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Unformatted text preview: Physics 505 Fall 2005 Homework Assignment #5 — Solutions Textbook problems: Ch. 3: 3.14, 3.26, 3.27 Ch. 4: 4.1 3.14 A line charge of length 2 d with a total charge Q has a linear charge density varying as ( d 2 z 2 ), where z is the distance from the midpoint. A grounded, conducting, spherical shell of inner radius b > d is centered at the midpoint of the line charge. a ) Find the potential everywhere inside the spherical shell as an expansion in Leg endre polynomials. We first ought to specify the charge density ρ ( ~x ) corresponding to the line charge. By symmetry, we place the line charge along the z axis. In this case, it is specified by cos θ = ± 1. As a slight subtlety, in order to get a uniform charge density in spherical coordinates, we need to divide out by r 2 . Hence for charge density varying as ( d 2 z 2 ) we end up with ρ ( ~x ) = ρ r 2 ( d 2 r 2 )[ δ (cos θ 1) + δ (cos θ + 1)] with the caveat that r < d . (This can be specified with a Heaviside step function Θ( d r ), but we will not bother with that.) The constant ρ is specified by evaluating the total charge Q = Z ρ ( ~x ) d 2 x = Z ρ r 2 ( d 2 r 2 )[ δ (cos θ 1) + δ (cos θ + 1)] r 2 dr dφd (cos θ ) = 2 π · 2 · ρ Z d ( d 2 r 2 ) dr = 8 π 3 ρ d 3 Thus ρ = 3 Q/ (8 πd 3 ). Since the spherical shell is grounded, the potential inside the shell is given by Φ( ~x ) = 1 4 π Z ρ ( ~x ) G ( ~x,~x ) d 3 x where G ( ~x,~x ) = X l,m 4 π 2 l + 1 r l < 1 r l +1 > r > l b 2 l +1 Y * lm ( θ ,φ ) Y lm ( θ, φ ) is the Dirichlet Green’s function inside a sphere of radius b . Because of spherical symmetry, we see that only m = 0 terms will contribute in the integral. This indicates that the expression for Φ( ~x ) reduces to one with ordinary Legendre polynomials Φ( ~x ) = 1 4 π ∞ X l =0 P l (cos θ ) Z ρ r 2 ( d 2 r 2 )[ δ (cos θ 1) + δ (cos θ + 1)] × r l < 1 r l +1 > r l > b 2 l +1 P l (cos θ ) r 2 dr dφ d (cos θ ) = 2 πρ 4 π ∞ X l =0 P l (cos θ )[ P l (1) + P l ( 1)] Z d ( d 2 r 2 ) r l < 1 r l +1 > r l > b 2 l +1 dr = ρ X l even I l ( r ) P l (cos θ ) where I l ( r ) = Z d ( d 2 r 2 ) r l < 1 r l +1 > r l > b 2 l +1 dr The reason only even values of l contribute is simply because the source is an even parity one. We are now left with evaluating the integral I l ( r ). There are two cases to consider. Case 1: r < d . This is the more involved computation, as the integral has to be divided into two segments I l ( r ) = 1 r l +1 r l b 2 l +1 Z r ( d 2 r 2 ) r l dr + r l Z d r ( d 2 r 2 ) 1 r l +1 r l b 2 l +1 dr = 1 r l +1 1 r b 2 l +1 d 2 l + 1 r 2 l + 3 r l +1 + r l " 1 r l d 2 l r 2 l 2 1 r l r b 2 l +1 d 2 l + 1 r 2 l + 3 # d r = d 2 2 l + 1 l ( l + 1) + 2 l ( l 2) r d l 2 ( l + 1)( l + 3) r d l d b l +1 r 2 2 l + 1 ( l 2)( l + 3) (1) Note that for either l = 0 or l = 2 we end up with a log I ( r ) = d 2 1 2 2 3 d b ln r d + 1 6 r 2 I 2 ( r ) = 5 6 d 2 r 2 7 10 + 2 15 d b 2 ln r d (2) Case 2: r > d .....
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 Fall '08
 Stephens,P
 Charge, Magnetism, Work, Cos, Boundary value problem, iL, charge density, Boundary conditions

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