Gauss' Law question

  • Thread starter EricVT
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  • #1
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Homework Statement



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.

Homework Equations



div E = p/eo

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

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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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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
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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
learningphysics
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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) + ...
you have the formula wrong here... check the divergence formula for spherical coordinates carefully....
 
  • #5
Dick
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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
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[tex] \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}[/tex]


This is what is written in the reference section of my text for the divergence in spherical coordinates. Is this wrong?
 
  • #7
Dick
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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
learningphysics
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Is the problem from the text?
 

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