FinalExam_Feb_2006_Answers - Θ µα 1(α p p[x0 sin θ(z0...

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Unformatted text preview: Θ µα 1 (α) p p [x0 sin θ + (z0 − h) cos θ] + [−x0 sin θ + (z0 + h) cos θ] 4π 0 R3 4π 0 R 3 Φ(x0, z0 ) = oπoυ R = [x2 + (z0 − h)2 ]1/2 , 0 R = [x2 + (z0 + h)2 ]1/2 0 (β) σ(x) = 3ph(−x0 sin θ + h cos θ) 2π(x2 + h2 )5/2 (γ) T = −ˆy i p2 sin 2θ 64π 0 h3 Θ µα 2 (α) Φ1 (φ) =A1 φ + B1 (0 ≤ φ < φ1 ) Φ2 (φ) =A2 φ + B2 (φ1 ≤ φ < φ0 ) oπoυ A1 = 2U ( 2 − 1 )φ1 + 1 φ0 , B1 = 0, A2 = 1U ( 2 − 1 )φ1 + 1 φ0 , B2 = ( 2 − 1 )U . ( 2 − 1 )φ1 + 1 φ0 (β) E1 = − 1 i A1ˆφ , rT E2 = − 1 i A2ˆφ. rT (γ) ρb1 = 0, ρb2 = 0, σb (φ1 ) = (A1 − A2 ) 0 , rT σb (φ = 0+ ) = ( 1 − 0 )A1 , rT (δ) σ(φ = 0+ ) = − A1 1 , rT σ(φ = φ− ) = + 0 A2 2 . rT σb (φ− ) = − 0 ( 2 − 0 )A2 , rT Θ µα 3 (α) σ0 a( 1 + σ0 Φ2 (x, y) = a( 1 + Φ1 (x, y) = (β) σ0 1 + σ0 E2 (x, y) = − ( 1+ E1 (x, y) = − ( 2) 2) 2) 2) eax sin(ay) e−ax sin(ay) (−∞ < x ≤ 0) (0 ≤ x < +∞) i i eax [sin(ay)ˆx + cos(ay)ˆy ] i i e−ax [− sin(ay)ˆx + cos(ay)ˆy ] (γ) E(x, y) = −ˆy i σ0 ( 1+ 2) e−a|x| (−∞ < x ≤ 0) (0 ≤ x < +∞) (−∞ < x < +∞) Θ µα 4 (α) H= 1 4π 2π φ =0 a rT =b K(φ ) × ˆR (rT , φ ) i rT dφ drT R2 oπoυ ˆR = 1 [(rT 0 cos φ0 − rT cos φ )ˆx + (rT 0 sin φ0 − rT sin φ )ˆy + z0ˆz ] = axˆx + ayˆy + azˆz , i i i i i i i R 2 2 R = [rT 0 + rT2 − 2rT 0 rT cos(φ0 − φ ) + z0 ]1/2, K × ˆR = ˆx (az K0 cos φ ) + ˆy (az K0 sin φ ) + ˆz (−K0 sin φ ay − ax K0 cos φ ). i i i i (β) √ b a + a2 + z 2 ˆz K0 − √ a √ H(0, 0, z) = i +√ + ln 2 a2 + z 2 b2 + z 2 b + b2 + z 2 ...
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FinalExam_Feb_2006_Answers - Θ µα 1(α p p[x0 sin θ(z0...

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