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Potential energy of an hemispheric shell

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Given that the hemisphere has a charge +Q distributed through its surface with radius a. Find the electric potential on any point on z axis (the plane of the hemisphere is oriented in positive z direction).


    2. Relevant equations
    [itex]\phi[/itex] = [itex]\int[/itex][itex]\frac{kQ}{|R-R'|}[/itex]*ds
    (surface integral)

    3. The attempt at a solution
    So i decided to break this problem into little pieces of the surface using some trig.

    I made a angle θ from the plane of the hemisphere around the cross section (so there is semi circles) and then an angle [itex]\varphi[/itex] on the xy-plane.

    I found ds to be a[itex]^{2}[/itex]d[itex]\varphi[/itex]d[itex]\theta[/itex] (using the arc lengths of my angles and radii).

    Now my problem is finding an equation for the vector R-R'. I tried making it a function of [itex]\theta[/itex] (i.e. i got (sqrt(z[itex]^{2}[/itex] + a[itex]^{2}[/itex]sin[itex]^{2}[/itex]([itex]\theta[/itex])
    but the integral was not a nice one and it lead me to believe that it wasn't correct. if someone can help me along that would be great!
     
  2. jcsd
  3. Oct 10, 2011 #2

    ehild

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    Is the charge evenly distributed on the surface?

    Make a picture and indicate R and R'!


    ehild
     
  4. Oct 10, 2011 #3
    oh yeah evenly distributed on surface. I made a paint object to help clarify. R' is to the ds and R is to the point on the z axis. I know R has to be z but im having a problem figuring out R'. Thanks for this and the last problem you help me with.
     

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  5. Oct 10, 2011 #4

    ehild

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    Nice picture! :rofl: You need the magnitude of the vector difference of R-R'. What about using the Law of Cosines?
    To make your task more clear I draw an other picture with Paint (using the option to draw circle, ellipse, straight line )

    ehild
     

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  6. Oct 10, 2011 #5
    ok so should my theta be from the z axis and i think that changes my whole set up oh wait i can use 90 + theta and it would work. Ok thanks time to go to work on it. Plus i only really care about the z direction.
     
  7. Oct 10, 2011 #6
    I get my |R- R'| = [itex]\sqrt{ a^2 +z^2 + 2azsin(θ)}[/itex]. look right? Umm this isn't turning out to be a easy integral. Im thinking mabye go back to Cartesian.
     
    Last edited: Oct 10, 2011
  8. Oct 10, 2011 #7

    ehild

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    Last edited: Oct 10, 2011
  9. Oct 10, 2011 #8
    I just need to add 90 degrees to my theta so it restricts it to my hemisphere not a whole sphere. The ds stays the same except for that sin(θ) term which i added.
     
  10. Oct 10, 2011 #9

    ehild

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    It will not be sin(theta) in the surface element of your special spherical system of coordinates.

    ehild
     
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