stunner5000pt
Oct27-06, 04:56 PM
Griffith's EM Problem 3.26
A sphere of radius R centered at the origin carries a cahrge density
\rho(r,\theta) = k \frac{R}{r^2} (R - 2r) \sin \theta
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
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'
so about taht integral
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
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
\pi kR \frac{1 \pi}{8} \int_{0}^{R} (r')^n \frac{R-2r'}{r^2} r^2 dr
\pi kR \frac{1 \pi}{8} \left[ \frac{-R(R)^{n+1}}{n+1} + \frac{2 R^{n+2}}{n+2} \right]
now looking at hte sum intself
we consider only n = 2 since others are too small
\frac{-1 \pi^2}{8} \frac{KR^{n+3}}{r^{n+1}} \left( \frac{1}{n+1} - \frac{2}{n+2} \right)
\frac{kR^5}{r^3} \pi^2 \frac{1}{48}
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!
A sphere of radius R centered at the origin carries a cahrge density
\rho(r,\theta) = k \frac{R}{r^2} (R - 2r) \sin \theta
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
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'
so about taht integral
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
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
\pi kR \frac{1 \pi}{8} \int_{0}^{R} (r')^n \frac{R-2r'}{r^2} r^2 dr
\pi kR \frac{1 \pi}{8} \left[ \frac{-R(R)^{n+1}}{n+1} + \frac{2 R^{n+2}}{n+2} \right]
now looking at hte sum intself
we consider only n = 2 since others are too small
\frac{-1 \pi^2}{8} \frac{KR^{n+3}}{r^{n+1}} \left( \frac{1}{n+1} - \frac{2}{n+2} \right)
\frac{kR^5}{r^3} \pi^2 \frac{1}{48}
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!