FinalExam_EM_B_Jul_2007_Answers - Θ µα 1 (α) σa2 4π 0...

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Unformatted text preview: Θ µα 1 (α) σa2 4π 0 Φ(r, θ, φ) = 2 2π π/2 φ =0 θ =0 sin θ dθ dφ R(r, θ, φ, a, θ , φ ) R(r, θ, φ, a, θ , φ ) = [r + a − 2ra(cos θ cos θ + sin θ sin θ cos(φ0 − φ ))]1/2 (β) Φ(z) = 2 √ σa2 z 2 + a2 − |z − a| 20 az 2 √ σa z 2 + a2 − |z + a| 20 −az (γ) Φ(z = a) − Φ(z = 0) = z>0 z<0 σa √ [ 2 − 1] 20 Θ µα 2 (α) Φ1 = 2V −1 + σ1 /σ2 1 φ+V , π 1 + σ1 /σ2 1 + σ1 /σ2 Φ2 = 2V σ1 /σ2 φ, π 1 + σ1 /σ2 0<φ< π 2 <φ<π π 2 (β) R= π(1 + σ1 /σ2 ) 2σ1 d ln(a/b) Θ µα 3 (α) K = rT σωˆφ i (β) H= σω 4π 2π φ =0 2 rT a rT =0 2 rT ˆ (iφ × ˆR ) drT dφ , i R2 R = [r2 + − 2rrT sin θ cos(φ − φ )]1/2, ˆR = 1 [(r sin θ cos φ − rT cos φ )ˆx + (r sin θ sin φ − rT sin φ )ˆy + r cos θˆz ], i i i i R ˆφ = − sin φ ˆx + cos φ ˆy i i i (γ) i H(0, 0, z) = ˆz σω a2 + 2z 2 √ − 2|z| 2 z 2 + a2 (δ) H(0, 0, z) = ˆz i ρaω 2 π sin θ θ=0 a2 sin2 θ + 2(z − a cos θ)2 (z − a cos θ)2 + a2 − 2|z − a cos θ| dθ 2 sin θ Θ µα 4 Φ(rT , φ) = 4V0 π 4V 0 π ∞ 1 rT m a m=1,3,5,··· ∞ m sin mφ 0 < rT < a −m m 1 rT − b2m rT sin mφ a < rT < b m am (1 − (b/a)2m) m=1,3,5,··· ...
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This note was uploaded on 10/02/2009 for the course G 001 taught by Professor Shmmygr during the Spring '07 term at National Technical University of Athens, Athens.

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FinalExam_EM_B_Jul_2007_Answers - Θ µα 1 (α) σa2 4π 0...

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