# Gauss' Law question

1. Sep 20, 2007

### EricVT

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

An electric field in the region r > a is given by

E,r = 2*A*cos(theda)/r^3
E,theda = A*sin(theda)/r^3
E,phi = 0
A = constant

Find the volume charge density in this region.

E,r and E,theda and E,phi are the components of E in the r, theda, and phi directions, respectively.

2. Relevant equations

div E = p/eo

p = volume charge density
E = electric field
eo = permittivity of free space
div is divergence

3. The attempt at a solution

Well, I started by calculating the divergence of E in spherical coordinates and got that it is equal to zero. So p/eo = 0 ==> p = 0?

This isn't correct, so I'm confused on what I need to do. Does r > a have some significance? It gives no information about what sort of volume this is or what a is, and there are no diagrams or anything.

Thanks for any help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 20, 2007

### Dick

I tried calculating the divergence, and I didn't get zero. I won't guarantee I'm right, but it sure doesn't look like it can cancel to zero unless you are working at a particular point?

3. Sep 20, 2007

### EricVT

Well let me write out my work here and maybe I can catch my mistake.

div E = (1/r^2)*d(r^2*E,r)/dr + (1/[r*sin(theda)])*d(sin(theda)*E,theda))/d(theda) + ...

I won't write out the phi term because E,phi = zero so it is just zero.

div E = (1/r^2)*d(2Acos(theda)/r))/dr + (1/[r*sin(theda)])*d(Asin^2(theda)/r^3))/d(theda)

div E = (1/r^2)(-2*A*cos(theda)/r^2) + (1/[r*sin(theda)])(2Asin(theda)cos(theda)/r^3)

div E = -2Acos(theda)/r^4 + 2Acos(theda)/r^4 = 0

Hmm, I'm not catching my mistake if I did make one.

4. Sep 20, 2007

### learningphysics

you have the formula wrong here... check the divergence formula for spherical coordinates carefully....

5. Sep 20, 2007

### Dick

Ah, ha. Ok, so theta is the polar angle and phi is the equatorial one? I was taking it the other way around (I've seen it both ways). If so then I agree it is zero. But if you say that isn't correct, then are you sure you have them right?

6. Sep 20, 2007

### EricVT

$$\nabla \cdot A = \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 A_{r}) \ +\ \frac {1}{r sin \theta} \frac {\partial}{\partial \theta} (sin \theta A_{\theta}) \ +\ \frac {1}{r sin \theta} \frac {\partial A_{\phi}}{\partial \phi}$$

This is what is written in the reference section of my text for the divergence in spherical coordinates. Is this wrong?

7. Sep 20, 2007

### Dick

If that's from the same text as the problem, then we can assume the same conventions are being used. In that case, yes, the divergence and the local charge density are zero. Whatever is saying that isn't correct, isn't correct. I think the r>a part just means you can assume you are not at the origin where things become singular.

8. Sep 20, 2007

### learningphysics

Is the problem from the text?