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Parallel plate capacitor problem

  1. Feb 27, 2006 #1
    A capacitor is constructed from 2 square metal plates. A dielectric k= 4.93 fills the upper half of the capacitor and a dielectric k=10 fills the lower half of the capacitor. Neglect edge effects. The plate are separated by a distance of 0.37 mm and the length of the plates is 27 cm. Calculate the capacitance C of the device. Answer in units of pF.
    I used the equation for C of a parallel plate capacitor.
    [tex] kE_o A/d + kE_o A/d [/tex]
    Where the area= 3.7 x 10^-4 * .135 = 4.995 x 10^-5
    So (4.93)(8.85 x 10^-12)(4.995 x 10^-5)/ 3.7 x 10^-4 + (10)(8.85 x 10^-12)(4.995 x 10^-5)/ (3.7 x 10^-4)
    = 1.78 x 10^-11 F.
    Converting to pF gave me 1.78 x 10^-23 which isn't right... help please?
     
  2. jcsd
  3. Feb 27, 2006 #2

    Astronuc

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    Staff: Mentor

    Last edited: Feb 27, 2006
  4. Feb 27, 2006 #3
    I think d has to be different for each one, but I'm not sure how to do that.
     
  5. Feb 27, 2006 #4
    consider two capacitors in series each having distance d = half of 0.37 mm.The first has dielectric k=4.93,second has k=10.

    Or you may use the following equation-

    C=EA/[d/k + d'/k'] here d=d'=half of 0.37 mm
     
  6. Feb 27, 2006 #5
    So would I do,
    (8.85 x 10^-12)(.27)/ (3.7 x 10^-4/4.93) + ((1.85 x 10^-4)/10) = 2.55 x 10^-8 ?
     
  7. Feb 27, 2006 #6
    area is (.27 m)^2;also d=1.85 x 10^-4 m.
     
  8. Feb 28, 2006 #7
    Ok so it would be:
    (8.85 x 10^-12)(.0729) / ((1.85 x 10^-4/4.93) + (1.85 x 10^-4/10)) =1.15 x 10^-8
    Am I doing this right now?
     
  9. Feb 28, 2006 #8

    Astronuc

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    Staff: Mentor

  10. Feb 28, 2006 #9
    yes,this should be the answer in farad.

    If you are not allowed to use the formula directly,then it would be better to consider two capacitors in series rather than to deduce the formula.
     
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