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Homework Help: One of those problems that looks simple but probably isn't

  1. Feb 24, 2005 #1
    A 2.7 uF capacitor is charged by a 12 V battery. It is disconnected from the battery and then connected to an uncharged 4 uF capacitor. Determine the total stored energy before the connection, after the two are connected, and the difference between the two energies.

    OK. So for the first question, obviously, you just find E by using 1/2 Q^2/C. Sounds good. I think the second part is where it gets tricky.

    So I have this system that now has a total capacitance of 6.7 uF. My instinct is to just say, well, I have a 6.7 uF capacitance system with a charge of whatever I found in the first part, and I can just solve E = 1/2 Q^2/C. There is no voltage, right? Because neither capacitor is connected to a battery.

    So then the difference is just the second minus the first. I am assuming the second part is wrong because it seems too easy.
  2. jcsd
  3. Feb 24, 2005 #2
  4. Feb 24, 2005 #3
    OK, yeah, it seems similar enough. So I do have to take voltage into account? I will look into that and post my earnest attempt at an answer.
  5. Feb 24, 2005 #4
    OK, I'm lost. This is what I did.

    Knowing the total capacitance (6.7 uF) and the total charge (from the first part, which is 3.24E -5 since it comes only from the first capacitor) we can find the total voltage. Now that we know all three things, can't we just use them to find energy, or do I have to find the value of each capacitor's energy separately? If I do have to do this, how? I guess you could set up two equations like:

    V = V1 + V2
    Q = C1V1 + C2V2

    but I'm pretty sure that doesn't work.
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