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(LC Circuit) Finding cap charge after connecting to inductor

  1. Apr 22, 2017 #1
    1. The problem statement, all variables and given/known data
    A capacitor with a capacitance of C = 5.95×10−5 F is charged by connecting it to a 12.0 −V battery. The capacitor is then disconnected from the battery and connected across an inductor with an inductance of L = 1.50 H .

    What is the charge on the capacitor after a time interval of 2.30×10−2 s after the connection to the inductor is made?

    2. Relevant equations
    q(t)=Qcos(ωt+∅)

    3. The attempt at a solution
    I have answered a couple questions prior to this question.
    I found these:
    ω = 106 rad/s
    Period of the electrical oscillations
    T = 5.94*10^-2 s
    Initial charge of the cap
    Q = 7.14*10^-4 C
    Energy initially stored in the cap
    4.28*10^-3 J

    For the question where i had to find the charge, I used
    q(t)=Qcos(ωt+∅)
    and got 7.13*10^-4 which is wrong. I plugged 2.30*10^-2 for time t and the ∅=0.
     
  2. jcsd
  3. Apr 22, 2017 #2

    TSny

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    Is the argument of the cosine function in radians or degrees? How about your calculator?
     
  4. Apr 22, 2017 #3
    My calculator is in degrees. I just changed it to radians. The cos function would be in radians. But i would get a negative number for q?
     
  5. Apr 22, 2017 #4

    TSny

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    Can you interpret the negative sign?
     
  6. Apr 22, 2017 #5
    Is it the capacitor discharging? I'm not exactly sure.
     
  7. Apr 22, 2017 #6

    TSny

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    When the capacitor was initially charged to 12 V, did both plates of the capacitor initially have a positive charge or did the plates have opposite charge?
     
  8. Apr 22, 2017 #7
    both plates had opposite charge
     
  9. Apr 22, 2017 #8
    i meant to say the plates had opposite charge* sorry
     
  10. Apr 22, 2017 #9

    TSny

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    Yes. So, suppose you consider the plate that is initially positively charged. When the capacitor is attached to the inductor to make an LC circuit, what happens to the charge on this plate. Does it just discharge to zero and remain zero? Or, ....?
     
  11. Apr 22, 2017 #10
    The capacitor starts discharging through the inductor but it doesn't just instantly discharge to zero. It discharges as current builds up to its maximum since current is not instantaneous. So if the current doesn't change automatically, the capacitor will start absorbing charge in the opposite polarity ??
     
  12. Apr 22, 2017 #11
    I had something like this in my notes but I didn't understand what it was saying. I'm just tryng to make sense out of it and I don't know if I'm explaining it the right way.
     
  13. Apr 22, 2017 #12

    TSny

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    You are right that the polarity of the charge on either plate alternates over and over again. At any time, the plates have equal but opposite charge. The cosine function with Φ = 0 describes the charge on the plate that was initially positively charged. So, when the cosine function gives you a negative value at some instant of time, that means the charge on that plate is negative at that time.

    Unfortunately, the question is a little ambiguous. It asks for "the charge on the capacitor" at a certain time. Usually, "charge on the capacitor" means the absolute value of the charge on either plate. So, it is not clear if they want you to include the negative sign in your answer.
     
  14. Apr 22, 2017 #13
    That's why I was thinking the negative wouldn't work. Thanks for clearing that up! Now I have to find the current in the inductor at that time given in part (e)

    I tried using i = -ωQmax(sin(ωt+∅)
    but it's telling me that it's not the right answer. I got -4.896*10^-9
     
  15. Apr 22, 2017 #14

    TSny

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    The 10-9 part of your answer does not look correct.
     
  16. Apr 22, 2017 #15
    I recalculated and got -0.0739A but it's still wrong
     
  17. Apr 22, 2017 #16

    TSny

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    Hmm. I think the 4.896 part of your answer in post #13 is correct. Just the 10-9 is wrong.
     
  18. Apr 22, 2017 #17
    Sadly i missed the question. For some reason im getting different numbers everytime i try to calculate. I've been doing homework for hours so i must jst be tired. The final answer is -4.91*10^-2, but i don't know how they got that answer
     
  19. Apr 22, 2017 #18

    TSny

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    The -4.91 agrees with your -4.896. The slight difference is just "round off" differences. I don't know how you got the 10-9. If you show your calculation with all of the numbers shown, then we might be able to spot the error.
     
  20. Apr 22, 2017 #19

    gneill

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    It's not clear how they intend you to interpret the direction of the current (and so be able to assign a sign to it). Did they include a circuit diagram with the current direction defined?
     
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