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Dielectric constant of the materials, k

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data

    C1 = 12uF, C2=20uF, and the capacitors are charged so that the voltage across each capacitor is 6V. When a dielectric material is inserted between the plates of C1, the charge on C2 changes by 32 uC. Calculate the dielectric constant of the material, k. (PS. C1 is parallel to C2)

    2. Relevant equations

    C=q/v

    3. The attempt at a solution

    This is my method. 6V = (12x6+20x6+32)/(12k + 20), then solve for k, but it seems like it won't give me the right answer.
     
  2. jcsd
  3. Mar 17, 2009 #2

    LowlyPion

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    Welcome to PF.

    Is the voltage disconnected before the dielectric is added?
     
  4. Mar 17, 2009 #3
    Hi LowlyPion,

    The question didn't mention that but I think it's not disconnected.

    Thanks
     
  5. Mar 17, 2009 #4

    LowlyPion

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    Well if the Voltage is still the same across C2, and the dielectric goes only in C1, C2 is still the same ...

    Q2 = C2*V
     
  6. Mar 17, 2009 #5
    Hi LowlyPion,

    oops, my bad. The voltage is disconnected...
     
  7. Mar 17, 2009 #6

    LowlyPion

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    Before the dielectric was inserted then you had a total charge on the 2 capacitors. It doesn't change that total, but it does redistribute.

    (12 μf + 20 μf ) * 6 V = 192 μ C

    Now you have a new capacitor that is k*12 μf and you have redistributed the charge between 20 μf and k*12 μf such that there is a transfer of 32 of the 192 μC from C1 to C2.

    Before
    Q1 = 72 μC
    Q2 = 120 μC

    after
    Q1 = ?
    Q2 = ?

    When you determine the new C1 (+ dielectric) New/Old = k right?
     
  8. Mar 17, 2009 #7
    so Q1 becomes 72+32 and Q2 becomes 120-32?

    Then how do I find the new voltage?
     
  9. Mar 17, 2009 #8

    LowlyPion

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    They didn't ask that. They want k.
     
  10. Mar 17, 2009 #9

    LowlyPion

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    Whatever the voltage

    Q1/C1 = Q2/C2
     
  11. Mar 17, 2009 #10
    ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, i got it now!!! so Q1 after = 72+32 = 104 uC and Q2 after = 120-32 = 88 uC


    104/12*k = 88/20, then k=1.97???


    Thanks!!!!
     
  12. Mar 17, 2009 #11

    LowlyPion

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    Eureka always sounds sweet.

    Good luck.
     
  13. Mar 17, 2009 #12
    haha, indeed! Thank you!
     
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