homework_5_solution

homework_5_solution - x = −a 2 x= a 2 a 0(x < − ) 2...

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Unformatted text preview: x = −a 2 x= a 2 a 0(x < − ) 2 ￿ = σ ex (− a < x < a ) E ￿ ￿0 2 2 a 0(x > ) 2 a 0(x < − ) 2 σ a a Φ = x( − < x < ) ￿0 2 2 a 0(x > ) 2 ∆Φ = σa 2￿0 ∂2Φ 1 ∂Φ 1 ∂2Φ ∂2Φ + +2 + =0 ∂ρ2 ρ ∂ρ ρ ∂φ2 ∂ z2 Φ(ρ, φ, z ) = R(ρ)Q(φ)Z (z ) d2 Z − k2 z = 0 dz 2 d2 Q + ν2Q = 0 dφ 2 d2 R 1 dR ν2 + + (k 2 − 2 )R = 0 dρ2 ρ dρ ρ Φ(z )|z=0 = 0 Q( φ ) = A Z (z ) = sinh kz R(ρ) = CJ0 (k ρ) + DN0 (k ρ) D=0 ρ=0 ρ=a k kn = xn xn a J0 ( x) Φ(ρ, φ, z ) = +∞ ￿ Cn J0 (kn ρ) sinh kn z n=1 V= +∞ ￿ z=L Cn J0 (kn ρ) sinh kn L n=1 Cn = csc kn L · V · 2π 2 π a2 J1 ( k n a ) ˆ 0 a dρρJ0 (kn ρ) d [xJ1 (x)] = xJ0 (x) dx Cn = csc kn L kn aJ1 (kn a) 2 csc kn L · V · 2π =V · 2 2 π a2 J1 ( kn a) kn akn J1 (kn a) ...
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This note was uploaded on 12/29/2011 for the course PHYSICS 606 taught by Professor Bedaque during the Spring '11 term at Maryland.

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