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CHAPTER 18 Complex Analysis and Potential Theory This is perhaps the most important justification for teaching complex analysis to engineers, and it also provides for nice applications of conformal mapping. SECTION 18.1. Electrostatic Fields, page 750 Purpose. To show how complex analysis can be used to discuss and solve two-dimensional electrostatic problems and to demonstrate the usefulness of complex potential, a major concept in this chapter. Main Content, Important Concepts Equipotential lines Complex potential (2) Combination of potentials by superposition The main reason for using complex methods in two-dimensional potential problems is the possibility of using the complex potential, whose real and imaginary parts both have a physical meaning, as explained in the text. This fact should be emphasized in teaching from this chapter. SOLUTIONS TO PROBLEM SET 18.1, page 753 2. F5 0.4 x 1 6.0 kV, F 5 0.4 z 1 6.0 4. F is expected to be linear because of the boundary (two parallel straight lines). From the boundary conditions and by inspection, F ( x , y ) 5 100( y 2 x )/ k . This is the real part of the complex potential F ( z ) 52 100(1 1 i ) z / k . The real potential can also be obtained systematically, starting from F ( x , y ) 5 ax 1 by 1 c . By the first boundary condition, for y 5 x this is zero: (1) ax 1 bx 1 c 5 0. By the second boundary condition, for y 5 x 1 k this equals 100: (2) ax 1 b ( x 1 k ) 1 c 5 100. Subtract (1) from (2) to get bk 5 100, b 5 100/ k . Substitute this into (1) to get ax 1 x 1 c 5 ( a 1 ) x 1 c 5 0. 100 } k 100 } k 289 im18.qxd 9/21/05 1:09 PM Page 289

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Since this is an identity in x , the coefficients must vanish, a 52 100/ k and c 5 0. This gives the above result. 6. We have F ( r ) 5 a ln r 1 b and get from this and the two boundary conditions F (1) 5 b 5 100 F (10) 5 a ln 10 1 b 5 a ln 10 1 100 5 1000. Hence a 5 900/ln 10, so that the answer is F ( r ) 5 (900 ln r )/ln 10 1 100. 8. F ( r ) 5 100 2 (50/ln 10) ln r , F ( z ) 5 100 2 (50/ln 10) Ln z 10. w 5 . w 5 0 gives z 5 z 0 1 1 i , Arg z 0 5 3 p /4. Hence at z 0 the potential is (1/ )3 /4 5 3/4. By considering the three given points and their images we see that the potential on the unit circle in the w -plane is 0 for the quarter-circle in the first quadrant and 1 for the other portion of the circle. This corresponds to a conductor consisting of two portions of a cylinder separated by small slits of insulation at w 5 1 and w 5 i , where the potential jumps. 12. CAS Experiment. (a) x 2 2 y 2 5 c , xy 5 k (b) xy 5 c , x 2 2 y 2 5 k ; the rotation caused by the multiplication by i leads to the interchange of the roles of the two families of curves. (c) x /( x 2 1 y 2 ) 5 c gives ( x 2 1/(2 c )) 2 1 y 2 5 1/(4 c 2 ). Also, 2 y /( x 2 1 y 2 ) 5 k gives the circles x 2 1 ( y 1 1/(2 k )) 2 5 1/(4 k 2 ). All circles of both families pass through the origin. (d) Another interchange of the families, compared to (c), ( y 2 1/(2 c )) 2 1 x 2 5 1/(4 c 2 ), ( x 2 1/(2 k )) 2 1 y 2 5 1/(4 k 2 ).
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ch18 - im18.qxd 1:09 PM Page 289 CHAPTER 18 Complex...

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