1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential inside NON-Conducting hollow spherical shell

  1. Sep 15, 2011 #1
    Hi Guys,
    Suppose we have a spherical shell with charge density on the surface [itex]\sigma[/itex] and radius R. The potential inside the shell is given by:

    V_(x,y,z) = [itex]\frac{V0}{R^{2}}(6z^2+ax^2+by^2)[/itex]
    It is assumed, that the potential is rotational symmetric around the z-axis inside and outside the shell, and goes to 0 far away from the shell. There's no charge inside and outside the shell and no outer field.

    How do i determine the constants a and b?

    Mabye change to spherical coordinates and solve the equation:
    [itex]\frac{\partial{V}}{\partial{\theta}}=0[/itex]
    for a or b. But i can't figure out any other conditions if this is right.
     
  2. jcsd
  3. Sep 15, 2011 #2

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    The Laplacian of the voltage is related to the charge. So your boundary conditions would be:

    [tex] \nabla^2 V = -\frac{\sigma}{\epsilon} [/tex]

    On the surface of the shell. So for ease of analysis you would want to do this in spherical coordinates.
     
  4. Sep 15, 2011 #3
    I can't really see that this is a boundary condition? And would solving possions equation on the surface help me determine a and b, i can't really follow?
     
  5. Sep 15, 2011 #4
    Aah sorry im really tired now. I take the laplacian, and this would relate it all, beautiful, Thanks. Just so used to try finding the potential :)
    And sorry for the language. I'm from Denmark.
     
  6. Sep 15, 2011 #5

    clem

    User Avatar
    Science Advisor

    If the pot inside is axially symjmetric, and b must be equal.
    Since DEl^2 V=0 inside, the angular distribution must be a Legendre polynomial.
    Since the power of x,y,and z is 2, it must be
    [tex]P_2(\cos\theta)=(3\cos^2\theta-1)/2[/tex].
    This means a=b=-3.
     
    Last edited: Sep 15, 2011
  7. Sep 15, 2011 #6
    That's gold thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Potential inside NON-Conducting hollow spherical shell
Loading...