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Association of capacitators

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    ;-------| |---------;
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    '-------| |---------'

    Consider a capacitator of capacity C1 charged with a charge Q. Suppose you connect it in parallel to another capacitator, C2, initially uncharged.

    Find the final charge and difference of potencial at each condensator.

    2. Relevant equations



    3. The attempt at a solution

    I know that the initial charge Q will be equal to:

    Q = q1+q2 , where q1 and q2 is the final charge of capacitator 1 and 2.
    The difference of potential, V, is equal in both the capacitators. Therefore I can say that

    q1 = Q - q2 = Q - C2/V = [ QV - C2 ] / V

    I don't know where to go from here. I don't think I can get anywhere... If anyone could point me in the right direction I'd really appreciate!

    Thanks.
     
  2. jcsd
  3. Nov 13, 2012 #2
    You're thinking in the right direction.

    Initially there is a charge on C1 equal to Qi1. There is no charge on C2.

    In the end, the voltage across both capacitors is equal, so:

    Vf1 = Vf2; Qf1/C1 = Qf2/C2

    I'm using Qi as initial charge and Qf as final charge.

    Between the initial state and the final state, charge moves from capacitor C1 to capacitor C2. If you assume a charge dQ moves from C1 to C2, you should be able to say something about Qf1 and Qf2 in terms of Qi1.
     
  4. Nov 13, 2012 #3

    ehild

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    Homework Helper
    Gold Member

    Q=CV. AS the potential difference is the same across the parallel connected capacitors, q1=C1V and q2 = C2V. q1+q2=Q. Can you find V in terms of Q and C1, C2?


    ehild
     
  5. Nov 13, 2012 #4
    Oh, so simple... I got to the correct result. Thank you very much!
     
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