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Unformatted text preview: Physics 505 Fall 2007 Homework Assignment #4 — Solutions Textbook problems: Ch. 3: 3.1, 3.2, 3.4, 3.7 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 cases b → ∞ , and a → 0. The general expansion in Legendre polynomials is of the form Φ( r,θ ) = X l [ A l r l + B l r l 1 ] P l (cos θ ) (1) Since we are working in the region between spheres, neither A l nor B l can be assumed to vanish. To solve for both A l and B l we will need to consider boundary conditions at r = a and r = b Φ( a,θ ) = X l [ A l a l + B l a l 1 ] P l (cos θ ) = n V cos θ ≥ cos θ < Φ( b,θ ) = X l [ A l b l + B l b l 1 ] P l (cos θ ) = cos θ > V cos θ ≤ Using orthogonality of the Legendre polynomials, we may write A l a l + B l a l 1 = 2 l + 1 2 V Z 1 P l ( x ) dx A l b l + B l b l 1 = 2 l + 1 2 V Z 1 P l ( x ) dx = 2 l + 1 2 V ( 1) l Z 1 P l ( x ) dx where in the last expression we used the fact that P l ( x ) = ( 1) l P l ( x ). Since the integral is only over half of the standard interval, it does not yield a particularly simple result. For now, we define N l = Z 1 P l ( x ) dx (2) As a result, we have the system of equations a l a l 1 b l b l 1 A l B l = 2 l + 1 2 V N l 1 ( 1) l which may be solved to give A l B l = 2 l + 1 2 V N l 1 b 2 l +1 a 2 l +1 ( 1) l b l +1 a l +1 ( ab ) l +1 ( b l + ( 1) l +1 a l ) Inserting this into (1) gives Φ( r,θ ) = 1 2 V X l (2 l + 1) N l 1 ( a b ) 2 l +1 " ( 1) l 1 + ( 1) l +1 a b l +1 r b l + 1 + ( 1) l +1 a b l a r l +1 # P l (cos θ ) (3) We now examine the integral (2). First note that for even l we may actually extend the region of integration N 2 j = Z 1 P 2 j ( x ) dx = 1 2 Z 1 1 P 2 j ( x ) dx = 1 2 Z 1 1 P ( x ) P 2 j ( x ) dx = δ j, This demonstrates that the only contribution from even l is for l = 0, correspond ing to the average potential. Using this fact, the potential (3) reduces to Φ( r,θ ) = V 2 + V 2 ∞ X j =1 (4 j 1) N 2 j 1 1 ( a b ) 4 j 1 " 1 + a b 2 j r b 2 j 1 + 1 + a b 2 j 1 a r 2 j # P 2 j 1 (cos θ ) Physically, once the average V/ 2 is removed, the remaining potential is odd under the flip z →  z or cos θ →  cos θ . This is why only odd Legendre polynomials may contribute. Note that an alternative method of solution would be to use linear superposition Φ = Φ inner + Φ outer where Φ inner is the solution where the inner sphere has potential V a ( θ ) and the outer sphere is grounded, and where Φ outer is the solution where the outer sphere has potential V b ( θ ) and the inner sphere is grounded. To calculate Φ inner...
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 Fall '08
 Stephens,P
 Electrostatics, Magnetism, Work, Spherical Harmonics, Cos, Electric charge, AL BL

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