In Ex. 3.9, we derived the exact potential for a spherical shell of radius R, which carries a surface charge σ = k cos θ.
(a) Calculate the dipole moment of this charge distribution.
(b) Find the approximate potential, at points far from the sphere, and compare the exact answer (Eq. 3.87). What can you conclude about the higher multipoles?
V(r, \theta)=\frac{k R^{3}}{3 \epsilon_{0}} \frac{1}{r^{2}} \cos \theta \quad(r \geq R) (3.87)
(a) By symmetry, p is clearly in the z direction: p =p \hat{ z } ; p=\int z \rho d \tau \Rightarrow \int z \sigma d a .
p=\int(R \cos \theta)(k \cos \theta) R^{2} \sin \theta d \theta d \phi=2 \pi R^{3} k \int_{0}^{\pi} \cos ^{2} \theta \sin \theta d \theta=\left.2 \pi R^{3} k\left(-\frac{\cos ^{3} \theta}{3}\right)\right|_{0} ^{\pi}
=\frac{2}{3} \pi R^{3} k[1-(-1)]=\frac{4 \pi R^{3} k}{3} ; p =\frac{4 \pi R^{3} k}{3} \hat{ z } .
(b) V \cong \frac{1}{4 \pi \epsilon_{0}} \frac{4 \pi R^{3} k}{3} \frac{\cos \theta}{r^{2}}=\frac{k R^{3}}{3 \epsilon_{0}} \frac{\cos \theta}{r^{2}} .
This is also the exact potential. Conclusion: all multiple moments of this distribution (except the dipole) are exactly zero.