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Find the Capacitance -- Plates

  1. Mar 26, 2017 #1
    1. The problem statement, all variables and given/known data
    In attachments Adsız 1.png

    2. Relevant equations


    3. The attempt at a solution
    K.png This is my charge distrubition.From there I draw electric field lines.And then from potential difference I wrote
    ##E=\frac {V} {d}## and then ##υ=\frac 1 2ε_0E^2## then ##U=\frac {CV^2}{2}##

    I am doing something wrong..I guess I should decide to Are these plates connected in parallel or in series ? But I dont know how to decide that

    Thanks
     
  2. jcsd
  3. Mar 26, 2017 #2

    TSny

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    So far, I don't see anything wrong. What did you do from this point?

    Using the approach of this problem, you do not need to decide whether the plates are in parallel or in series. Your answer will tell you if the plates are in series or parallel.
     
  4. Mar 26, 2017 #3

    mfb

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    Which part are you working on?

    You have a relation between U and C, but you have to determine both.
    In the same way you determine it for every circuit, but you don't necessarily have to determine that.
     
  5. Mar 26, 2017 #4
    Welli I found (a) and it is ##E=\frac {V} {d}## then Energy density will be ##υ=\frac 1 2ε_0E^2## .
    I am stucked here I wrote for (b) ##U=\frac {CV^2}{2}## but I dont know whats the "U".Its like saying ##U=U## (I found like this) Thats why something seemes wrong.

    Thats where I am stucked I think here ##E## will be different then ##\frac {V} {d}## otherwise as I said it becomes ##U=U## and ##C=C## which sounds nonsense in some sense
     
  6. Mar 26, 2017 #5
    Or its U=3U cause theres 3 segments ? The sum of energy of plates ?
     
  7. Mar 26, 2017 #6

    TSny

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    How would you express the total U of the system in terms of electric field energy?
     
  8. Mar 26, 2017 #7

    mfb

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    Something cannot be 3 times itself (unless it is 0 - it is not).
    See TSny's post for a good hint.
     
  9. Mar 26, 2017 #8
    Not like that I meant ##U_{tot}=3U=3(Ad)υ=3(Ad)\frac 1 2 ε_0E^2=\frac {3CV^2} {2}##
     
  10. Mar 26, 2017 #9

    mfb

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    That part is fine.
    That would need a "capacitance per layer" or something like that. Don't do that.
    You have the total energy, you know the voltage, that is sufficient to determine the total capacitance.
     
  11. Mar 26, 2017 #10
    ##C_{tot}=3ε_0 \frac A d## ?
     
  12. Mar 26, 2017 #11

    TSny

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    Yes. But I can't tell how you got that.
     
  13. Mar 26, 2017 #12
    ##3(Ad)\frac 1 2 ε_0E^2=\frac {CV^2} {2}## and ##E=\frac V d##
     
  14. Mar 26, 2017 #13

    TSny

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    OK.
     
  15. Mar 26, 2017 #14
    Thanks
     
  16. Mar 26, 2017 #15

    TSny

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    Can you tell from your answer if the three gaps in the system act like capacitors in series or do they act like capacitors in parallel?
     
  17. Mar 26, 2017 #16
    parallel
     
  18. Mar 26, 2017 #17

    mfb

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    Right.

    Another way to see this: All three gaps have the same electrodes.
     
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