How to compute the volume integral for the potential of an arbitrary point outside a uniformly charged ball?

Question

$$\frac{\rho}{4\pi\epsilon_0}\iiint_{D}^{}\frac{1}{\left\| \mathbf{r}-\mathbf{r'} \right \| }dV'$$

• $D$ is a ball of radius $R$
• $\mathbf{r}$ is the position vector of the point where we want to calculate the potential
• $\mathbf{r'}$ is the position vector of the parametrisation of the ball

When $$\mathbf{r}=z\hat{\bf{z}}$$ $$ \Rightarrow \left\| \mathbf{r}-\mathbf{r'} \right \|^2=z^2+r'^2-2r'z\cos\vartheta'$$ the integral can be computed using spherical coordinates since

$$\int\frac{r'^2\sin\vartheta'}{\sqrt{r'^2+z^2-2r'z\cos\vartheta'}}d\vartheta'=\frac{r'\sqrt{r'^2+z^2-2r'z\cos\vartheta'}}{z}$$

David J. Griffiths on Introduction to Electrodynamics states that when $$\mathbf{r}=x\hat{\bf{x}}+y\hat{\bf{y}}+z\hat{\bf{z}}$$ $$ \Rightarrow \left\| \mathbf{r}-\mathbf{r'} \right \|^2=r'^2+x^2+y^2+z^2-2r'(x\sin\vartheta' \cos\varphi' + y\sin\vartheta' \sin\varphi' + z\cos\vartheta')$$ the integral should have the same value - by symmetry - since we can rotate the system and align the point with the $z$ axis but I cannot figure a way to compute it as is or make a suitable substitution (maybe in terms of three-dimensional rotation).

$$\int\frac{r'^2\sin\vartheta'}{\sqrt{r'^2+x^2+y^2+z^2-2r'(x\sin\vartheta' \cos\varphi' + y\sin\vartheta' \sin\varphi' + z\cos\vartheta')}}d\vartheta'=\ ??$$

Answer 1

$\textbf{Hint:}$ The easiest way to do this would be change your coordinate system to shift the center of the sphere from the origin to $\mathbf{r}$. This gives a new integral

$$\frac{\rho}{4\pi\epsilon_0}\iiint_{B(\mathbf{r},R)}\frac{dV'}{||\mathbf{r}'||}$$

Then there are two cases, for when $|\mathbf{r}|\equiv r$ is greater than or less than $R$. To get you started on the greater than case, convert the equations to spherical coordinates

$$||\mathbf{r}||^2-2\mathbf{r}\cdot\mathbf{r}'+||\mathbf{r}'||^2=R^2$$

$$\implies r^2-R^2+r'^2=2rr'\cos\theta'$$

by choosing to align $\mathbf{r}$ along the $z'$ axis which we are free to do. The easiest way to set up the integral is with $d\theta'$ first

$$\frac{\rho}{4\pi\epsilon_0}\int_0^{2\pi}\int_{r-R}^{r+R}\int_{0}^{\cos^{-1}\left(\frac{r^2-R^2+r'^2}{2rr'}\right)}r'\sin\theta'\:d\theta' \:dr'\:d\varphi'$$

$$= \frac{\rho}{2\epsilon_0}\int_{r-R}^{r+R}r'-\frac{(r^2-R^2)}{2r}-\frac{r'^2}{2r}\:dr'$$

