I want to show the following equality
$$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\frac{\vec{p}}{3\epsilon_0}$$
where $\vec{p}$ is the dipole moment of a charge distribution $\rho\left(\vec{r}^{\prime}\right)$ with respect to the origin. The charge distribution is located inside a sphere with radius $R$.
I started with Gauss's Theorem:
$$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\int_{\left|\vec{r}\right|<R}d^3r\nabla\Phi\left(\vec{r}\right)=-\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{R}R^2 \Phi\left(\vec{r}\right)$$
Now I can use the Poisson integral:
$$\Phi\left(\vec{r}\right)=\frac{1}{4\pi\epsilon_0}\int d^3r^{\prime}\frac{\rho\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}$$
And thus:
$$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\frac{R}{4\pi\epsilon_0}\int d^3r^{\prime}\rho\left(\vec{r}^{\prime}\right)\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}$$
I know that the dipole moment is:
$$\vec{p}=\int d^3r^{\prime}\vec{r}^{\prime}\rho\left(\vec{r}^{\prime}\right)$$
Therefore, to obtain the top equation, the solid angle integration has to be:
$$\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}\overset{!?}{=}\frac{4\pi}{3R}\vec{r}^{\prime}$$
Does anyone know how to solve this integral?
Best Answer
We start with the integral $$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$ Since we are integrating over $\vec{r}$, we can without loss of generality, arrange for $\vec{r}'$to lie along the $+Z$ axis, so that $\vec{r}'=r'\hat{z}$. Then the angle between $\vec{r}$ and $\vec{r}'$ is the standard angle (in spherical coordinates) $\theta$ measured from the $+Z$ axis. Then we can write \begin{align} \mathrm{d}\Omega &= \sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\\ \vec{r} &= x\hat{x} + y\hat{y} + z\hat{z}\\ &=r\sin\theta\cos\varphi\hat{x}+r\sin\theta\sin\varphi\hat{y}+r\cos\theta\hat{z}\\ \mbox{and by the law of cosines}\\ |\vec{r}-\vec{r}'| &= \sqrt{r^{2}+r'^{2}-2rr'\cos{\theta}}\\ \end{align} So our integral has become: \begin{align} \oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|} &= \oint_{|\vec{r}|=R}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\frac{r\sin\theta\cos\varphi\hat{x}+r\sin\theta\sin\varphi\hat{y}+r\cos\theta\hat{z}}{\sqrt{r^{2}+r'^{2}-2rr'\cos{\theta}}} \end{align} But when doing the $\varphi$ integrals, the $\cos\varphi\,\hat{x}$ and $\sin\varphi\,\hat{y}$ terms go to zero, so all you are left with is $$\oint_{|\vec{r}|=R}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\frac{r\cos\theta\hat{z}}{\sqrt{r^{2}+r'^{2}-2rr'\cos{\theta}}}$$ We choose $u=-2rr'\cos{\theta}$, covert the integral into a standard form, and do it, keeping in mind while evaluating that that $r'<r=R$, so that $\sqrt{r^{2}+(r')^{2}-2rr'}=r-r'$, to finally get: $$\frac{4\pi r'\hat{z}}{3r}|_{|\vec{r}|=R} = \frac{4\pi r'\hat{z}}{3R}$$. But we had chosen $\vec{r}'$ to lie along the $+Z$ axis, so $r'\hat{z}=\vec{r}'$. So our final answer is: $$\frac{4\pi \vec{r}'}{3R}$$