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Homework Help: Find the unknown capacity of two capacitors

  1. Jun 10, 2018 #1
    1. The problem statement, all variables and given/known data

    We consider two capacitors with the unknown capacitances C1 and C2. If
    one connects the capacitors in series and applies a voltage of 100 V, finds
    that two positively charged plates in total have the charge 1.5 · 10-4 C. If
    one switches the capacitors in parallel and again applies a voltage of 100 V, finds
    that two positively charged plates, in total have the charge charge 4 · 10-4 C. Which
    capacities have the two capacitors?

    2. Relevant equations
    Q=C*V

    Q=Charge
    C=Capacitance
    V=Voltage

    3. The attempt at a solution
    My attepmpt to solve this is attached in the photos bellow. In general , I was trzing to apply Q=C*V to the first and the second case and then tried to combine my answers
     

    Attached Files:

  2. jcsd
  3. Jun 10, 2018 #2

    TSny

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    Hello.

    I don’t think you are using the correct charge for each capacitor when they are in series. Read the second sentence of the problem statement carefully. Otherwise, your work looks good so far.
     
  4. Jun 10, 2018 #3

    epenguin

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    It seems to me OK. You realise that in series the charge is the same on both capacitors; and that the potentials add up to the known totaL potential. The rest is an algebraic excercise. I would now express in terms of C1 and the ratio r of capacitances. You should be able to eliminate C1 and find two capacitances, but from these measurements you will not know capacitor has which capacitance.
     
  5. Jun 10, 2018 #4

    TSny

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    Hi, epenguin.

    The problem states that when the two capacitors are in series, the "two positively charged plates in total have the charge 1.5 · 10-4 C".
    To me, this is different than saying that "each positively charge plate has a charge of 1.5 · 10-4 C".

    [If each positively charged plate has 1.5 · 10-4 C, then I get that there is no real solution to the set of equations for C1 and C2.]
     
  6. Jun 11, 2018 #5
    Thank you for your replies. I indeed missed a part concerning a total charge. Attached is the new calculations that I did. So in my calculations instead of 1.5 · 10-4 C i used 7.5 · 10-5 C ( which I got by deviding a total charge when capacitors were connected in series by 2 ). Then all calculations work out just fine. Thank you again for your help.
     

    Attached Files:

  7. Jun 14, 2018 #6

    epenguin

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    You have to be right about that what the question means. I guess I had resistance in accepting that because in that case that part of the question seemed so artificially contrived - testing examsmanship :devil: rather than physics understanding. Why would anyone add those charges, to what physics, or what straightforward measurement would that correspond? The pair combines as one equivalent capacitance of 0.75 μF.

    More constructively at least this made me realise something I had never heard of or thought, nor had occasion to: that two capacitors In series cannot have more than a quarter of the capacitance as the same two in parallel (the ≤ becoming = when the two capacitances are equali. Likewise two resistors in parallel cannot have more than a quarter the resistance of the two in series, which I don't remember having particularly heard of. )
    1.5 is more than quarter of 4 so without needing to solve equations we should have been able to see immediately that the physically meaningful charge is not meant to be 1.5.
     
    Last edited: Jun 14, 2018
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