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Homework Help: Dipole moment of sphere with uniform volume charge Q

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine the dipole moment, [tex]\mathbf{p}[/tex], of a sphere of radius R with a uniform volume charge, total Q, with respect to its center.


    2. Relevant equations
    [tex]\mathbf{p}=\int \mathbf{r} \rho(\mathbf{r}) d\tau[/tex]


    3. The attempt at a solution
    I know that [tex]\mathbf{p}=\mathbf{0}[/tex], but I have a hard time finding a rigorous argument to prove it. Looking at the definition of [tex]\mathbf{p}[/tex] given above, all I can see is that [tex]\rho(\mathbf{r})[/tex] is in fact constant for r<R, but this doesn't seem to get me anywhere. Other than "it's not a dipole", I'm stuck. Can anyone point me in the right direction?
     
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  3. Jun 14, 2010 #2

    phyzguy

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    It's zero because rho is even about the origin ( i.e. rho(-r) = rho(r)), while the components of r are odd (i.e.z(-r) = -z(r)), so when you carry out the integral there are equal parts + and -, so it all cancels. If you don't believe it, try carrying out the integral for one of the components, say z. Can you write the integral for Pz in a form that you can evaluate it?
     
  4. Jun 14, 2010 #3
    I think so, but I'm not sure how to argue from one dimension to three dimensions. I do get the explanation intuitively, though.

    I think the integral you are suggesting is as simple as:

    [tex]p_z=\int_{-R}^{R} z\rho dz=\rho(\frac{1}{2}R^2-\frac{1}{2} R^2)=0[/tex]

    Is this correct?
     
  5. Jun 14, 2010 #4

    phyzguy

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    No, not really. You still have to integrate over the whole volume of the sphere, so it's really:
    [tex]p_z = \int_0^{2\pi}\int_0^{\pi}\int_0^Rz r^2 sin(\theta)dr d\theta d\phi[/tex]
    Since z = r cos(theta), this can be written:
    [tex]p_z = \int_0^{2\pi}\int_0^{\pi}\int_0^R r^3 cos(\theta)sin(\theta)dr d\theta d\phi[/tex]
    Can you evaluate this?
     
  6. Jun 14, 2010 #5
    Aha, yes of course! I was having a problem just replacing r for z in the triple integral, so I thought you meant the other way, even though I couldn't find my way from there. I now see that by superposition, I just evaluate the integral for each of r's components, then sum them.

    So the idea is that evaluating one of those triple integrals and getting zero really shows that it will be zero for the other components as well, by symmetry. Thank you!
     
  7. Jun 14, 2010 #6

    phyzguy

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    Yes, or you could explicitly evaluate the other components as well (x = r sin(theta)cos(phi), y = r sin(theta)sin(phi)), and show that they are zero.
     
  8. Jun 14, 2010 #7
    Thanks, I feel silly for even asking the question now, it just seems obvious. ^_^
     
  9. Jun 14, 2010 #8

    phyzguy

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    Glad to help. As they say, "The only stupid question is the one that is not asked."
     
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