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Homework Help: The potential difference in a capacitor with dielectric

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Two identical capacitors are connected in parallel, charged by a battery ov Voltage V, and then isolated from the battery. One of the capacitors is then filled with a material of relative permittivety E.

    2. Relevant equations

    What is the final potential difference across it's plates?

    3. The attempt at a solution

    Potential difference?

    Q = Charge

    2C in parallel

    Q1 = C1V
    Q2 = C2V

    C1=C2=C

    Q=Q1+Q2 = CV+CV=2CV (on each capacitor)

    Q=2C0V -> C0 =( Q / (2V) )

    The battery is removed, one capacitor is filled with dielectric with relative permittivety E.

    C = EC0

    Q = CfVf = CiVi

    Vf = (Ci/Cf)*Vi = (CiVi)/Cf = (2C0V)/EC0 = 2V/E

    The correct solution should look like: (2V)/(E+1)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 27, 2010 #2

    ehild

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    Homework Helper

    The sum of the charge on both capacitors stays the same as before: Q=2VCo, but the new resultant capacitance is (E+1)Co, so the new voltage is V'= Q/Cresultant= 2VCo/((E+1)Co)

    ehild
     
  4. Jan 27, 2010 #3
    So what they mean by isolating the circuit from the battery is that it still remains a closed circuit, hence we have to take both capacitors into account when solving this?
     
    Last edited: Jan 27, 2010
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