Can you finish this computation, and compute the limits for the $r<R$ case?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{{\rho \over 4\pi\epsilon_{0}}\iiint_{D}{1 \over \verts{\vec{r} - \vec{r}\,'}}\dd V'\ \color{black}{\mbox{where}}\ D\ \color{black}{\mbox{is a ball of radius}}\ R}:\ {\LARGE ?} \end{align} Because $\ds{\vec{r}}$ is a $\ds{given\ vector}$, I can choose $\ds{\vec{r}}$ along the $\ds{z\mbox{-}axis}$. At the end of the whole calculation, I can restore the general characteristic of $\ds{\vec{r}}$. Namely, \begin{align} & \color{#44f}{{\rho \over 4\pi\epsilon_{0}}\iiint_{D}{1 \over \verts{\vec{r} - \vec{r}\,'}}\dd V'} = {\rho \over 4\pi\epsilon_{0}}\iiint_{D}{\,r'^{2}\, \dd r'\ \dd\Omega_{\,\vec{\,r}\,'} \over \root{r_{>}^{2} -2r_{>}r_{<}\cos\pars{\theta\,'} + r_{<}^{2}}} \\[2mm] & \ \mbox{where}\quad {r_{<} \atop r_{>}} = {\min \atop \max}\braces{r,r'}\mbox{. Therefore,} \\[5mm] & \color{#44f}{{\rho \over 4\pi\epsilon_{0}}\iiint_{D}{1 \over \verts{\vec{r} - \vec{r}\,'}}\dd V'} \\[5mm] = & \ {\rho \over 4\pi\epsilon_{0}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}{\,r'^{2}\sin\pars{\theta'}\,\dd r'\,\dd\theta\,'\dd\phi' \over r_{>}\root{1 -2\pars{r_{<}/r_{>}}\cos\pars{\theta\,'} + \pars{r_{<}/r_{>}}^{2}}} \\[5mm] = & \ {\rho \over 4\pi\epsilon_{0}}\, 2\pi\int_{0}^{R}r'^{2}\int_{-1}^{1}{\,\,\dd x\, \over r_{>}\root{1 -2\pars{r_{<}/r_{>}}x + \pars{r_{<}/r_{>}}^{2}}} \dd r' \\[5mm] = & \ {\rho \over 2\epsilon_{0}} \int_{0}^{R}r'^{2}\int_{-1}^{1} {1 \over r_{>}}\sum_{\ell = 0}^{\infty}\pars{r_{<} \over r_{>}}^{\ell}\on{P}_{\ell\,}\pars{x}\,\dd x\,\dd r' \end{align} $\ds{\on{P}_{\ell}}$ is a $\ds{Legendre\ Polynomial}$. Then, \begin{align} & \color{#44f}{{\rho \over 4\pi\epsilon_{0}}\iiint_{D}{1 \over \verts{\vec{r} - \vec{r}\,'}}\dd V'} \\ = & \ {\rho \over 2\epsilon_{0}}\int_{0}^{R}r'^{2} {1 \over r_{>}}\sum_{\ell = 0}^{\infty} \pars{r_{<} \over r_{>}}^{\ell}\ \overbrace{\int_{-1}^{1}\on{P}_{\ell\,}\pars{x}\,\dd x} ^{\ds{2\delta_{\ell 0}}}\ \,\dd r' \\[5mm] = & \ {\rho \over \epsilon_{0}}\int_{0}^{R}r'^{2}\ {1 \over r_{>}}\,\dd r' \\[5mm] = & \ \left\{\begin{array}{rclc} \ds{{\rho \over \epsilon_{0}}\pars{% \int_{0}^{r}{r'^{2} \over r}\dd r' + \int_{r}^{R}{r'^{2} \over r'}\dd r'}} & \ds{=} & \ds{\color{#44f}{{\rho \over 6\epsilon_{0}}\pars{3R^{2} - r^{2}}}} & \mbox{if} & \ds{r < R} \\[2mm] \ds{{\rho \over \epsilon_{0}} \int_{0}^{R}{r'^{2} \over r}\dd r'} & \ds{=} & \ds{\color{#44f}{{\rho \over 3\epsilon_{0}}{R^{3} \over r}}} & \mbox{if} & \ds{r > R} \end{array}\right. \end{align}

Answer 3

For an independent check, the integral formulas are solutions to the following problem: find the spherically symmetric, diffenrtiably continuous solution of $$\mathbf d \star \mathbf d \Phi =\cases{0 \quad |\mathbf x|\gt R\\ \rho \quad\ |\mathbf x| \le R}$$ for a piecewise continuous charge distribution.

For a function of radial coordinate $r$, independent of the spherical angles, the metric Laplacian $\Delta$-term has the form $$\mathbf \Delta=\mathbf d \star \mathbf d = \frac{1}{\text{dvol}} \nabla_r^+ \ \text{dvol} \ \nabla_r = \frac{1}{h_r h_\theta h_ \phi}\partial_r \frac{1}{h_r}\ \ h_r h_\theta h_\phi \frac{1}{h_r}\partial_r$$ with the three metric scale factors $h_k dx_k$ for the coordinate differentials. Only the two $r$ scale factors of the angles $r d \theta,r \sin \theta d \phi$ survive:

$$\frac{1}{r^2} \partial_r \left( r^2 `\Phi'(r)\right)=\cases{0,\quad r\gt R\\ \rho , \quad\ r\le R}$$

The general external solution is $$r^2 \Phi'(r) = c'_1, \quad \Phi(r)= \frac{c_1}{r}+ c_2$$

The value $c_2$ at $r=\infty$ can be set to zero (gauge invariance).

The internal solution is

$$\frac{1}{r^2}\ (r^2 \Phi')' = \rho, \quad (r^2 \Phi')' =\rho \ r^2, \quad r^2 \Phi' = \frac{\rho \ r^3}{3} + c'_3, \quad \Phi = \frac{\rho r^2}{6} +\frac{c_3}{r}+c_4$$

In the internal solution $c_3$ represents a finite point charge at $r=0$, can be set to zero

At $r=R$ both solutions have to be coincide differentiably continuously (continuity of the normal electric field in absence of a surface charge distribution)

$$c_4+\frac{\rho R^2}{6}=\frac{c_1}{R} , \quad \frac{\rho R}{3}=-\frac{c_1}{R^2}$$

$$c_1 = -\frac{1}{3} \ \rho \ R^3 ,\quad c_4 =-\frac{1}{2} \ \rho \ R^2$$

$$\Phi(r)=\cases{ \rho \ \left(\frac{r^2}{6} - \frac{R^2}{2}\right) \quad r<R \\ -\rho \frac{R^3}{3}}$$