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!

Electric Potential

  1. Oct 27, 2006 #1
    Griffith's EM Problem 3.26
    A sphere of radius R centered at the origin carries a cahrge density
    [tex] \rho(r,\theta) = k \frac{R}{r^2} (R - 2r) \sin \theta [/tex]

    where k is constant and r and theta are teh usual spherical coordinates. Find teh approximate potentail for the points on the z axis, far from the sphere

    i know ican do this with Laplace's equation but i wnana do it with the multipole expansion formula

    [tex] V(\vec r) = \frac{1}{2 \pi \epsilon_{0}} \sum_{n=0}^{\infty} \frac{1}{r^{n+1}} \int (r')^n P_{n} (\cos \theta') \rho (r') d\tau' [/tex]

    so about taht integral

    [tex] kR \int_{0}^{R} (r')^n \frac{R-2r'}{r'^2} r'^2 dr' \int_{0}^{2\pi} P_{n} (\cos \theta) \sin^2 \theta d\theta \int_{0}^{\pi} \phi [/tex]

    after some integration i found that the integral with respect to theta for
    n =2 is -pi/8 dipole term
    n = 4 = -pi/64 quadropole term

    for n = 1, 3, and 5 is zero
    so we end up with
    [tex] \pi kR \frac{1 \pi}{8} \int_{0}^{R} (r')^n \frac{R-2r'}{r^2} r^2 dr [/tex]

    [tex] \pi kR \frac{1 \pi}{8} \left[ \frac{-R(R)^{n+1}}{n+1} + \frac{2 R^{n+2}}{n+2} \right] [/tex]

    now looking at hte sum intself

    we consider only n = 2 since others are too small
    [tex]\frac{-1 \pi^2}{8} \frac{KR^{n+3}}{r^{n+1}} \left( \frac{1}{n+1} - \frac{2}{n+2} \right) [/tex]

    [tex] \frac{kR^5}{r^3} \pi^2 \frac{1}{48} [/tex]

    is this fine?

    should the answer include both dipole and quadropole terms?? Since this a sphere it makes sense to have that.
    thank you for your help!
     
    Last edited: Oct 27, 2006
  2. jcsd
  3. Oct 28, 2006 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    It did ask for an approximation, and the quadrapole term is way smaller than the dipole at large z, so I would leave it.

    (I did not check your calculations. I assume you did it OK)
     
  4. Oct 28, 2006 #3
    Your integration limits for [tex]\phi[/tex] and [tex]\theta[/tex] should be switched.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric Potential
  1. Electric Potential (Replies: 3)

  2. Electric potential (Replies: 1)

  3. Electrical Potential (Replies: 1)

  4. Electric Potential (Replies: 4)

Loading...