1. Mar 30, 2007

### FulhamFan3

I tried to do a post here but it didn't take so I'm going to put something much shorter that asks the same thing.

I tried to calculate the electric field produced by a sphere of uniform charge using this formula:

$$E(r)=\frac{1}{4\pi\epsilon_0} \iiint\rho(r') \frac{r-r'}{|r-r'|^3} d^3r'$$

I know I could could use gauss's law but I wanted to do it directly just to see. I tried to do it in cartesian coordinates. Well as you probably know the math gets super tedious for this problem. So what I did is I tried to simplify the problem in lower dimensions. I thought about doing a circle of uniform charge but instead I went for the one-dimensional case of a line charge on the x-axis from -R to R. I came up with this formula to solve it.

$$E(x)=\frac{\lambda}{4\pi\epsilon_0} \int_{-R}^{R} \frac{x-x'}{|x-x'|^3} dx'$$

After I solved this problem I found something unusual. As I approached R from the x>R side the value of E became infinity. When I know for a uniform sphere it becomes a finite value. Does this make sense to anybody?

2. Mar 30, 2007

### lalbatros

That's quite normal, by taking a plane geometry, you introduce a much large charge on the scene. The total charge will be infinite.

3. Mar 30, 2007

### ZapperZ

Staff Emeritus
I don't get it.

If all you wanted to do is prove to yourself that you could get the identical answer using Gauss's Law and Coulomb's Law, then why are you using cartesian coordinates and not spherical coordinates for the Coulomb's law method? What do cartesian coordinates accomplish that spherical coordinates do not? Just more pain and suffering? Then why not torture yourself even more and shift the origin to somewhere else other than the high symmetry location of the problem?

Zz.

4. Mar 30, 2007

### StatMechGuy

ZZ has a point. It's one thing to work in generalities, it's another to make the problem much more painful with no gain in generality.

5. Mar 30, 2007

### FulhamFan3

If I did it in polar coordinates then the $$\abs(r-r')$$ term becomes $$\sqrt(r^2+r'^2-2rr'cos(\theta-\theta'))$$. The limits become constants instead of functions but I wouldn't really say this is easier to integrate than $$\sqrt((x-x')^2+(y-y')^2+(z-z')^2)$$.

Regardless, the point wasn't really to integrate the first problem but more to understand why the one dimensional case goes to infinity. I have a mild obsession with being able to solve a problem in any coordinate system and I wasn't really asking how to solve it.

Say you had a point charge at (R,0,0). If you approach that charge from the right side you get an infinite E-field due to the inverse-square relation. However say you add charge until you get a sphere of radius R. The potential is then finite at (R,0,0). Why does the addition of positive charge make something infinite finite

6. Mar 30, 2007

### FulhamFan3

The total charge isn't infinite. I'm integrating from R to -R with a charge/length of $$\lambda$$. So the total charge is $$2R\lambda$$.

7. Mar 30, 2007

### ZapperZ

Staff Emeritus
You have simple encountered a mathematical artifact.

When you have a volume of charge, as r -> 0, the volume enclosed also goes to zero and you do not get into trouble.

But when we make an assumption of a "point charge" sitting at r=0, we no longer have the luxury of saying that the "enclosed volume" is zero and so, the value of the charge enclosed is also zero. This is because we have made an mathematical construction of something have no size having some charge sitting at r=0.

I still do not see solving in cartesian coordinates would give you any better insight into this than spherical coordinates.

Zz.

8. Mar 30, 2007

### ObsessiveMathsFreak

Today's Fun Math Fact

It turns out that if you have an n dimensional vector $$\mathbf{x}$$

Then the integral

$$\int_{\mathbf{x} \in \mathbb{R}^n} \frac{1}{(1+|\mathbf{x}|)^s} d\mathbf{x}$$

Is convergent if and only if s>n

And since $$\frac{1}{(1+|\mathbf{x}|)^s}<\frac{1}{(|\mathbf{x}|)^s}$$, this has implications for the "convergence" of $$\frac{1}{(|\mathbf{x}|)^s}$$.

I hope I remembered that correctly. In fact, I'm going to fish out my notes on this and write up a proof, because I think this is a useful theorem to know.

9. Mar 30, 2007

### FulhamFan3

I wasn't doing it for any insight. I was just doing it to do it. I know this is a science board and everything should have some sort of purpose but I did it for the pure absurdity of it. It's like saying why bike when you can drive there. Maybe I felt like riding a bike.

thanks mathfreak.