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Electric potential and field of a thick spherical shell

  1. Feb 27, 2008 #1
    Question: I have a thick spherical shell (inner radius a and outer b) with charge density = kr^2. I need to find the field and potential at all points in space and calculate the energy.

    Is it possible to take the equation for the electric field of a solid sphere, and integrate from a to b? Or should I find a way to relate the volume charge density to a surface charge density and then integrate the equation for the field of a thin hollow sphere over the radii of various shells?

    Any advice would be appreciated....

    W. =)
  2. jcsd
  3. Feb 27, 2008 #2


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    Use Gauss's law to get E.
  4. Feb 27, 2008 #3
    ok...tried that, but the thing that is confusing me is the r in the charge density. If the density is changing throughout the shell, then wouldn't that somehow have to be taken into account when I use Gauss's law? the r in the charge density is different from the r in da, one represents the radius of the gaussian surface, the other is the radius of the shell. Wouldn't I need to integrate over both?

    I was thinking that if the charge density is changing according to the radius of the shell, then couldn't I say that, if I were to divided the thick shell into an infinite number of thin shells, the surface charge density would be the kr^2 for each r-shell? And then I could integrate over r to get the total field because of the principle of superposition?
  5. Feb 28, 2008 #4
    It is easy to show that a sphere of constant surface density causes exactly the same electric field (outside the sphere) as a point charge in the center (use Gauss law and symmetry).
    Similary Gauss law predicts that constant surface density sphere does not cause any field inside it.
    So if you calculate E(r) you can imagine all charge inside the ball with radious r lies in the center.
  6. Feb 28, 2008 #5
    I'm thinking that what I need to do is simply find q enclosed and use gauss's law to get the field. For q enclosed, however, since it's not a constant volume charge density, I would have to integrate rho, from the inner radius to the outer over spherical coordinates, to find the volume of charge in the shell...

    if that's right, then my q enclosed would be q = 4*pi*k*(b^3-a^3)/3

    which would lead to an electric field of k(b^3-a^3)/3*epsilon-naught*r^2 in the r hat direction.

    does that sound about right?
    Last edited: Feb 28, 2008
  7. Feb 28, 2008 #6
    Charge density is ussualy charge per unit of volume and not charge per unit of radious.
  8. Feb 28, 2008 #7
    but this particular charge density isn't constant...it's defined as rho = kr^2 so it's dependent on the radius of the shell.
  9. Feb 28, 2008 #8


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    You have the right idea but be careful. Recall that dV = [tex] r^2 \sin \theta d\theta d\phi dr [/tex] Integrated over the angles, it gives [tex] 4 \pi r^2 dr [/tex] You seem to have used 4 pi dr.
  10. Feb 28, 2008 #9
    you're right...I did forget the r^2.

    thanks. =)
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