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Determine the charge now on each capacitor

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Two identical capacitors are connected in parallel and each acquires a charge Q0 when connected to a source of voltage V0. The voltage source is disconnected and then a dielectric (K=3.1) is inserted to fill the space between the plates of one of the capacitors. Assume that the capacitor without the dielectric is the first and the capacitor with the dielectric is second. Determine the charge now on each capacitor and the voltage now across each capacitor.


    2. Relevant equations
    Q = KQ0
    V = V0/K

    3. The attempt at a solution
    The answers are wanted in terms of Q0 and V0, but I can't figure out what to do. The only solution I could come up with is Q = 3.1Q0 and V=(1/3.1)V0, both of which are wrong. I am also not sure how to approach the capacitor without the dielectric.
     
  2. jcsd
  3. Feb 10, 2009 #2

    AEM

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    Re: Dielectrics

    Just a quick question: Does your text use [tex] \epsilon = \kappa
    \epsilon_o [/tex] as the expression for the dielectric constant where [tex] \epsilon_o [/tex] is the permittivity of free space?
     
  4. Feb 10, 2009 #3
    Re: Dielectrics

    Yes it does. Do I need to use that?
     
  5. Feb 10, 2009 #4

    LowlyPion

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    Homework Helper

    Re: Dielectrics

    How did the charge change if the voltage source was disconnected?
     
  6. Feb 10, 2009 #5
    Re: Dielectrics

    I considered that and tried to just put 1Q0 as the answer, but that was wrong.
     
  7. Feb 10, 2009 #6

    LowlyPion

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    Re: Dielectrics

    The charge on both will still be whatever charge they acquired when the voltage was present. The charge didn't go anywhere. It couldn't because it was disconnected.

    But the capacitors are still connected in ||
     
  8. Feb 10, 2009 #7

    LowlyPion

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    Re: Dielectrics

    That means that the charge will redistribute itself over the two capacitors.

    Figure the new equivalent capacitance and that total charge 2Qo is now distributed according to the values of the new capacitance.

    The capacitor with a dielectric has an increased capacity of 3.1Co so total capacitance is 4.1Co ...
     
  9. Feb 10, 2009 #8
    Re: Dielectrics

    Ok, so I found the two values for Q1 and Q2 (.49 and 1.51, respectively) and used that to solve for voltage. Thanks!
     
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