28 we find that the complex potential is just linear

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Unformatted text preview: cle of radius 0.5 centered at the origin. We now find the potential for the nice coaxial cylinders (Example (B) in the previous section): ∞ = A ln(r) + B 0 = Aln(1)+ B 110 = A ln(0.5) + B This gives B = 0 and A = 110/ln(0.5) ‡ -158.7. So, ∞ = 110 ln(r)/ln(0.5) The associated complex potential (See Example (B) above) is 110 F(z) = Ln(z) ln(0.5) Therefore, the complex potential on the non-coaxial region is È2z - 1 ˘ È2z - 1 ˘ 110 ˙ Í ˙ F(r(z)) = LnÍ Î z - 2 ˚ ‡ -158.7 LnÎ z - 2 ˚ ln(0.5) The real potential is Ô2z - 1 Ô Ô ∞ = -158.7 lnÔ Ôz - 2 Ô We can now express the RHS in terms of (x, y) if we really want, Question What are the equipotentials: Ô2z - 1 Ô Ô Answer ∞ = constant iff Ô Ô z - 2 Ô = |w| = constant. But these are just circles in the nice coaxial region, which correspond to off-centered circles in the non-coaxial region. Question What are the lines of force? Answer Set the imaginary part of the complex potential equal to constants: 35 È2z - 1 ˘ ˙ -158.7 ArgÍ Î z - 2 ˚ = const giving Arg(w) =...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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