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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!

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!

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