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Spherical Shell-Potential Energy, Energy density

  1. Mar 27, 2017 #1
    1. The problem statement, all variables and given/known data
    Shell.png

    2. Relevant equations


    3. The attempt at a solution
    I fount these

    Part(a)its ##E=\frac {ρa^3} {3ε_0}## and
    ##υ=\frac 1 2ε_0E^2##
    Part (b)
    ##dU=4πr^2drυ##
    Part (c)
    ##U=\int_0^a 4πr^2udr## but it gives me ##U=\frac {-Q^2} {8πε_0a}##

    This"-" bothers me.
     
  2. jcsd
  3. Mar 27, 2017 #2

    TSny

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    How is the charge Q distributed for a metallic sphere?
    What is ##E## for points inside the sphere?
    What is ##E## for points outside the sphere?

    [EDIT: Note that your expression for ##E## does not have the correct dimensions for an electric field.]
    The "-" sign should bother you. Your integrand is positive (including the dr). So, the integral must be positive. But are you sure you want to integrate from 0 to ##a##? If you can answer my questions above, it should help you see what you should use for the range of integration.
     
  4. Mar 27, 2017 #3
    Someone was faster :P
     
  5. Mar 27, 2017 #4
    Just in surface
    E=0
    ##E=\frac {ρa^3} {3ε_0r^2}## where ##r≥a## I know in the upside I forget ##r^2##
    What else it could be.. ?
    I thought about that but it still give me nothing...
    I am thinking there should be minus cause I am thinking like bringing charges from ##r=0## to ##r=a## but I should bring charges ##r=∞## to ##r=a## but still there appears "-" sign also in ##r=∞## case what should I use the volume...? Or is it make sense ?
    The result of integral will be ##U=Constant\int_∞^a\frac {1} {r^2}dr=constant\frac {-1} {r}## I cant get rid of "-".I think its cause of potential energy case.
     
  6. Mar 27, 2017 #5
    Is it ##U=Constant\int_a^r\frac {1} {r^2}dr## where r goes to ∞ ?
     
  7. Mar 27, 2017 #6

    TSny

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    So, all the charge Q is on the surface. So, there is no volume charge density ##\rho##.

    Yes, ##E = 0## inside the sphere. So, what is the energy density inside the sphere? What would you get if you integrated the energy density over the volume of the inside of the sphere (##0<r<a##)?

    There is no volume charge density. For points outside the sphere, try to express E in term of Q and r.

    I'm not sure what you are doing here. Follow the outline given in the problem. How would you express the energy contained in a spherical shell of radius r and thickness dr?
     
  8. Mar 27, 2017 #7
    Zero ?
    ##E=\frac {Q} {4πε_0r^2}##
    ##dU=4πr^2drυ##
     
  9. Mar 27, 2017 #8

    TSny

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    Yes Yes Yes
     
  10. Mar 27, 2017 #9
    soo?
     
  11. Mar 27, 2017 #10

    TSny

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    What is preventing you from finishing the problem?
     
  12. Mar 27, 2017 #11
    Limits of integral
     
  13. Mar 27, 2017 #12

    TSny

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    You need to add up the energy in every spherical shell for which ##E \neq 0##.
     
  14. Mar 27, 2017 #13
    from a to ∞ ?
     
  15. Mar 27, 2017 #14
    a to ∞ ?
     
  16. Mar 27, 2017 #15

    TSny

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    Yes
     
  17. Mar 27, 2017 #16
    İnteresting... ok thanks a lot
     
  18. Mar 27, 2017 #17

    TSny

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    Yes, it is very interesting. The potential energy stored in the system can be thought of as stored in the field that extends from the surface of the sphere all the way out to infinity!
     
  19. Mar 27, 2017 #18
    Thats just amazing....
     
  20. Mar 27, 2017 #19
    I used gaussian law and surface charge density etc to find E
     
  21. Mar 27, 2017 #20

    TSny

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    OK
     
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