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Paralell Plate Capacitors and Dielectric

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Two parallel plates, each having area A = 3118cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.35cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

    A dielectric having dielectric constant κ = 2.6 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3118 cm2 and thickness equal to half of the separation (= 0.175 cm) . What is the charge on the top plate of this capacitor?

    2. Relevant equations

    C=Q/V

    C(new) = KC(old)

    3. The attempt at a solution

    I really have no idea.

    I tried dividing the new capacitance by (because the dielectric is d/2 long) and calculating Q from that, but that didn't work.
     

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  3. Feb 12, 2012 #2

    gneill

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    Staff: Mentor

    The resulting configuration will behave like two capacitors in series: One without dielectric and one with dielectric.
     
  4. Feb 12, 2012 #3
    Okay, that makes sense.

    So, what I have so far now is

    1/C(without) + 1/C(with) = 1/C

    C(without)= εA/(d/2)

    C(with)= εAK/(d/2)

    Unfortunately, this does not give me the right answer. What am I doing wrong?
     
  5. Feb 12, 2012 #4

    gneill

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    Staff: Mentor

    Your equations look okay. Perhaps a calculation issue? What numerical values did you find?
     
  6. Feb 12, 2012 #5
    Yep! Got it!

    The next part asks for the voltage when the battery is disconnected and the dielectric is removed. Does this have to do with the difference in capacitance between when the dielectric is present and when it is not?
     
  7. Feb 12, 2012 #6

    gneill

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    Staff: Mentor

    Yes. What will be conserved?
     
  8. Feb 12, 2012 #7
    Charge, apparently. Got it!
     
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