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Continuing where this problem left off, I'd like to actually evaluate the triple integral, but I'm having difficulty with it. My suspicion is that in this form, common integral techniques (from Calculus I-III) aren't quite sufficient, and something else needs to be done. If that's the case, I don't know what to try.

Poisson's Equation:

$$\Delta u(r,\theta,\phi) = -\frac{q(r,\theta,\phi)}{\epsilon} \ \ \ \ 0<\theta<\pi,0<\phi<2\pi , 0<r<\infty \\ \cases{u(r,\theta,0) = u(r,\theta,2\pi) \\ u'(r,\theta,0) = u'(r,\theta,2\pi) \\ u(a,\theta,\phi) = u(\theta,\phi)}$$

From the solution of Laplace's equation, we utilize Green's second identity build the fundamental Green's function: $$G(r,0) = -\frac{1}{4\pi r}$$

The solution (incorporating the inhomogeneous part via the surface integral as well) is then: $$u(r,\theta,\phi) = \frac{1}{\epsilon}\bigg(\int_0^{2\pi}\int_0^\pi\int_{-\infty}^\infty G(r,r')q(r',\theta',\phi')r'^2\sin\theta' dr'd\theta'd\phi' + \oint\oint u(\theta,\phi)\nabla_{r'}G(r,r')\cdot d\textbf{S}'\bigg); d\textbf{S}' = \hat{\textbf{n}}r'^2\sin\theta' d\theta'd\phi'$$

We'll ignore the surface integrals and the integrals wrt $\theta$ and $\phi$ for now as these aren't what I'm stuck on (yet).

If our forcing term is at a specified point such that $q(\textbf{r}) = \delta(\textbf{r})= \frac{\delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')}{r^2\sin(\theta)} = \delta(r-1)$, the radial part of our integral $-\frac{1}{4\pi\epsilon}\int_{-\infty}^\infty \frac{\delta(r'-a)}{||r-r_0||}r'^2 dr'$ works just fine.

What I'd like to solve (or really just figure out how to integrate), is this uniformly charged solid ball or radius $r = 1$ example:

$$\tag{1}q(r,\theta,\phi) = \cases{\frac{2}{\epsilon}\ \ \ \ 0 < r-r_0 \le 1 \land 0 \le \theta \le \pi \land 0 \le \phi \le 2\pi \\ 0 \ \ \ \ \text{otherwise}}$$

Again, I'm only hung up on the radial integral (currently, will most certainly have issues when you I attempt the other integrals later on).

Researcher R
  • 807

1 Answers1

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Someone was kind enough to provide me the steps offsite.

First off, we can't actually integrate this problem, there needs to be some sort of workaround that involves a $\delta()$ function. This is something that has to be obtained via physical reasoning, and in the case of the electrostatic potential problem, the substitution is actually quite simple. To start with, we'll present the equation for total charge due to N charged particles.

$$Q = \sum_{n=1}^N \frac{q_n}{|r-r_n|}; q_n = \text{nth charge},r_n = \text{position of nth charge}$$

And the equation for the electric field is always: $$u(r) = -\frac{Q}{4\pi \epsilon} $$

We can reconfigure our sphere such that all the charges are moved to a single point at the origin of the ball and center the ball to $(x,y,z) = (0,0,0)$ and this is because the electrostatic field lines point outwardly normal from the origin of the sphere to the surface, so taking all the charges and collecting them to a single point would result in the exact same electric field generated from the surface. Another thing we can do is move the charges to the surface of the sphere so that $r_n = 1$. In doing this (and understanding that our problem has been set up such that all $q_n$ are the same), the total number of charges would then be $\rho_{vol}*(\frac{4}{3}\pi r_0^3) = \rho_{surf}*{4\pi r_0^2}k \rightarrow k = \frac{r_0}{3} \rightarrow Q_{surf} = \rho_{vol} * 4\pi r_0^2 \frac{r_0}{3}$ so the number of charges along our surface at $r_0=1$ is now the same as the total number of charges all throughout our sphere. The electrostatic potential should then be:

$$u(r) = -\frac{2}{3\epsilon|r-1|}$$

Which is the same as what we get from the full integral when $q(r_0)=\frac{2}{\epsilon}$ is distributed all along the surface of the sphere $\delta(r_0-1)$:

$$-\frac{1}{4\pi\epsilon}\int_0^{2\pi}\int_0^\pi \int_{-\infty}^\infty \frac{2\delta(r_0-1)}{||r-r_0||}*\Big(\frac{r_0}{3}\Big)r_0^2\sin\theta_0dr_0d\theta_0d\phi_0 \\ = -\frac{1}{4\pi\epsilon}\frac{4\pi}{3} * \bigg(\int_{-\infty}^{\infty}\frac{2\delta(r_0-1)}{||r-r_0||}r_0^3dr_0\bigg) \\ =-2\frac{1}{4\pi\epsilon} \frac{\rho_{vol}*4\pi(1)^2*\frac{(1)}{3}}{|r-(1)|} = -\frac{2}{3\epsilon|r-1|}$$

This answer is more of a work-in-progress to my own question, I didn't explicitly get the answer from the offline source, but I'll leave this here for now as I work to verify the result.

Researcher R
  • 807