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Capacitors - charged in parallel and reconnected to one another.

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Capacitors of 7.6 µF and 2.1 µF are charged as a parallel combination across a 333V battery. The capacitors are disconnected from the battery and each other. They are then connected positive plate to negative plate and negative plate to positive plate.
    Find the resulting charge on the first capacitor.
    Answer in units of µC.
    2. Relevant equations
    C = Q/V
    3. The attempt at a solution
    I'll preface this by saying that I looked at other threads about this, but it's still just not working out for me so I think I need to actually converse about it.
    Knowing that parallel capacitors have the same voltage:
    q1 = (7.6e-6)(300V) = .00228C
    q2 = (2.1e-6)(300V) = 6.3e-4C
    Qtotal = q1 + q2 = .00291C
    To check:
    Qtotal = CtotalV
    (7.6e-6C + 2.1e-6c)(300V) = .00291C
    It's just that from here, I don't know where to go at all. When the charges are connected one end to another, are they in a series or are they parallel?
    Hints aren't working very well for me. I don't know if that's because I'm behind in physics or what, but nothing that's being thrown against the wall is sticking.
  2. jcsd
  3. Sep 23, 2014 #2


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    Before we go any further, is that supposed to be 300 V, or 333 V?
    After correcting any possible minor mistakes mistakes so far (e.g., 300 V vs. 333 V), here is my advice as to where to go from there:

    (a) When a given parallel capacitor is charged to some amount q, it really has +q on its positive plate and -q on its negative plate. So we say that it is charged to "q" even though it really has net zero charge with +q on one plate and -q on the other. So for the next step, work out what these values are for each capacitor and then label each capacitor plate with +q on one and -q on the other. (Substitute the actual numbers in for the qs.) Actually draw it out and label them on paper.

    (b) Now connect the capacitors together. [Edit: after flipping one of them upside down.] Actually draw this result out on your paper too. It looks like one big capacitor now doesn't it? Find the charge on each of the "big combination"'s capacitor plates. You should notice that on a given plate, it has positive and negative charge. So some of the charge cancels out [on each "big" plate]. But now knowing the total charge, and the capacitance (treat this big, combo capacitor as the parallel of the smaller capacitors), find the voltage.

    (c) Now you can separate the capacitors from each other. And you know the voltage and capacitance of each of the smaller capacitors, so you can find the charge on each, individual capacitor.
    Last edited: Sep 23, 2014
  4. Sep 24, 2014 #3


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    Hi Casey, welcome to PF!

    Look at the figure. In the first, the capacitor are connected to the battery. There is q1 charge on the first one and q2 charge on the second one, positive on the left plates and negative on the right ones.

    You disconnect the capacitors from the battery, and connect them again, but you connect the negative plate of the second one to the positive plate of the first one. And they are still parallel capacitors. What is the net charge on the combination of them? What is the voltage on the combination of the capacitors?


  5. Sep 24, 2014 #4
    Oh fricking heck. Well, that makes a lot of sense as to why what I was trying wasn't working - yes, it's supposed to be 333V, not 300V. Being tired silly does that to a person, I guess.

    But thank you! I'll try that, now that's it's morning and I'm (relatively) fresh. Just one question: what do you mean by 'one big capacitor'? That's kind of perplexing me, and I don't know how important that bit is to understanding the concept as a whole.

    Thank you! I appreciate it.

    Looking at the image you provided does really help me to visualize it.
  6. Sep 24, 2014 #5


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    Well, what I mean by that is if you connect the terminals of one capacitor to the terminals of the other capacitor, it is almost like connecting the plate of one capacitor to the plate of another, and the opposite plate of the first capacitor to the other plate of the second.

    It might help to draw it out though. Look at ehild's second picture. Those two capacitors are about to act like like one big capacitor. Imagine not only connecting the terminals, but imagine the connections are brought closer to the plates until the plates themselves are connected. But what's not so obvious in ehild's second picture, is that as soon as you connect the two smaller capacitors together like that, charge will flow through the terminals until equilibrium of reached. Just remember that when this happens, total charge is conserved.
  7. Sep 24, 2014 #6
    Ahhh, thanks. That makes more sense.

    Also I got the right answer! :D Thanks dude.
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