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## Homework Statement

We want to calculate the field of a uniformly polarized sphere of radius=R

## Homework Equations

[tex]

V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau'

[/tex]

## The Attempt at a Solution

i)I know that

[tex]\sigma_{b} = P \cos\theta[/tex]. I know that [tex]\rho_{b}=0[/tex]

Now using the cosine law I find that

[tex]

\tilde{r}

= \sqrt{R^2 + r^2 - 2Rr\cos\theta}[/tex]

and using spherical coordinates I find that

[tex]da' = R^2 \sin\theta d\theta d\phi[/tex]

So here comes the problem :When I try to use the voltage equation I can't find the right answer.I've searched everywhere on this matter and I can't understand why this [tex]

V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}} \cos\theta' R^2 \sin\theta' d\theta' d\phi[/tex] is correct and not this [tex]

V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}}R^2 \sin\theta' d\theta' d\phi[/tex] and where does this extra [tex]\cos\theta'[/tex] come from and why?

ii) Secondly I don't understand why when I integrate from 0 to pi we don't have to multiply the whole integral with the number 2.Do we perceive it only as the half sphere and if yes why?

Thanks In advance and please help..this is baffling me for waay too long now [/B]

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