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Energy of Spherical Distribution

  1. Feb 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the energy required to generate a spherical charge distribution
    of radius R and uniform density ρ0 . In class, we went over two different
    methods. For this assignment, you need to work out the following two
    integrals over a volume (and surface) of a concentric sphere of radius a,
    where a > R.

    U = [tex]\epsilon//2[/tex] [ [tex]\int[/tex]E^2 +[tex]\oint[/tex] VE ]
    here the first integral is a volume integral over the sphere and the second integral is a surface integral over the surface of the sphere.
    this is equation 2.44 in Griffiths


    2. Relevant equations
    I started with the equation for electric field inside a sphere of uniform distribution

    which is E = kQr/R^3 in radially outward, where k is the 1/4 pi epsilon constant and r is some some distance from the center less than R

    Then I use the equation for Potential inside the sphere as V= (kQ/2R)(3-r^2/R) again where r is less than R




    3. The attempt at a solution

    I am just stuck with the integration I guess. The first integral uses the square of E which is easily found by just squaring the equation above, so that now E^2 is not a vector but just the scalar square of the equation above. then dTau for the volume of the sphere is just the volume element for spherical coordinates? (r^2)sin(phi)drd(phi)d(theta)

    The second integral uses the V equation, multiplied by the E equation and integrates over the surface where the da = area element for area of surface of sphere = (r^2)sin(phi)d(phi)d(theta)



    I apologize if this is hard to read, I am terrrrrrible at using the latex stuff, it ever seems to work out

    Edit: I don't yet have a complete solution but I just realized my confusion was coming form mistranslation in the coordinate systems. I always forget what theta and phi correspond to as i have had teachers switch them up before. I saw the sin(phi) thinking the integrals were going to integrate sin (phi) from 0---2Pi which would yield zero
     
    Last edited: Feb 20, 2008
  2. jcsd
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