1. The problem statement, all variables and given/known data We want to calculate the field of a uniformly polarized sphere of radius=R

  2. Relevant equations

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

  1. The attempt at a solution i)I know that $$\sigma_{b} = P \cos\theta$$ $$\rho_{b}=0$$

Now using the cosine law I find that $$\tilde{r} = \sqrt{R^2 + r^2 - 2Rr\cos\theta}$$ and using spherical coordinates I find that $$da' = R^2 \sin\theta d\theta d\phi$$

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 $$V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}} \cos\theta' R^2 \sin\theta' d\theta' d\phi$$ is correct and not this $$V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}}R^2 \sin\theta' d\theta' d\phi$$ also where does $$\cos\theta'$$ 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 o_O

  • $\begingroup$ The integral in $\phi$ from 0 to $2\pi$ takes care of the factor of two for you: imagine that the integral in $\theta$ from 0 to $\pi$ sweeps out a half circle, which you then rotate in $\phi$ to get a full sphere. $\endgroup$
    – Chris
    Jul 5, 2016 at 19:12

2 Answers 2


1) Maybe this could provide you with some insight. This is from Griffiths Electrodynamics 3rd Edition. I'm not sure why the solutions you've found require the cosθ′ but this one doesn't include it.

Griffiths electrodynamics

2) When using spherical coordinates and trying to outline a full sphere, theta outlines a half-circle (existing in the xz plane), phi then rotates and projects this half-circle aroud the z axis. θ then must range from 0 to 2π, ϕ from 0 to 2π for the entire sphere.


If you try to find it for any arbitrary $\vec{r} = (r, \theta)$, you'll get the following integral:

$$V(r, \theta) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta' \frac{P \cos\theta'}{\sqrt{R^2 + r^2 - 2Rr\cos (\theta' - \theta)}} R^2 \sin\theta' $$

This is not too much fun to compute by hand, so textbooks want you to fix $\vec r$ along the z-axis and set $\theta = 0$. That gives the following integral:

$$V(z) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta' \frac{P \cos\theta'}{\sqrt{R^2 + z^2 - 2Rz\cos \theta'}} R^2 \sin\theta' $$

Which is the same as the one Keanu shows in his answer.


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