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

In summary, Vela found that the direct integration to solve for the electric field inside a sphere is quite difficult. She also notes that there is a singularity at the source of the field, which can be solved analytically without the use of Cauchy's principle.
  • #1
klondike
125
2

Homework Statement



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.

Homework Equations


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

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.pdfIn 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?

Thank you for your time.

P.S. This is not a homework question. I've done my degree 17 years ago.
 
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  • #2
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
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
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
Antiphon said:
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.

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

1. What is the formula for calculating the electric field inside a uniformly charged solid sphere?

The formula for calculating the electric field inside a uniformly charged solid sphere is given by:E = (kQr) / (R^3)

Where E is the electric field, k is the Coulomb's constant, Q is the charge of the sphere, r is the distance from the center of the sphere, and R is the radius of the sphere.

2. How do you determine the direction of the electric field inside a uniformly charged solid sphere?

The direction of the electric field inside a uniformly charged solid sphere is always radially inward, towards the center of the sphere. This is because the electric field lines always point from positive to negative charges.

3. Can the electric field inside a uniformly charged solid sphere be negative?

No, the electric field inside a uniformly charged solid sphere cannot be negative. This is because the electric field is always directed towards the center of the sphere, and the direction of the field is determined by the sign of the charge. A negative charge inside the sphere would create an electric field in the opposite direction, but this does not affect the field inside the sphere itself.

4. How does the electric field vary as you move closer to the center of the uniformly charged solid sphere?

The electric field inside a uniformly charged solid sphere decreases as you move closer to the center. This is because the amount of charge enclosed within a certain radius decreases as you move closer to the center, causing the electric field to decrease according to the inverse square law.

5. Can you use brute force to find the electric field inside a uniformly charged solid sphere?

Yes, you can use brute force to find the electric field inside a uniformly charged solid sphere by using the formula mentioned in the first question and plugging in values for the variables. However, this method may not always be the most efficient or accurate, and there are other methods, such as using Gauss's law, that can be used to find the electric field more easily.

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