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# For this we take z0 b a point somewhere on the x axis

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Unformatted text preview: and region gives the desired complex potential: 6 È1 + z˘ ˙ G(z) = -iπLnÍ Î1 - z˚ kV È1 + z˘ 6 ˙ Its real part, πArgÍ Î1 - z˚ kV, is our desired potential function. Question What are the equipotentials: È1 + z˘ ˙ Answer ∞ = constant iff ArgÍ Î1 - z˚ = Arg(w)= constant. But these are just rays from the origin in the right--hand region. Since these rays extend from 0 to infinity, they must give circular arcs in the left-hand region extending from -1 to 1. Question What are the lines of force? Answer Setting the imaginary part of the complex potential equal to constants gives § = constand iff |w| = const, giving semicircles centered at the origin on the right-hand side, corresponding to circular arcs, roughly as shown: 33 Note that the y-axis itself is one of those arc, corresponding to the unit semicircle on the right. Question What is the vector form of the electric field? Answer We use E = F'(z) ; actually G'(z) in this case. 6 È1 + z˘ 6 ˙ G(z) = -iπLnÍ Î1 - z˚ = -iπ[Ln(1+z) - Ln(1-z)] 12z G'(z) = 2 π(1 - z ) 12z– G'(z) = 2 π(1 - z– ) T...
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## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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