/tex]
    and [tex]V_{1}=\frac{q}{C_{1}}[/tex] [tex]V_{2}=\frac{q}{C_{2}}[/tex]
    and [tex]V=V_{1}+V_{2}[/tex]
    so [tex]\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]
     
  6. May 28, 2008 #5
    I appreciate your attempt at helping me Phizyk, but unfortunately we're no closer to understanding this problem than before...
     
  7. May 28, 2008 #6

    tiny-tim

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    Hi phyzmatix! :smile:

    Thanks for the PM …

    Did you divide by 2 instead of multiply?
     
  8. May 28, 2008 #7
    Thanks for joining me here tiny-tim :smile:

    Yes, I divided the total area given by 2 (I wish I had the means to show you the figure given with the question, on the figure they show that the left side is A/2 and the right side is also A/2)

    So my understanding is that the left half of the parallel-plate capacitor has capacitance [tex]C_{left}[/tex] (calculated with K1 and A/2) and the right half has capacitance [tex]C_{right}[/tex] (calculated with K2 and A/2) but how to combine the two values into [tex]C_{tot}[/tex]?

    (I'm not sure my assumption that the two can be treated as separate parallel capacitors is correct)
     
  9. May 28, 2008 #8

    tiny-tim

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    Hi phyzmatix! :smile:

    Well, they're side-by-side, not one-after-the-other, so they're "in parallel" rather than "in series" (plus, the question calls it a parallel-plate capacitor), so I think you just add them.

    Isn't 8.41 one of the given answers?

    (if not, what are the given answers? :wink:)
     
  10. May 28, 2008 #9
    *DOH!!!*

    I feel like such a tw@t :redface:

    No wonder I didn't get the right answer, I was using the equation for capacitors in series...

    *banging head against wall*

    Cheers Tim! You legend! :biggrin:

    (oh, by the by, 8.41 is indeed amongst the possibilities :wink:)
     
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