Potential inside NON-Conducting hollow spherical shell

  • Thread starter dikmikkel
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  • #1
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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.
 

Answers and Replies

  • #2
Born2bwire
Science Advisor
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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.
 
  • #3
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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?
 
  • #4
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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.
 
  • #5
Meir Achuz
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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.
 
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  • #6
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That's gold thank you.
 

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