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hw02_solutions - Solutions to Problem Set 2 1 We can use...

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Solutions to Problem Set 2 1. We can use the method of images to solve this problem on z-axis. The positive charges are located at the center, z = 0 (original charge) and at positions z = ± 4 an where n is an integer while the negative charges are at z = ± 2 a (2 n + 1). The potential is then Φ( x, y, z ) = q 4 π 0 ( 1 p x 2 + y 2 + z 2 + X n =1 1 p x 2 + y 2 + ( z + 4 na ) 2 + X n =1 1 p x 2 + y 2 + ( z - 4 na ) 2 - X n =1 1 p x 2 + y 2 + ( z + 2(2 n + 1) a ) 2 - X n =1 1 p x 2 + y 2 + ( z - 2(2 n + 1) a ) 2 ) which can be rewritten in a compact form as Φ( x, y, z ) = q 4 π 0 X n = -∞ ( - 1) n p x 2 + y 2 + ( z - 2 an ) 2 2. a) General expansion for z 6 = 0 (azimuthal symmetry): φ ( ρ, z ) = X k ( e kz + A k e - kz )( J 0 ( ) B k + N 0 ( ) C k ) Boundary conditions: φ ( ρ = a, z ) = 0; φ ( ρ, ±∞ ) = 0; φ ( ρ = 0 , z 6 = 0) finite. So φ ± ( ρ, z ) = X k n e k n z J 0 ( k n ρ ) B ± k n , φ ± φ ( z 0) where k n a is the n th zero of J 0 . B + k n = B - k n since φ is continuous at z = 0. - ∂φ + ∂z - ∂φ - ∂z z =0; ρ = 4 πσ ( ρ ) = 4 πqδ 2 ( ρ ) Z a 0 h 2 X k n k n J 0 ( k n ρ ) B k n i ρJ 0 ( k m ρ ) = 4 π Z a 0 σ ( ρ ) ρJ 0 ( k m ρ ) LHS = (2 k m B k m
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