# Use brutal force to find E field inside a uniformly charged solid sphere

1. Jun 14, 2012

### klondike

1. The problem statement, all variables and given/known data

Use direct integration to find electric field inside a uniformly charged non-conducting solid sphere. The radius of the sphere is R, observing point is at a way from center of the sphere while a<R.

2. Relevant equations
Use Coulomb's law only. No Gauss law is allowed. You may use Gauss law to verify your result.

3. The attempt at a solution
Performing the triple integral is quite sweating. I managed to carry it out. My answer agrees with Gauss's law. I'm not completely satisfied with the way I do it.

My "brutal force" solution can be found at: http://gradovec.com/scratches/solid-sph.pdf

In the final integration with respect to r, I found that there is a jump discontinuity(also a pole) in the integrand at r=a. hence need to handle it carefully.

My questions are:
1) Did I handle the improper integral correctly?
2) Is there any textbook demonstrate how to carry out the direct integration? I can't find any in popular undergrad and graduate level textbook.. Why this is not popular?
3) I'm thinking of using dirac delta function, but don't see how to setup the integral. Can the integral be simplified by using elementary distribution theory? If yes, how?
4) Is there other simpler way to handle the singularity in the integrand?

P.S. This is not a homework question. I've done my degree 17 years ago.

Last edited: Jun 14, 2012
2. Jun 14, 2012

### vela

Staff Emeritus
I didn't spot any errors. As you noted, it's kind of painful to get the answer by direct integration, so everyone just uses Gauss's law instead. This is really a math problem, so if it's worked out anywhere in gory detail, it's probably in a mathematical methods book. I don't see how pedagogically useful it would be to include it in a electromagnetism book.

Another way to do the integral is to note that
$$\frac{a-r\cos\theta}{(r^2+a^2-2ar\cos\theta)^{3/2}} = -\frac{\partial}{\partial a}\frac{1}{\sqrt{r^2+a^2-2ar\cos\theta}}$$ so you get
$$-\int_0^R \frac{\partial}{\partial a}\int_{-1}^{1} \frac{1}{\sqrt{r^2+a^2-2ar\cos\theta}}r^2 \,d(\cos\theta)\,dr.$$ Expand the integrand in terms of the Legendre polynomials $P_l(\cos\theta)$. You have to treat the case for r<a and r>a separately. In either case, only the $l=0$ term will survive the integration over the angle. (Sorry, I used the physics convention: my $\theta$ is your $\phi$.)

3. Jun 15, 2012

### klondike

Hi Vela,

Thank you for checking my result and reminding me the multipole expansion method.

The reason I did such exercise is I was having trouble writing a computer program to convolute a given dyadic Green's function with a source distribution to find the field at source. The fundamental problem is handling the singularity at source in numerical integration. People in the last few decades spend a lot of raw intelligence to tackle this problem. I read quite a few papers but haven’t fully understood their mechanics detail enough to get my algorithm converged.

I took a step back to see how simple "singularity at source" is resolved analytically since I remembered some field at source problem can be found by direct integration. After exercised and reviewed some of the classical textbook examples and homework problems, I found that most them have the source singularity as a _removable_ discontinuity in the integrand or in the final anti-derivative, _and_ the integral “accidentally” converges which agrees with physical phenomenon. There is no need to explicitly invoke Cauchy PV limiting process.

Yep, this indeed more of a math problem.

4. Jun 16, 2012

### Antiphon

P.S. Brute force is used to describe a solution which is not elegant or efficient but relies on sheer "strength" to arrive at the result.

Brutal force is the application of physical strength in an inhumane way such as hitting someone until they agree to work out the solution to a physics problem.

5. Jun 16, 2012

### klondike

Exactly what my dad did to me when I was a child