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Homework Help: Charge across a switch (almost solved)

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Two resistors, R1 = 8.00 Ω and R2 = 4.00 Ω and two uncharged capacitors, C1 = 0.480 μF and C2 = 0.240 μF are arranged as shown in the figure below.

    With a potential difference of 21.1 V across the combination, what is the potential at point a with S open? (Let V = 0 at the negative terminal of the source.)
    7.03 V
    What is the potential at point b with the switch open?
    14.1 V
    When the switch is closed, what is the final potential of point b?
    7.03 V
    How much charge flows through the switch S after it is closed?

    2. Relevant equations
    I know that the potential is the same across the switch when the switch is closed, but I don't know how you can calculate the charge at the specific point B.

    3. The attempt at a solution

    I tried Q=CV for V=7.03 and C=.24 microF, and I got Q = 1.69 microC (I thought that because the potential drop at C2 is equal to the potential at B), but that didn't work. Everything else is right, but I just don't know how you do the last part.

    Other answer I tried, but were wrong, were 3.38 microC, 6.75 microC, and 0 C (just to see).
  2. jcsd
  3. Jul 11, 2008 #2
    Okay, so I did some reading, but I still didn't get the answer. I know that when the switch is open, the charge on both C1 and C2 is the same (3.38 micro C). But after the switch is closed, the charge on C1 becomes (6.75 microC) to account for the same potential drop, and the charge on C2 becomes (1.69 microC).

    I'm am completely confused.
  4. Jul 11, 2008 #3


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    Staff: Mentor

    With your numbers in post #2, I got 1.68uC. How accurate does the answer have to be to pass the question? You said in post #1 that you got 1.69uC. Your reasoning seems okay to me -- calculate the difference in total charge before and after, and it seems like that charge delta has to flow as current through the switch...
  5. Jul 11, 2008 #4


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    Homework Helper

    I think you are pretty much on top of it.

    What is the net charge at point b before the switch is closed? Think about how much charge + and - there is on each plate of each capacitor. If you are assuming that there is no potential difference between the inner plates because of their being connected that would be a clue.

    What happens with the closing of the switch? How much charge is now on each plate of each capacitor? If it is not the same as before the switch closed wouldn't there have had to be a charge transfer?
  6. Jul 11, 2008 #5
    When the switch is open, the charge Q is constant (3.38e-6 C) for both plates. When the switch closes, the charge Q1 is 6.76e-6 C and Q2 is 1.68e-6 C.

    I'm confused on how you find what B actually is, because you don't have a capacitance for that point. I know the voltage difference before the switch is closed is 7.03 V (C1=7.03 and C2=14.1). But after the switch is closed, C1=14.1 and C2=7.03, so the potential difference is -7.03 V?

    The net charge on B before the switch is closed is zero isn't it, because there is an equal charge on both C1 and C2. But after, does B become 5.064e-6 C, since that is the difference in charges afterward?

    If I were to take that 5.064e-6 C difference, and subtract that from the original charge at point B (3.38e-6 C), I would still get 1.68e-6 C, which was wrong.

    Is any of what I said right?
    Last edited: Jul 11, 2008
  7. Jul 11, 2008 #6


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    Homework Helper

    You have a value for the net charge before the switch was closed that you established as 0. (+ on one, - on the other)

    You identify that there would be a net charge now at the node of 5.06 x 10^-6 C.

    Are you missing that the charge on the plates taken together was 0 after you established that already?

    Didn't the charge came from somewhere? Wouldn't flow through the switch be a good candidate?
  8. Jul 11, 2008 #7
    Thanks. Your right. Before my first post, I put 5.06e-6 C as the answer, but that was wrong. Turns out that the answer was -5.06e-6 C, which makes sense now. Thanks again.
  9. Jul 11, 2008 #8


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    Homework Helper

    Happy sailing then.

    Good Luck
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