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Capacitance & Dielectrics

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A parallel plate capacitor of area 2 cm^2 is filled with three dielectrics with constants k1=3.5, k2=1.7 and k3=8.5. The dielectrics are placed as shown in the figure, each occupying half the total area, while the volume of the first double of the other two. If the separation distance between the plates is 2 mm.

    a. What is the capacitance before including the dielectrics?
    b. What is the capacitance after including the dielectrics?
    c. If we connect the capacitor to a battery of 12 V, What is the energy stored?

    Image of the exercise: http://i41.tinypic.com/jhg7l1.jpg


    2. Relevant equations

    Part a. C=(E0)*(Area)/(distance)
    Part b. C=(Keq)*(E0)*(Area)/(distance)
    Part c. V = 1/2 * E0 * E^2



    3. The attempt at a solution

    Part a. I solved this part of the exercise by just plugging in the value given by the exercise.

    Part b. I am assuming I have to use the formula I denoted as Part b. I looked at the exercise and assumed it was a circuit... When I looked at it that way, I calculated the Equivalent Capacitance. I had two capacitance in Series then that went to be in parallel with the other capacitance. I calculated the Keq like that. Then since we applied dielectrics, I am assuming the Area would half of what it used to be. This is the part where I am stuck... trying to understand clearly how to solve part b.

    Part c. I think I need the capacitance of part b to solve part c using the formula presented above?

    Thanks for all the help that can be provided!
     
    Last edited: Feb 27, 2012
  2. jcsd
  3. Feb 27, 2012 #2

    gneill

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    Staff: Mentor

    Hi wilito92, welcome to Physics Forums.

    Yes, you can solve part b in the manner you've indicated. You need to be a bit careful about the areas (which you've figured out) as well as the separations. The capacitor with dielectric κ1 retains separation d = 2mm, but the other two have their separations cut in half.
     
  4. Feb 27, 2012 #3
    Hi,

    Thanks for the prompt response. I found the way to calculate the Capacitance after the dielectric.

    I calcualted C1 by using this formula: K1*E0*(A/2)/d then C2 and C3 using the same formula with there respectable K values and dividing the distance by two.

    However, now I am stuck on how to calculate part 3... any help on that?
     
  5. Feb 27, 2012 #4

    gneill

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    Staff: Mentor

    For part 3, what formula are you planning to use for the energy stored on a capacitor? You've got the capacitance and you're given a voltage...
     
  6. Feb 27, 2012 #5
    I was using U=(1/2)*C*(V)^2 but apparently the result I get is wrong...

    Is this the right formula?
     
  7. Feb 27, 2012 #6

    gneill

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    Staff: Mentor

    Yes, that formula is correct. What value did you calculate?
     
  8. Feb 27, 2012 #7
    UPDATE: I was using the right formula.. just not using the correct Capacitance.

    U=(1/2)*(2.80*10^-12)*(12)^2=2.02*10^-10

    Thanks for all the help!
     
  9. Feb 27, 2012 #8

    gneill

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    Staff: Mentor

    Okay, glad it worked out. :smile:
     
  10. Feb 27, 2012 #9
    Just though I'd leave some details here on how I solved it so anyone else having doubts with it can solve it too :)

    Part a: C=(Eo)*(A)/(d) = (8.85*10^-12)*(2.0*10^-4)/(2.0*10^-3)= 8.85*10^-13 F

    Part b:
    C1=(K1)*(Eo)*(A/2)/d=(3.5)*(8.85*10^-12)*((2.0*10^-4)/2)/(2.0*10^-3)=1.5487*10^-12
    C2=(K2)*(Eo)*(A/2)/(d/2)=1.5045*10^-12​
    C3=7.52*10^-12​

    Then C_tot = C1 + (1/C2 + 1/C3)^-1 since C2 and C3 are in series and then in parallel with C1. C_tot=2.80 pF

    Part c:
    U=(1/2)*(2.80*10^-12)*(12)^2=2.02*10^-10 J
     
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