Electrodynamics - finding potential of a non conducting shell

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To find the potential of a charged non-conducting spherical shell, one must relate the given surface potential to the radius R and the distance from the shell. Solving Laplace's equation is necessary, using the surface potential as a boundary condition. The potential can be expressed using Legendre polynomials, which simplifies the problem. Comparing this to the general solution of Laplace's equation in spherical coordinates will aid in finding the solution. Understanding these concepts is crucial for solving the problem effectively.
jerry222
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Homework Statement
Consider a spherical, charged, non-conducting shell of radius R. Given "surface potential", find potential at any distance.

I do realise there might be such a thing as a surface potential but how can i relate it to R, the distance? Am i supposed to solve the laplace equation with the given surface potential as a solution? I'm a bit stuck, appreciate any hint
Relevant Equations
$\Del V = 0$
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jerry222 said:
Homework Statement:: Consider a spherical, charged, non-conducting shell of radius R. Given "surface potential", find potential at any distance.

I do realise there might be such a thing as a surface potential but how can i relate it to R, the distance? Am i supposed to solve the laplace equation with the given surface potential as a solution? I'm a bit stuck, appreciate any hint
Relevant Equations:: $\Del V = 0$

View attachment 323688
Have you tried part (b) first?
From the answer to that you should be able to get the answer to (a) if the integral is not too nasty.
 
jerry222 said:
Am i supposed to solve the laplace equation with the given surface potential as a solution?
Yes. My hint would be to notice that the potential on the surface of the sphere, that you are given, can be expressed as the sum of just a few Legendre polynomials with certain coefficients. Then compare to the general solution of Laplace's equation in spherical coordinates for problems with azimuthal symmetry. Hopefully, you're familiar with equation (14) here.
 
Last edited:
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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