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fluidistic

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## Homework Statement

I must use Green's function [itex]\Phi (x)=\int _{\mathbb{R}^3 } \frac{\rho (x')}{|x-x'|}d^3x'[/itex] to calculate the potential and the electric field due to a spherical shell of radius R and unfiform charge density [itex]\sigma _0[/itex].

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.

## Homework Equations

Already given but in my case it reduces to (I believe): [itex]\int _S \frac{\sigma _0 r'^2 \sin \theta d\theta d\phi }{\sqrt{ r^2+r'^2-2rr' \cos \gamma }}[/itex] where [itex]\cos \gamma = \cos \theta \cos \theta '+ \sin \theta \sin \theta ' \cos (\varphi ' - \varphi )[/itex].

## The Attempt at a Solution

I assume the potential function phi to be radially symmetrical, in other words I can calculate [itex]\Phi (r)[/itex] 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 [itex]\cos \gamma = \cos \theta '[/itex].

Therefore [itex]\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 ' } }[/itex][itex]=2 \pi \sigma _0 \int _0 ^{\pi }\frac{r'^2 \sin \theta ' d \theta ' }{\sqrt {r^2+r'^2 -2rr' \cos \theta ' } } [/itex]. Now I make the change of variable [itex]x=\cos \theta ' \Rightarrow dx=-\sin \theta '[/itex].

And thus [itex]\Phi (r)=2\pi \sigma _0 \int _{-1}^1 \frac{r'^2 dx}{\sqrt {r^2+r'^2-2rr'x} }[/itex]. 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 [itex]-\frac{\sqrt {r^2-2rr'x+r'^2 } }{rr'} \big | _{-1}^1[/itex].

Giving me [itex]\Phi (r)=\frac{ 2\pi \sigma _0}{rr'} (\sqrt {r^2+r'^2+2rr'}- \sqrt {r^2+r'^2-2rr'} )[/itex]. I'm sure there's an error somewhere, the square roots are "too ugly". By the way [itex]r'=R=\text {constant} [/itex].

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