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Homework Help: Dipole moment of non-uniform surface charged sphere

  1. Jan 27, 2005 #1
    I'm trying to find the dipole moment of a non-uniform surface charge distribution on a sphere of radius a:

    The surface charge distribution is:
    [tex] \sigma = \sigma_{0} cos \theta [/tex]
    where theta is the polar angle.

    Here is what I did:
    [tex] \vec{p} = \int\vec{r}\sigma dA [/tex]
    [tex] = \int r \sigma_{0} cos \theta (2\pi r dr d\theta) [/tex]

    and I'm thinking r should be integrated from 0 to a and theta integrated from -pi/2 to pi/2, but I'm not sure. Please help; thanks.
     
  2. jcsd
  3. Jan 27, 2005 #2

    dextercioby

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    Hold on.
    What is the surface element in spherical coordinates??

    Daniel.
     
  4. Jan 27, 2005 #3
    Ok I see. Should I use: [tex] r dr d\theta [/tex] ?
     
  5. Jan 27, 2005 #4
    dA is not [itex] 2 \pi r dr d \theta [/itex], how could the surface element depend on r ???

    try [tex] dA = 2 \pi R^2 sin \theta d \theta [/tex]
    and go do some reading on 3D integral
     
  6. Jan 27, 2005 #5
    I get 0 for the answer if I do it like you said:

    [tex] \int a \sigma_{0} cos\theta (2\pi a^2 sin\theta d\theta) [/tex]
    [tex] = 2\pi a^3 \sigma_{0} \int sin\theta cos\theta d\theta = 0 [/tex] theta ranges from -pi/2 to pi/2
     
  7. Jan 27, 2005 #6
    people measure theta from 0 to pi

    if your question define theta like you said, you will get zero anyway because the surface charge is symmetric....(save you some time to do the integral)
     
  8. Jan 27, 2005 #7

    dextercioby

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    Good answer... :smile:

    Daniel.
     
  9. Jan 29, 2005 #8

    ehild

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    You miss the z component of the position vector from your integrand and you have to integral from 0 to pi. Theta is the angle of the position vector with respect to the z axis.


    [tex] p_z=\int_{0}^{\pi}{\sigma* z* dA } = \int_{0}^{\pi}{(\sigma_0 \cos{\theta})*(acos{\theta})*(2\pi a^2 \sin{\theta}d\theta)}[/tex]

    ehild
     
  10. Jun 20, 2005 #9
    may you explain it. I didn't get why z component is here.
     
  11. Jun 21, 2005 #10

    ehild

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    Remember the definition of dipole moment.

    [tex] \vec{p} = \int {\sigma \vec{r} dA }[/tex]

    Because of symmetry, the dipole moment has got only z component here. The z component of the eq. above is:

    [tex] p_z = \int {\sigma z dA}[/tex]

    ehild
     
  12. Jun 21, 2005 #11
    ok, thank you.
     
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