hw02_solutions

Hw02_solutions - 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

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Unformatted text preview: 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 ( 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 ∞ 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 ( kρ ) B k + N ( kρ ) 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 ( k n ρ ) B ± k n , φ ± ≡ φ ( z ≷ 0) where k n a is the n th zero of J . B + k n = B- k n since φ is continuous at...
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This note was uploaded on 02/04/2012 for the course PHY 501 at Princeton.

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Hw02_solutions - 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

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