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The solution of Poisson equation is given by $$ \mathbf E = \int \frac{\rho (\mathbf r )(\mathbf r_{0} - \mathbf r )}{|\mathbf r_{0} - \mathbf r|^{3}}d^{3}\mathbf r. $$ I tried to use this term for a field of uniformly charged ball and got incorrect result: for the field inside the ball integration gives $$ \int \frac{\rho (\mathbf r )(\mathbf r_{0} - \mathbf r )}{|\mathbf r_{0} - \mathbf r|^{3}}d^{3}\mathbf r = \rho \int \frac{\mathbf r_{0}}{|\mathbf r_{0} - \mathbf r|^{3}}d^{3}\mathbf r - \rho \int \frac{\mathbf r}{|\mathbf r_{0} - \mathbf r|^{3}}d^{3}\mathbf r = $$

$$ = |\mathbf r_{0} = r_{0}\mathbf e_{z}| = \rho r_{0}\mathbf e_{z} \int \limits_{0}^{R} \int \limits_{0}^{\pi} \int \limits_{0}^{2 \pi} \frac{r^{2}dr sin(\theta )d\theta d \varphi}{(r^{2} + r_{0}^2)^{\frac{3}{2}}} + 0 = 4 \pi \rho r_{0}\mathbf e_{z} \left( \frac{1}{\sqrt{2}} - ln(\sqrt{2} + 1)\right), $$ which isn't correct. Where did I make the mistake?

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What is your justification for rewriting $|\mathbf{r}_0-\mathbf{r}|^3 as (r^2+r_0^2)^{\frac{3}{2}}$? – KDN Mar 1 '13 at 18:54
Oh, of course. So I get $$ \rho \mathbf r_{0}\int \frac{dV}{x^2 + y^2 + (z - r_{0})^2}. $$ After substituting $z -> z - r_{0}$ without using spherical coordinates I got integral which doesn't express in terms of standart functions. So do you know, how to transite to spherical coordinate system correctly? – user8817 Mar 1 '13 at 22:15
up vote 2 down vote accepted

Suppose the radius of the sphere is $R$. If you'll permit me, let's calculate the electrostatic potential $\Phi$ inside of the sphere, and then let's use the definition $$ \mathbf E = -\nabla\Phi $$ to determine the electric field. If you want, I can directly do the integral for $\mathbf{E}$, but it's just a bit messier. In any case, we have $$ \Phi(\mathbf x) = \frac{1}{4\pi\epsilon_0}\left(\int_{|\mathbf x'|<|\mathbf x|}d^3x'\,\frac{\rho}{|\mathbf x - \mathbf x'|}+\int_{R>|\mathbf x'|>|\mathbf x|}d^3x'\,\frac{\rho}{|\mathbf x - \mathbf x'|} \right) $$ Let's choose $\mathbf x = r\mathbf e_z$ as you did, for simplicity, then in spherical coordinates we have $$ d^3x' = dr'd\theta'd\phi'\,r'^2\sin\theta', \qquad |\mathbf x - \mathbf x'|=r^2+r'^2-2rr'\cos\theta' $$ So we have $$ \Phi(\mathbf x) = \frac{2\pi\rho}{4\pi\epsilon_0}\int_0^r dr'r'^2\int_0^\pi d\theta'\sin\theta'(r^2+r'^2-2rr'\cos\theta')^{-1/2}+ \Big(\int_r^R\cdots\Big) $$ Where the $2\pi$ came from the integration in $\phi'$. Now, we make the substitution $$ u = r^2+r'^2-2rr'\cos\theta',\qquad \frac{du}{2rr'} = \sin\theta' d\theta' $$ so the integral becomes $$ \Phi(\mathbf x) = \frac{\rho}{4\epsilon_0r}\int_0^r dr' r'\int_{(r-r')^2}^{(r+r')^2}du\, u^{-1/2} + \Big(\int_r^R\cdots\Big) $$ Performing the integral in $u$ gives \begin{align} \Phi(\mathbf x) &= \frac{\rho}{2\epsilon_0r}\int_0^r dr' r'(\sqrt{(r+r')^2}-\sqrt{(r-r')^2}) + \Big(\int_r^R\cdots\Big)\\ &= \frac{\rho}{\epsilon_0r}\int_0^r dr' r'^2 +\frac{\rho}{\epsilon_0}\int_r^R dr'r'\\ &= \frac{\rho}{\epsilon_0}\left[\frac{r^2}{3} - \frac{r^2}{2} + \frac{R^2}{2}\right] \end{align} where we have used the fact that $\sqrt{(r^2-r'^2)}$ equals $r-r'$ for $r>r'$ and $r'-r$ for $r<r'$. Finally, taking the negative gradient of the potential in spherical coordinates gives $$ \mathbf E(\mathbf x) = \frac{\rho }{3\epsilon_0}r\,\mathbf e_r $$ which is correct as you can check via Gauss's Law.

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Thank you! I'll try to make similar calculations with the integral for $\mathbf E$. $$ $$ "...I've used the fact that there is no contribution to the potential by charge..." $$ $$ How to prove this? – user8817 Mar 2 '13 at 10:30
@Maxim_Ovchinnikov I was actually wrong about that. Thanks for checking my calculation closely. I edited it appropriately. Cheers! – joshphysics Mar 2 '13 at 16:07

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