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Homework Help: Combination of capacitors

  1. Sep 16, 2018 at 12:21 PM #1
    1. The problem statement, all variables and given/known data
    Between the plates of parallel plate condenser having charge Q,a plate of thickness t1 and dielectric constant k1 is placed.In the rest of the space,there is another plate of thickness t2 and dielectric constant K2.The potential difference across the condenser will be.

    2. Relevant equations
    C=kAe0/(d-t+t/k)
    k----->dielectric constant
    A--->Area of the plates
    e0---->permittivity
    d---->distance between the plates
    t------>thickness of the plate

    V=Qnet/C


    3. The attempt at a solution

    Combination of capacitors with materials of dielectric constant K1 and K2 is obtained.i am not sure if this parallel or series connection.

    My logic says they are in series.

    So Cnet=C1C2/C1+C2

    But i am not arriving at a final answer

     
  2. jcsd
  3. Sep 16, 2018 at 1:01 PM #2

    kuruman

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    Your logic is correct, they are in series. Also your formula for the equivalent capacitance is correct. So now what are C1 and C2?
     
  4. Sep 16, 2018 at 1:12 PM #3
    Yeah but isnt the distance between the plates of C1 d/2 ?
     
  5. Sep 16, 2018 at 1:27 PM #4

    Merlin3189

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    Not on my reading of the Q.
    the dielectric plates are thickness t1 and t2 and they filled the gap d between the plates, so that t1 + t2 = d
    It is not stated that t1 = t2
     
  6. Sep 16, 2018 at 1:42 PM #5
    IMHO, the question is both incomplete and ambiguous but, given the information provided, can only be solved as 'C in Series'.
     
  7. Sep 16, 2018 at 2:59 PM #6
    Is there any combination in which the area of both plates become A/2(not related to this question)
     
  8. Sep 16, 2018 at 5:07 PM #7

    kuruman

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    You can have the gap in the bottom half of the capacitor be completely filled with one kind of dielectric and the top half filled with a different kind. Is that what you mean? (Not the case here).
     
  9. Sep 16, 2018 at 6:14 PM #8

    Merlin3189

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    I thought that's exactly what we have here? EDIT: Except that it's not halves
    capacitor_double.png
     
  10. Sep 16, 2018 at 6:18 PM #9

    kuruman

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    That's exactly my interpretation.
     
  11. Sep 16, 2018 at 6:45 PM #10

    gneill

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    Only if the dielectric slabs are arranged side by side and so divide the plate area between them. The width of the dielectrics would have to be identical in order to divide the area of the plates equally.

    upload_2018-9-16_19-41-48.png

    In this case you could consider the resulting device to be two capacitors with the same plate area (A/2) but different dielectrics and connected in parallel.
     
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