Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: SEPARATION OF VARIABLES AND OTHER METHODSvariable solution of Laplace’s equation in spherical polars. It isu(r, θ, φ) =(Ar l + Br −(l+1) )(C cos mφ + D sin mφ)[EP m l (cos θ)+FQ m l (cos θ)],(21.49)where the three bracketted factors are connected only through the integer parametersl and m, 0≤|m| ≤l. As before, a general solution may be obtainedby superposing solutions of this form for the allowed values of the separationconstants l and m. As mentioned above, if the solution is required to be finite onthe polar axis then F = 0 for all l and m.◮An uncharged conducting sphere of radius a is placed at the origin in an initially uniformelectrostatic field E. Show that it behaves as an electric dipole.The uniform field, taken in the direction of the polar axis, has an electrostatic potentialu = −Ez = −Ercos θ,where u is arbitrarily taken as zero at z = 0. This satisfies Laplace’s equation ∇ 2 u =0,asmust the potential v when the sphere is present; for large r the asymptotic form of v muststill be −Er cos θ.Since the problem is clearly axially symmetric, we have immediately that m =0,andsincewerequirev to be finite on the polar axis we must have F = 0 in (21.49). Thereforethe solution must be of the form∞∑v(r, θ, φ) = (A l r l + B l r −(l+1) )P l (cos θ).l=0Now the cos θ-dependence of v for large r indicates that the (θ, φ)-dependence of v(r, θ, φ)is given by P1 0 (cos θ) =cosθ. Thus the r-dependence of v must also correspond to anl = 1 solution, and the most general such solution (outside the sphere, i.e. for r ≥ a) isv(r, θ, φ) =(A 1 r + B 1 r −2 )P 1 (cos θ).Theasymptoticformofv for large r immediately gives A 1 = −E and so yields the solution(v(r, θ, φ) = −Er + B )1cos θ.r 2Since the sphere is conducting, it is an equipotential region and so v must not depend onθ for r = a. This can only be the case if B 1 /a 2 = Ea, thus fixing B 1 . The final solution istherefore)v(r, θ, φ) =−Er(1 − a3cos θ.r 3Since a dipole of moment p gives rise to a potential p/(4πɛ 0 r 2 ), this result shows that thesphere behaves as a dipole of moment 4πɛ 0 a 3 E, because of the charge distribution inducedon its surface; see figure 21.6. ◭Often the boundary conditions are not so easily met, and it is necessary touse the mutual orthogonality of the associated Legendre functions (and thetrigonometric functions) to obtain the coefficients in the general solution.734