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Calculating Charge with Capacitors in Series

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the charge on the right
    plate of C2 long after the switch is
    closed?

    1. CV
    2. CV/2
    3. CV/3
    4. 2CV/3
    5. 0

    2. Relevant equations

    Q = CeqV

    3. The attempt at a solution

    Q = 1/3c * 2V
    so I thought
    Q = 2V/3C
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2012 #2

    gneill

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    Staff: Mentor

    Is that (2V)/(3C) or (2V/3)C ?
     
  4. Oct 6, 2012 #3
    I wanted (2V/3)C but written that way it looks like

    2CV/3 which is one of the answers but I'm trying to understand why "C" is in the numerator. Is it because Ceq = 1 / C?
     
  5. Oct 7, 2012 #4

    gneill

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    Staff: Mentor

    What's the net (equivalent) capacitance of the two given capacitors?
     
  6. Oct 7, 2012 #5
    Ceq = 1/C + 1/2C = 3/2C
     
  7. Oct 8, 2012 #6

    gneill

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    Staff: Mentor

    Not quite. Two capacitors Ca and Cb in series yield:

    $$C_{eq} = \frac{1}{\frac{1}{Ca} + \frac{1}{Cb}}$$

    You've forgotten to take the overall reciprocal.
     
  8. Oct 8, 2012 #7
    I see what I did incorrectly now. My final answer should have been 2C/3.
     
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