# Evaluation of potential, problem reduced to an integral

1. Mar 23, 2012

### fluidistic

1. The problem statement, all variables and given/known data
I must use Green's function $\Phi (x)=\int _{\mathbb{R}^3 } \frac{\rho (x')}{|x-x'|}d^3x'$ to calculate the potential and the electric field due to a spherical shell of radius R and unfiform charge density $\sigma _0$.

P.S.:I know I could use Dirac's delta to get the solution faster but I'm asked to do so in another exercise.
2. Relevant equations
Already given but in my case it reduces to (I believe): $\int _S \frac{\sigma _0 r'^2 \sin \theta d\theta d\phi }{\sqrt{ r^2+r'^2-2rr' \cos \gamma }}$ where $\cos \gamma = \cos \theta \cos \theta '+ \sin \theta \sin \theta ' \cos (\varphi ' - \varphi )$.

3. The attempt at a solution
I assume the potential function phi to be radially symmetrical, in other words I can calculate $\Phi (r)$ and that would be the solution for whatever phi and theta.
Let's say I calculate Phi along the z-axis so that theta=0. This gives $\cos \gamma = \cos \theta '$.
Therefore $\Phi (r)=\int _0^{2\pi } \int _0^{\pi } \frac{r'^2 \sin \theta ' d\theta ' d \varphi ' }{\sqrt {r^2+r'^2-2rr' \cos \theta ' } }$$=2 \pi \sigma _0 \int _0 ^{\pi }\frac{r'^2 \sin \theta ' d \theta ' }{\sqrt {r^2+r'^2 -2rr' \cos \theta ' } }$. Now I make the change of variable $x=\cos \theta ' \Rightarrow dx=-\sin \theta '$.
And thus $\Phi (r)=2\pi \sigma _0 \int _{-1}^1 \frac{r'^2 dx}{\sqrt {r^2+r'^2-2rr'x} }$. This is where I'm stuck. Basically r' is a constant... oh wait then I guess I should look into an integral table?
If you see any error feel free to help me.

Edit: The integral is worth $-\frac{\sqrt {r^2-2rr'x+r'^2 } }{rr'} \big | _{-1}^1$.
Giving me $\Phi (r)=\frac{ 2\pi \sigma _0}{rr'} (\sqrt {r^2+r'^2+2rr'}- \sqrt {r^2+r'^2-2rr'} )$. I'm sure there's an error somewhere, the square roots are "too ugly". By the way $r'=R=\text {constant}$.

Last edited: Mar 23, 2012
2. Mar 23, 2012

### sunjin09

Complete the square in the square root in your final step, the answer looks clean. The dimension of your answer is sigma/meter, while before you evaluate the integral it is sigma*meter, somewhere you missed some r^2.

3. Mar 23, 2012

### fluidistic

Oops you are right about the r'² term, I forgot it. Ok about completing the square...

4. Mar 23, 2012

### fluidistic

I get $$\Phi (r)=4\pi \sigma _0 \frac{R^2}{r}$$ if $r>R$.
And $\Phi (r)=0$ if $r<R$ which, if I recall well is false. I think it should be constant rather than null.

5. Mar 24, 2012

### sunjin09

remember potential is not a physical quantity, you can add arbitrary constant to it and still valid. In order to make your potential look nice, you can afford to make it continuous across the sphere.

6. Mar 24, 2012

### fluidistic

Thanks for helping me.
So here I could "define" $\Phi (r)=4 \pi \sigma _0 R$ when r<R. Is this right?