Classical Electrodynamics · Problem 3.8

Multipoles, Electrostatics of Macroscopic Media, Dielectrics — John David Jackson

Physics — E&M · legendre-polynomials-spherical-harmonics

Problem

There is a puzzling aspect of the solution (3.136) for the potential inside a grounded sphere with a uniformly charged wire along a diameter. Very close to the wire (i.e., for ρ=rsin⁡θ≪b\rho = r\sin\theta \ll b), the potential should be that of a uniformly charged wire, namely, Φ=(Q/4πϵ0b)ln⁡(b/ρ)+Φ0\Phi = (Q/4\pi\epsilon_0 b)\ln(b/\rho) + \Phi_0. The solution (3.136) does not explicitly have this behavior. (a) Show by use of the Legendre differential equation (3.10) and some integration by parts, that ln⁡(csc⁡θ)\ln(\csc\theta) has the appropriate expansion in spherical harmonics to permit the solution (3.136) to be written in the alternative form, Φ(x)=Q4πϵ0b{ln⁡(2brsin⁡θ)−1−∑j=1∞4j+12j(2j+1)(rb)2jP2j(cos⁡θ)}\Phi(\mathbf{x}) = \frac{Q}{4\pi\epsilon_0 b}\left\{\ln\left(\frac{2b}{r\sin\theta}\right) - 1 - \sum_{j=1}^{\infty} \frac{4j+1}{2j(2j+1)}\left(\frac{r}{b}\right)^{2j} P_{2j}(\cos\theta)\right\} in which the expected behavior near the wire is manifest. Give an interpretation of the constant term Φ0=−Q/4πϵ0b\Phi_0 = -Q/4\pi\epsilon_0 b. Note that in this form, for any r/b<1r/b < 1 the Legendre polynomial series is rapidly convergent at all angles. (b) Show by use of the expansion (3.38) that 12(1sin⁡θ/2+1cos⁡θ/2)=2∑j=0∞P2j(cos⁡θ)\frac{1}{2}\left(\frac{1}{\sin\theta/2} + \frac{1}{\cos\theta/2}\right) = 2\sum_{j=0}^{\infty} P_{2j}(\cos\theta) and that therefore the charge density on the inner surface of the sphere, Eq. (3.137), can be expressed alternatively as σ(θ)=−Q4πb2{12(1sin⁡θ/2+1cos⁡θ/2)−∑j=0∞12j+1P2j(cos⁡θ)}\sigma(\theta) = -\frac{Q}{4\pi b^2}\left\{\frac{1}{2}\left(\frac{1}{\sin\theta/2} + \frac{1}{\cos\theta/2}\right) - \sum_{j=0}^{\infty} \frac{1}{2j+1}P_{2j}(\cos\theta)\right\} The (integrable) singular behavior at θ=0\theta = 0 and θ=π\theta = \pi is now exhibited explicitly. The series provides corrections in ln⁡(1/θ)\ln(1/\theta) as θ→0\theta \rightarrow 0.

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