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Homework Statement
A point charge [itex]q[/itex] is located a distance [itex]d[/itex] away from the centre of a grounded conducting sphere of radius [itex]R<d[/itex]. I need to find the charge density on the sphere and the total induced charge on the sphere.
This is very similar to example 2 here; http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/P435_Lect_06.pdf
Homework Equations
Gauss's Law in differential from; [itex]\rho=\epsilon_0(\nabla \cdot E)[/itex].
The electric field in terms of potential; [itex]E=-\nabla V[/itex]
Potential for a point charge; [itex] V=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|} [/itex]
The Attempt at a Solution
First I found the potential and obtained the same result as in the provided link, that is; [tex]
V(r,\theta)=\frac{q}{4\pi\epsilon_0}\left [ \frac{1}{\sqrt{d^2+r^2-2drcos(\theta)})}-\frac{R}{d\sqrt{(R^2/d^2)+r^2-2(R^2/d)rcos(\theta)}} \right ] [/tex].
Using an image charge of magnitude [itex]-\frac{R}{a}q[/itex] located at [itex]\frac{R^2}{d}[/itex]
Next I try to find the electric field via; [tex]E=-\nabla V=\left ( -\frac{\partial V}{\partial r},-\frac{1}{r}\frac{\partial V}{\partial \theta} \right )[/tex]
This is where things start to get really messy (the derivatives are simple so i won't type them out, they're just really messy). Also, this is in spherical coordinates.
Finally, I use this to find the charge density via Gauss's Law. Applied to the above, this yields;
[tex]
\rho=\epsilon_0\frac{1}{r^2}\frac{\partial}{\partial r}\left ( -r^2\frac{\partial V}{\partial r} \right )+\epsilon_0\frac{1}{rsin(\theta)}\frac{\partial}{\partial \theta}\left ( \frac{-sin(\theta)}{r}\frac{\partial V}{\partial \theta} \right )
[/tex]
And then evaluate this at [itex]r=R[/itex] to find the charge density on the sphere.
This expression ends up being incredibly nasty. As I also have to find the total induced charge I would then need to integrate this expression over the sphere. By rights this should give an induced charge of [itex]-q[/itex]. However, I have no idea how to uintegrate what I obtain, and Mathematica gives a different answer (which happens to be in terms of elliptic integrals).
So, my question has three parts. Firstly, is my reasoning/understanding in the above correct?
Secondly, in the provided link they claim that the charge density is simply given by; [tex]\rho=-\epsilon_0\frac{\partial V}{\partial r}[/tex]
Evaluated at the radius of the sphere. I have no idea how this is derived (I've seen it used a fair bit while searching around, but I don't understand how it is derived or where it is applicable).
Finally, can anybody point me in the right direction?
Thanks in advance for any help!