Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PDES: SEPARATION OF VARIABLES AND OTHER METHODS

variable solution of Laplace’s equation in spherical polars. It is

u(r, θ, φ)=(Ar+Br−(+1))(Ccosmφ+Dsinmφ)[EPm(cosθ)+FQm(cosθ)],

(21.49)

where the three bracketted factors are connected only through theintegerpa-

rametersandm,0≤|m|≤. As before, a general solution may be obtained

by superposing solutions of this form for the allowed values of the separation

constantsandm. As mentioned above, if the solution is required to be finite on

the polar axis thenF= 0 for allandm.

An uncharged conducting sphere of radiusais placed at the origin in an initially uniform

electrostatic fieldE. Show that it behaves as an electric dipole.

The uniform field, taken in the direction of the polar axis, has an electrostatic potential

u=−Ez=−Ercosθ,

whereuis arbitrarily taken as zero atz= 0. This satisfies Laplace’s equation∇^2 u=0,as
must the potentialvwhen the sphere is present; for largerthe asymptotic form ofvmust
still be−Ercosθ.
Since the problem is clearly axially symmetric, we have immediately thatm=0,and
sincewerequirevto be finite on the polar axis we must haveF= 0 in (21.49). Therefore
the solution must be of the form

v(r, θ, φ)=

∑∞

=0

(Ar+Br−(+1))P(cosθ).

Now the cosθ-dependence ofvfor largerindicates that the (θ, φ)-dependence ofv(r, θ, φ)
is given byP 10 (cosθ)=cosθ. Thus ther-dependence ofvmust also correspond to an
= 1 solution, and the most general such solution (outside the sphere, i.e. forr≥a)is

v(r, θ, φ)=(A 1 r+B 1 r−^2 )P 1 (cosθ).

Theasymptoticformofvfor largerimmediately givesA 1 =−Eand so yields the solution

v(r, θ, φ)=

(

−Er+

B 1

r^2

)

cosθ.

Since the sphere is conducting, it is an equipotential region and sovmust not depend on
θforr=a. This can only be the case ifB 1 /a^2 =Ea, thus fixingB 1. The final solution is
therefore

v(r, θ, φ)=−Er

(

1 −

a^3

r^3

)

cosθ.

Since a dipole of momentpgives rise to a potentialp/(4π 0 r^2 ), this result shows that the
sphere behaves as a dipole of moment 4π 0 a^3 E, because of the charge distribution induced
on its surface; see figure 21.6.

Often the boundary conditions are not so easily met, and it is necessary to

use the mutual orthogonality of the associated Legendre functions (and the

trigonometric functions) to obtain the coefficients in the general solution.