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Capacitors disconnected from the battery

  1. Oct 2, 2008 #1
    Capacitors C1 = 10 µF and C2 = 23 µF are each charged to 9 V, 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.
    Afterward, what are the charge on and the potential difference across each capacitor?

    The question ask for Q1, Q2(charge (µC)), and V1, V2(potential difference (V)).

    I used equation Q=CV which I think is the right equation to use... but there are some conceptual problems like what happens when "C1 connected to the negative plate of C2". When battery is disconnected from the circuit, how is the distribution of the voltage related to capacitors?
  2. jcsd
  3. Oct 2, 2008 #2


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    Yes, Q=CV is the right equation. You can use it to figure out the charges on each capacitor when they are at 9V.

    Charge will flow, until the two capacitors are at the same voltage.

    At that point, each capacitor has a 9V potential difference.
  4. Oct 2, 2008 #3
    I applied the equation...
    Q1 = C1*V1 = 23 µF * 9 V = 207 µC
    Q2 = C2*V2 = 10 µF * 9 V = 90 µC
    The results are incorrect.
    potential difference on each Capacitor is not 9V.
    Kind confusing... X(
  5. Oct 2, 2008 #4


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    Okay, good.
    So that means C1 has a charge of ______ on the positive plate, and ______ on the negative plate?
    Similary, C2 has ______ on the + plate and _____ on the - plate?

    Next, imagine what happens if the + plate of C1 is connected to the - plate of C2 ... i.e. what will be the net charge on those 2 plates?

    It is 9V before the capacitors are connected together. It is something different after they are connected.
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