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Calculating the divergence of r(arrow)/r^a and finding charge density

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    This is a three part problem. My first task is to calculate the divergence of [tex]\vec{r}/r^{a}[/tex]. Next, I am to calculate its curl. Then I'm supposed to find the charge density that would produce the field

    [tex]\vec{E}=\frac{q\vec{r}}{4\pi\epsilon_{0}r^{a}}[/tex]

    3. The attempt at a solution

    I calculated the curl by first calculating the surface integral through a sphere of r/r^a:

    [tex]\oint\frac{\vec{r}}{r^{a}}r^{2}d\phi sin\theta d\theta \hat{r}=\frac{4\pi\left|\vec{r}\right|}{r^{a-2}}[/tex]

    By the divergence theorem [tex]\frac{4\pi\left|\vec{r}\right|}{r^{a-2}}=\int_{v}\nabla\bullet\frac{\vec{r}}{r^{a}}d\tau=\int_{v}\frac{4\pi\left|\vec{r}\right|}{r^{a-2}}\delta^{3}(\vec{r})d\tau[/tex]

    Which implies that the divergence of r/r^a is [tex]\frac{4\pi\delta^{3}(\vec{r})}{r^{a-3}}[/tex].

    This would eventually give me the charge density as [tex]\frac{q\delta^{3}(r)}{r^{a-3}}[/tex]

    Does this look correct? Any help would be appreciated. Thanks in advance.
     
  2. jcsd
  3. Sep 13, 2009 #2

    kuruman

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    Look at the expression of the curl in spherical coordinates. What happens if you calculate the curl of a field that has only a radial component?

    I don't see how you get what you get "by the divergence theorem". Use the divergence in spherical coordinates to take the divergence of your field correctly. You don't get a Dirac delta function for all values of a.
     
  4. Sep 13, 2009 #3
    Sorry, I didn't post that earlier. I calculated the curl of the field to be 0.

    As for the dirac-delta function, I only learned about that a few days ago and I have a very un-intuitive, elementary understanding of it. That said, I don't see why the dirac-delta function isn't applicable for all values of 'a' since the divergence outside of the origin should be 0 for a field directed radially outward. I didn't have any test to check my answer except plugging 3 into the value for 'a', which reduces to the normal electric field, and my answer checks out at that point.

    According to my understanding of the divergence theorem the surface integral of a field through a surface, which I chose to be a sphere, become the volume integral of the divergence of that field. In the following equation:

    [tex]\int_{v}\nabla\bullet\frac{\vec{r}}{r^{a}}d\tau=\int_{v}\frac{4\pi\left|\vec{r}\right|}{r^{a-2}}\delta^{3}(\vec{r})d\tau[/tex]

    The integral on the right hand side when evaluated has the same value as the surface integral for my field. Since the volumes of both integrals are the same I reasoned that the integrands must be the same.

    Could you please point me in the right direction as to what I may be doing wrong.

    Thanks for the response.
     
    Last edited: Sep 13, 2009
  5. Sep 13, 2009 #4

    kuruman

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    The divergence theorem is not what you say it is. It relates a volume integral to a surface integral. Please look it up.

    Also look up how to express the divergence in spherical coordinates. In spherical coordinates the radial piece of the divergence is not just the derivative with respect to r. So look it up and take the correct divergence. Then you will see that if a = 2, you get something that is zero everywhere except at r = 0. That's where the Dirac delta function comes in.
     
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