Homework5

# Homework5 - Homework 5 Answers, 95.657 Electromagnetic...

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Homework 5 Answers, 95.657 Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Jackson 3.1 Two concentric spheres have radii a , b ( b > a ) and each is divided into two hemispheres by the same horizontal plane. The upper hemisphere of the inner sphere and the lower hemisphere of the outer sphere are maintained at potential V . The other hemispheres are at zero potential. Determine the potential in the region a r b as a series in Legendre polynomials. Include terms at least up to l = 4. Check your solution against known results in the limiting case b → ∞, and a → 0. SOLUTION: The geometry of the problem can be sketched as a cross-section of the spheres: 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: r = a = V 1 where V 1 = V if / 2 and V 1 = 0 if / 2 V V 0 0 x z a b

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V 1 = l A l a l B l a l 1 P l cos  Multiply both sides by P l ' cos  sin and integrate 0 V 1 P l ' cos  sin d = l A l a l B l a l 1 0 P l ' cos P l cos  sin d Use the orthogonality condition of Legendre polynomials: 0 P l ' cos  P l cos  sin d = 2 2 l 1 l ' l 0 V 1 P l ' cos  sin d = l A l a l B l a l 1 2 2 l 1 l ' l The Kronecker delta makes every term in the sum collapse to zero except when l = l ' 0 V 1 P l cos  sin d = A l a l B l a l 1 2 2 l 1 Flip the equation and plug in the explicit form of V 1 A l a l B l a l 1 = 2 l 1 2 V 0 / 2 P l cos  sin d Make a change of variables: x = cos , dx =− sin d A l a l B l a l 1 = 2 l 1 2 V 0 1 P l x dx Apply the boundary condition:  r = b = V 2 where V 2 = 0 if / 2 and V 2 = V if / 2 V 2 = l A l b l B l b l 1 P l cos
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## This note was uploaded on 03/21/2012 for the course PHY 2030 taught by Professor Avery during the Spring '11 term at University of Florida.

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Homework5 - Homework 5 Answers, 95.657 Electromagnetic...

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