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Capacitors are charged, disconnected from battery, and had plates swit

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Capacitors C1=13μF and C2=21μF are each charged to 10V, then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C1 connected to the negative plate of C2 and vice versa.


    2. Relevant equations
    Q=CV
    Ceq = C1 + C2

    3. The attempt at a solution

    Ceq = C1 + C2
    Ceq = 34μF

    Q=C/V

    Q1= 13μF*10V = 130μC

    Q2= 21μF*10V = 210μC

    In the new C1, the positive plate should be 130 μC and the negative -210μC.
    In the new C2, the positive plate should be 210μC and the negative -130 μC.

    I'm not sure what to do after this.

    I have Googled the problem and have found a Yahoo Answers thread with the correct method, but I'm not sure why the method was correct. If someone can explain his steps and/or help me with the next step, then that would be great.

    Edit: I think I may have figured it out. Is he taking the net charge of each capacitor, multiplying it with the first capacitor (to find the charge of the new C1) and dividing by Ceq because the net charge divided by Ceq is equal to the new value for V? Which is basically taking the fraction of one capacitor over the total capacitance and finding the charge in that one capacitor? And because in a parallel, voltage is the same in both capacitors? I'm not sure if that makes sense, but hopefully it does?
     
    Last edited: Feb 24, 2013
  2. jcsd
  3. Feb 25, 2013 #2

    ehild

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    Re: Capacitors are charged, disconnected from battery, and had plates

    What is the question at all?

    ehild
     
  4. Feb 25, 2013 #3
    Re: Capacitors are charged, disconnected from battery, and had plates

    I'm sorry, I forgot to state the questions. It's just asking for the charges in each of the capacitors after connecting the positive plate of one to the negative plate of the other and vice versa, then the potential difference in both.
     
  5. Feb 25, 2013 #4

    ehild

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    Re: Capacitors are charged, disconnected from battery, and had plates

    OK, your method looks correct. What voltage and charges have you got?

    ehild
     
  6. Feb 25, 2013 #5
    Re: Capacitors are charged, disconnected from battery, and had plates

    ~31 microCoulombs for Q1 and ~49 microCoulombs for Q2.

    V1 = V2 = V = 2.4 volts.
     
  7. Feb 25, 2013 #6

    ehild

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    Re: Capacitors are charged, disconnected from battery, and had plates

    Correct!

    ehild
     
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