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Mathematical Methods for Physics and Engineering - Matematica.NET

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25.2 APPLICATIONS OF CONFORMAL TRANSFORMATIONSyssxrr(a) z-plane (b) w-plane (c) w-planeFigure 25.3 The equipotential lines (broken) <strong>and</strong> field lines (solid) (a) <strong>for</strong> aninfinite charged conducting plane at y =0,wherez = x + iy, <strong>and</strong> after thetrans<strong>for</strong>mations (b) w = z 2 <strong>and</strong> (c) w = z 1/2 of the situation shown in (a).◮ Find the complex electrostatic potential associated with an infinite charged conductingplate y =0, <strong>and</strong> thus obtain those associated with(i) a semi-infinite charged conducting plate (r >0, s =0);(ii) the inside of a right-angled charged conducting wedge (r > 0, s = 0 <strong>and</strong>r =0, s>0).Figure 25.3(a) shows the equipotentials (broken lines) <strong>and</strong> field lines (solid lines) <strong>for</strong> theinfinite charged conducting plane y = 0. Suppose that we elect to make the real part ofthe complex potential coincide with the conventional electrostatic potential. If the plate ischarged to a potential V then clearlyφ(x, y) =V − ky, (25.6)where k is related to the charge density σ by k = σ/ɛ 0 , since physically the electric field Ehas components (0,σ/ɛ 0 )<strong>and</strong>E = −∇φ.Thus what is needed is an analytic function of z, ofwhichtherealpartisV − ky. Thiscan be obtained by inspection, but we may proceed <strong>for</strong>mally <strong>and</strong> use the Cauchy–Riemannrelations to obtain the imaginary part ψ(x, y) as follows:∂ψ∂y = ∂φ∂x =0 <strong>and</strong> ∂ψ∂x = − ∂φ∂y = k.Hence ψ = kx + c <strong>and</strong>, absorbing c into V , the required complex potential isf(z) =V − ky + ikx = V + ikz. (25.7)(i) Now consider the trans<strong>for</strong>mationw = g(z) =z 2 . (25.8)This satisfies the criteria <strong>for</strong> a con<strong>for</strong>mal mapping (except at z = 0) <strong>and</strong> carries the upperhalf of the z-plane into the entire w-plane; the equipotential plane y =0goesintothehalf-plane r>0, s =0.By the general results proved, f(z), when expressed in terms of r <strong>and</strong> s, will give acomplex potential whose real part will be constant on the half-plane in question; we877

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