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Homework5 - Homework 5 Answers 95.657 Fall 2011...

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Homework 5 Answers, 95.657 Fall 2011, Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Problem 1 Two concentric spheres are centered at the origin, the inner sphere with radius a and the outer sphere with radius b . The outer sphere is maintained at a potential of zero, and the inner sphere is maintained at a potential of V cos θ where θ is the polar angle of spherical coordinates. Find the potential everywhere between the spheres. SOLUTION: Because of the spherical geometry of the problem and the absence of charge, we choose to solve the Laplace equation in spherical coordinates: 2 = 0 1 r 2 r 2 r  1 r 2 sin ∂ sin ∂  ∂ 1 r 2 sin 2 2 ∂ 2 = 0 Using the method of separation of variables leads to the general solution:  r , , = l A l r l B l r l 1  A m = 0 B m = 0  P l m = 0 cos  m 0, l A l r l B l r l 1  A m e i m B m e im P l m cos  The boundaries in this problem are all azimuthally symmetric, so that the solution for the electric potential will not be a function of . The only way to make the general solution independent of is to set m = 0 and B m =0 =0.  r , , = l A l r l B l r l 1 P l m = 0 cos  In other problems where we must have a valid solution at the origin, B l must be zero to keep the solution from blowing up at the origin. But in this problem, we do not seek a valid solution at the origin, so we cannot use this restriction. Apply the boundary condition: Φ( r = b )= 0 0 = l ( A l b l + B l b l 1 ) P l ( cos θ) This must be true for all polar angles, so that the terms in the parentheses must be zero, leading to: B l =− A l b 2 l + 1 The solution now becomes: Φ( r , θ , ϕ)= l A l [ ( b r ) l + 1 ( r b ) l ] P l ( cos θ)
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Apply the boundary condition Φ( r = a )= V cos θ V cos θ= l A l [ ( b a ) l + 1 ( a b ) l ] P l m = 0 ( cos θ) The usual approach would be now to multiply both sides by a Legendre polynomial and integrate over all polar angles, then use the orthogonality of Legendre polynomial to pick out one term in the series sum, then solve for the A l . But this problem is simple enough that we recognize the fact that: P 1 cos = cos therefore the potential on the outer sphere follows the functional dependence of the l =1 Legendre polynomial. We don't need any other terms. V cos θ= A 1 [ ( b a ) 2 ( a b ) ] cos θ A 1 = V [ ( b a ) 2 ( a b ) ] 1 and A l = 0 for all other l The final solution then becomes: Φ( r , θ , ϕ)= ( b r ) 2 ( r b ) ( b a ) 2 ( a b ) V cos θ Problem 2 A spherical surface of radius R has charge uniformly distributed over its surface with a density Q /4 πR 2 , except for a spherical cap at the north pole, defined by the cone θ = α .
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