1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    Good answer... :smile:

    Daniel.
     
  9. Jan 29, 2005 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member


    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

    User Avatar
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Dipole moment of non-uniform surface charged sphere
Loading...