This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 505 Fall 2005 Homework Assignment #4 Solutions Textbook problems: Ch. 3: 3.4, 3.6, 3.9, 3.10 3.4 The surface of a hollow conducting sphere of inner radius a is divided into an even number of equal segments by a set of planes; their common line of intersection is the z axis and they are distributed uniformly in the angle . (The segments are like the skin on wedges of an apple, or the earths surface between successive meridians of longitude.) The segments are kept at fixed potentials V , alternately. a ) Set up a series representation for the potential inside the sphere for the general case of 2 n segments, and carry the calculation of the coefficients in the series far enough to determine exactly which coefficients are different from zero. For the nonvanishing terms, exhibit the coefficients as an integral over cos . The general spherical harmonic expansion for the potential inside a sphere of radius a is ( r, , ) = X l,m lm r a l Y lm ( , ) where lm = Z V ( , ) Y * lm ( , ) d In this problem, V ( , ) = V is independent of , but depends on the azimuthal angle . It can in fact be thought of as a square wave in 2 V V n =4 This has a familiar Fourier expansion V ( ) = 4 V X k =0 1 2 k + 1 sin[(2 k + 1) n ] This is already enough to demonstrate that the m values in the spherical harmonic expansion can only take on the values (2 k +1) n . In terms of associated Legendre polynomials, the expansion coefficients are lm = s 2 l + 1 4 ( l m )! ( l + m )! Z 2 V ( ) e im d Z 1 1 P m l ( x ) dx = 4 V s 2 l + 1 4 ( l m )! ( l + m )! X k =0 1 2 k + 1 Z 2 sin[(2 k + 1) n ] e im d Z 1 1 P m l ( x ) dx = 4 iV s 2 l + 1 4 ( l m )! ( l + m )! X k =0 m, (2 k +1) n m, (2 k +1) n 2 k + 1 Z 1 1 P m l ( x ) dx Using P m l ( x ) = ( ) m [( l m )! / ( l + m )!] P m l ( x ), we may write the nonvanishing coefficients as l, (2 k +1) n = ( ) n +1 l, (2 k +1) n = 4 iV 2 k + 1 s 2 l + 1 4 ( l (2 k + 1) n )! ( l + (2 k + 1) n )! Z 1 1 P (2 k +1) n l ( x ) dx (1) for k = 0 , 1 , 2 , . . . . Since l (2 k + 1) n , we see that the first nonvanishing term enters at order l = n . Making note of the parity of associated Legendre polyno mials, P l m ( x ) = ( ) l + m P m l (...
View
Full
Document
This document was uploaded on 02/10/2010.
 Spring '09
 Magnetism, Work

Click to edit the document details