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Parallel-plate capacitor with 2 dielectrics?

  1. Mar 13, 2015 #1
    1. The problem statement, all variables and given/known data

    The figure shows a parallel-plate capacitor with a plate area A = 5.29 cm^2 and plate separation d = 8.10 mm. The left half of the gap is filled with material of dielectric constant κ1 = 6.50; the right half is filled with material of dielectric constant κ2 = 11.4. What is the capacitance?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c25/fig25_48_wiley.gif


    (Given)

    A = 5.29 cm^2
    d = 8.10 mm
    k1 = 6.50
    k2 = 11.4
    e0 = 8.85E-12

    (unknowns)

    CNoDielectrics = ?
    CDielectrics = ?


    2. Relevant equations

    C
    Parallel-Plate = (e0*A)/d

    C
    AddedDielectric = CParallel-Plate*Kdielectricconstant


    3. The attempt at a solution

    CNo Dielectric/2 = (e0*((5.29E-2)/2))/(8.10E-3) = 2.89E-11

    Since the separated capacitor is in series (I think), you can use the law of inverse capcitance sums, after multiplying the CNo Dielectric/2 once by k1, and another time by k2, and add them, then inverse them, you get

    1/((CNo Dielectric/2)*k1) + 1/((CNo Dielectric/2)*k2) =

    = 3.04E-11 + 5.32E-11 = 8.36E-11

    1/8.36E-11 = 1.20E-12 = CDielectrics

     
  2. jcsd
  3. Mar 13, 2015 #2
    Apparently, my image couldn't be posted as it was a secure URL. I will provide a description. It is 1 parallel-plate capacitor split in half VERTICALLY, each half has a different dielectric. The one on the left is K1, the one of the right is K2. The Area value is for the whole parallel plate capacitor, and each half is A/2. d is the distance between plates.
     
  4. Mar 13, 2015 #3

    gneill

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    Staff: Mentor

    What is your logic for considering the two capacitances to be in series?
     
  5. Mar 14, 2015 #4
    I figured that it would just be because 1 came after the other, but I thought about it for awhile and came to the conclusion that isn't it just the voltage remaining constant throughout, or the charge remaining constant which determines they are in series, or parallel? So, would these actually be in parallel, since it is "Technically" the same capacitor, just with different dielectrics, then the voltage must be constant? Implying it would be parallel... ?
     
  6. Mar 14, 2015 #5

    gneill

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    Staff: Mentor

    They are parallel because they share the same two nodes at their connection points to any circuit: they share the same wire leads. That also means of course that they share the same potential difference.

    So re-work your calculation with the assumption that the two sections are in parallel.
     
  7. Mar 14, 2015 #6
    I reworked the calculations for the capacitance, the program is giving me the wrong answer, even though I calculated them in parallel.

    And yes, it wants the answer in just Farads.

    1zn7pz7.png
     
  8. Mar 14, 2015 #7

    gneill

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    Staff: Mentor

    Check your area value. The given units is square centimeters. What's that in square meters?

    Otherwise, your method looks fine.
     
  9. Mar 14, 2015 #8

    SammyS

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    Seems like the image finally showed up for me.

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c25/fig25_48_wiley.gif

    fig25_48_wiley.gif
     
    Last edited: Mar 14, 2015
  10. Mar 14, 2015 #9
    OH GOODNESS! I can't believe I overlooked that, It's always the units that get me. Thank you so much!
     
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