Understanding Image Charge and Potential in Electrostatics

[crimsonidol]

1. Homework Statement
We have a grounded conducting sphere of radius R and a charge q is placed a distance a from the center of the sphere. Show that the potential in the interior produced by q and the distributed induced charge is the same as that produced by q and its image charge q'.The image charge is at a distance a'=R^2/a fro the center collinear with q and the origin. Calculate the electrostatic potential for a<R<a'. Show that potential vanishes for r=R if we take q'=-qR/a


2. Homework Equations
In the course we are investigating legendre equations, legendre polynomials etc.


3. The Attempt at a Solution
I can find potential by simply using V=Sum (1/4pieps)q/r however there is no legendre polynomial or legendre series in it. I tried Laplace's Equation however I got confused. Because in Laplace's eqn I can only deal with r. Where have I gone wrong? How should i think?

 

[crimsonidol]

i just need a spark :(

 

[gabbagabbahey]

I'm not sure exactly what the questioner is expecting from you for this problem. The easy way to show that the actual potential is the same as the potential you get from the method of images is to simply show that they satisfy the same boundary conditions and differential equation (Laplace's equation) in the given region (interior of sphere) and then appeal to the uniqueness theorem. Another, more difficult way is to actually solve Laplace's equation using separation of variable (infinite sum of legendre polynomials) and compare that to the solution you get using method of images by expanding the latter in powers of $r$.

Given how easy the first method is (or should be), I suspect that the questioner expects you to do it the second way. However, the first solution should get you full credit given the ambiguous wording of the problem statement (assuming you've transcribed it word for word from your assignment sheet).

 

[crimsonidol]

thanks for the help. I got it and it was quite straightforward. I wonder why i didn't do it until now :)
thanks again.