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Reconfiguring Capacitors

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Two capacitors, 6uF and 14uF, ae conncted in series across an 18 V battery. They are carefully disconnected so that they are not discharged and they are then reconnected to each other, with postive plate to positive plate and negative plate to negative plate.

    Find the resulting potential difference across each capacitor after they are connected. Answer in units of V.

    2. Relevant equations

    C=Q*V

    For Plates in series:
    Q1 = Q2 = Q
    V1+V2=V(total)


    3. The attempt at a solution

    The part that confuses me is when it say that they reconnect the plates "positive plate to positive plate and negative plate to negative plate." I thought maybe it was recongifuring them into a parallel series which would mean the voltage was 18 V or 9 V if the battery was taken out of the loop but since both those answers are incorrect that doesn't seem to be the case.
     
  2. jcsd
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