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Energy in a volume

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    If V=2x^2+6y^2 V in free space, find the energy stored in a volume defined by -1<x<1,-1<y<y, and -1<z<1. (BTW the < are suppose to less than or equal to or greater than or equal to)


    2. Relevant equations
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    3. The attempt at a solution
    I am not really sure of the formula here but I assume it needed at triple integral as it gave me 3 sets of limits and a function. Furthermore it asked for a volume so it was the only thing I could think of I relize it is probably wrong but am I even close? If not what formula should I use?
     

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  3. Sep 14, 2013 #2

    mfb

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    A triple integral is good, but you have to find out what to integrate first.

    I'm not sure how to interpret "V=2x^2+6y^2 V", but it looks like an electric potential, so we have an electric field in that volume. What is the energy density of an electric field?
    Are you sure there are no units involved?
     
  4. Sep 14, 2013 #3
    Oh sorry yah the V stands or volts so yes it is electric potential. (1/2)(epsilon)E^2 equals energy density of an electric field
     
  5. Sep 14, 2013 #4

    mfb

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    Good, now you can use that to solve the problem.
     
  6. Sep 14, 2013 #5
    So what your saying is that E would equal electric potential here or should I use dV=-Es(dS)
     
  7. Sep 14, 2013 #6
    Thus I would take the derivative of the voltage
     
  8. Sep 14, 2013 #7

    mfb

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    That's the idea, right.
     
  9. Sep 14, 2013 #8
    Well what should I derive the function by x or y I'm guessing x
     
  10. Sep 14, 2013 #9

    gneill

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    x,y, and z, actually. You want to determine the electric field, a vector quantity, by finding the gradient of the potential. Presumably you'll then use the magnitude of this vector in your energy density evaluation (triple integral over the volume).
     
  11. Sep 15, 2013 #10
    Ah ok man thanks so partial derivative with respect to x y and z then integrate that three times with the limits given got it will do later on
     
  12. Sep 15, 2013 #11
    Well the gradient is (4x,12y) if I integrate it straight I get 0 which can't be right . I don't know if it is mathematically legal for me to e the agnitude with the x and y arable still in it but ill try anyway see if that does the trick I gets 101.19288
     
  13. Sep 15, 2013 #12

    mfb

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    Remember that you have to square its magnitude:
     
  14. Sep 15, 2013 #13
  15. Sep 15, 2013 #14
    I got the magnitude to be sqrt(4^2+12^2) which is sqrt(160)=E so then E^2=160 which when I integrate I get 6e-9
     
  16. Sep 15, 2013 #15
    Where epsilon is 8.85e-12
     
  17. Sep 15, 2013 #16

    mfb

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    (4x,12y) depends on x and y, and so does the magnitude. In addition, you are missing the z-component here (it is zero, but you have to include it to get a vector with 3 components).
     
  18. Sep 15, 2013 #17
    I get E=sqrt((4x)^2+(12y)^2)=sqrt(16x^2+144y^2) then E^2=16x^2+144y^2 after integrating I get (1/2)epsilon(1280/3)=2e-9
     
  19. Sep 16, 2013 #18
    Am I ok not trying to rush you just wondering?
     
  20. Sep 16, 2013 #19

    mfb

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    You can check calculations with a computer, so what are you waiting for?
    The answer has missing units, but the problem statement has the same problem.
     
  21. Sep 16, 2013 #20
    Ok thanks man yah u are right wolfram said it was ok. I guess I said it wrong did I have the integral set up correctly so that when I performed the integration it would be correct?
     
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