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Determine the charge on each and voltage across each.

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Two identical capacitors are connected in parallel. Each aquires a charge Q0 when connected to source voltage V0. The voltage source is the dissconnected and a dielectric(K=3.2) is inserted to one of the capacitors. Determine the charge on each and voltage across each.

    2. Relevant equations
    Given from problem:
    C10 = C20
    Q10 = Q20
    V10 = V20 = V0

    3. The attempt at a solution

    I have an example in my txtbook showing a capacitor is that charged by a battery. The battery is then disconnected and a dielectric inserted. The result is that the charge is held constant(because there is nowhere for it to go) and that the voltage drops by a factor of K.
    V = V0/K
    This is throwing me off because do i want to say the same thing about the voltage of the capacitor in my problem. They are in parallel so that would mean that both capacitors would have their voltages dropped by this factor of K? If this is true i have the problem solved but something doesn't feel right about it. Anyone know how to handle this problem already? Thanks for your time as always!
     
  2. jcsd
  3. Feb 21, 2009 #2

    LowlyPion

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    Re: dielectrics

    Consider then what the new equivalent capacitance is and proceed accordingly.
     
  4. Feb 21, 2009 #3
    Re: dielectrics

    I said that the dielectric is added to C2. So the new capacitance for C2 is
    C2 = KC20 and C1 is still
    C1 = C10

    Using the voltage drop result from my textbook, if i apply this i get that
    V2 = V1 = V0/K

    I also know that the equivalent charge Qeq of the initial setup with the battery connected will be the same Qeq after beacuse the the charge is conserved.
    Qeq0 = Q10 + Q20 = Q1 + Q2 = Qeq

    Is this information correct?
     
  5. Feb 21, 2009 #4

    LowlyPion

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    Re: dielectrics

    Not quite. Certainly the total charge doesn't change. But what about the Voltage?

    For instance what is your new equivalent capacitance? And what has that done to the voltage over both capacitors, and hence the charge distribution on each?
     
  6. Feb 21, 2009 #5
    Re: dielectrics

    My new equivalent capacitance is
    Ceq = C1 + C2
    = C10 + K*C20

    So the total capacitance has increased.

    And Ceq = Qeq / Veq
    where Qeq = Q1 + Q2 stays the same
    so Veq = V1 = V2 has to decrease to balance out the equation.

    I wanna say that Q1 = CV = C10*V0/k = Q10/k
    So for Qeq to stay the same
    Q2 would have to equal
    Q2 = Q20(3/2)K
    yes????
     
  7. Feb 21, 2009 #6

    LowlyPion

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    Re: dielectrics

    Isn't your new equivalent capacitance 4.2*C?

    Previously you had Ceq = C + C
    But after dialectric it's Ceq = C + 3.2*C = 4.2*C

    Q = V*C

    Means that Vo*2*C = V*4.2*C

    That means V = 2*Vo/4.2
     
  8. Feb 21, 2009 #7
    Re: dielectrics

    wow im an idiot haha that makes so much sense and is so much easier than thinking about it the way i was. thanks for the help!!!
     
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