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# FinalExam_Jan_2005_a_Answers - Θ µα 1(α Φ(x0 y0 z0 =...

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Θ μα 1 ( α ) Φ( x 0 , y 0 , z 0 ) = σ 0 4 π 0 a 2 + a - a + a - a | x y | dx dy [( x 0 - x ) 2 + ( y 0 - y ) 2 + ( z 0 - z ) 2 ] 1 / 2 ( β ) Φ(0 , 0 , z 0 ) = σ 0 3 π 0 a 2 [(2 a 2 + z 2 0 ) 3 / 2 - 2( a 2 + z 2 0 ) 3 / 2 + ( z 2 0 ) 3 / 2 ] ( γ ) Φ(0 , 0 , 0) = 2 σ 0 a 3 π 0 [ 2 - 1] ( δ ) E (0 , 0 , z ) = σ 0 π 0 a 2 z [2( a 2 + z 2 0 ) 1 / 2 - (2 a 2 + z 2 0 ) 1 / 2 - | z | ] ˆ i z ( ) Φ(0 , 0 , z 0 a ) σ 0 a 2 4 π 0 | z 0 | Θ μα 2 ( α ) P = ( - ρ 0 r T 2 ) ˆ i r , σ b = - ρ 0 a/ 2 ( β ) D = ˆ i r 0 ( r T < a ) σ o a r T ( r T > a ) E = ˆ i r ρ 0 r T 2 0 ( r T < a ) 0 0 r T ( r T > a ) Φ( r T ) = ρ 0 4 0 ( a 2 - r 2 T ) ( r T a ) - 0 0 ln( r T /a ) ( r T a ) Θ μα 3 ( α ) H ( z ) = ˆ i z K 0 R 3 2 π θ =0 sin 2 θ [ z 2 + R 2 - 2 zR cos θ ] 3 / 2

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( β ) H ( z = 0) = ˆ i z K 0 π 4 ( γ ) H ( z = 0) = ˆ i z J 0 π 4 ( R 2 - R 1 ) Θ μα 4 ( α ) Φ( x, y ) = n =1 4 V 0 π 1 2 n - 1 1 sinh( k n a ) sinh( k n x ) sin( k n y ) k n = (2 n - 1) π 2 b ( β ) J = - σ n =1 k n C n cosh( k n x ) sin( k n y ) ˆ i x + - σ n =1 k
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