Mathematical Methods for Physics and Engineering - Matematica.NET

APPLICATIONS OF COMPLEX VARIABLESdeduce thatF(w) =f(z) =V + ikz = V + ikw 1/2 (25.9)is the required potential. Expressed in terms of r, s and ρ =(r 2 + s 2 ) 1/2 , w 1/2 is given by[ ( ) 1/2 ( ) ] 1/2 ρ + r ρ − rw 1/2 = ρ 1/2 + i, (25.10)2ρ2ρand, in particular, the electrostatic potential is given byΦ(r, s) =ReF(w) =V −k √2[(r 2 + s 2 ) 1/2 − r ] 1/2. (25.11)The corresponding equipotentials and field lines are shown in figure 25.3(b). Using results(25.3)–(25.5), the magnitude of the electric field is|E| = |F ′ (w)| = | 1 2 ikw−1/2 | = 1 2 k(r2 + s 2 ) −1/4 .(ii) A transformation ‘converse’ to that used in (i),w = g(z) =z 1/2 ,has the effect of mapping the upper half of the z-plane into the first quadrant of thew-plane and the conducting plane y =0intothewedger>0, s =0andr =0,s>0.The complex potential now becomesF(w) =V + ikw 2= V + ik[(r 2 − s 2 )+2irs], (25.12)showing that the electrostatic potential is V −2krs and that the electric field has componentsE =(2ks,2kr). (25.13)Figure 25.3(c) indicates the approximate equipotentials and field lines. (Note that, in bothtransformations, g ′ (z) iseither0or∞ at the origin, and so neither transformation isconformal there. Consequently there is no violation of result (ii), given at the start ofsection 24.7, concerning the angles between intersecting lines.) ◭The method of images, discussed in section 21.5, can be used in conjunction withconformal transformations to solve some problems involving Laplace’s equationin two dimensions.◮A wedge of angle π/α with its vertex at z =0is formed by two semi-infinite conductingplates, as shown in figure 25.4(a). A line charge of strength q per unit length is positionedat z = z 0 , perpendicular to the z-plane. By considering the transformation w = z α ,findthecomplex electrostatic potential for this situation.Let us consider the action of the transformation w = z α on the lines defining the positionsof the conducting plates. The plate that lies along the positive x-axis is mapped onto thepositive r-axis in the w-plane, whereas the plate that lies along the direction exp(iπ/α) ismapped into the negative r-axis, as shown in figure 25.4(b). Similarly the line charge at z 0is mapped onto the point w 0 = z0 α.From figure 25.4(b), we see that in the w-plane the problem can be solved by introducinga second line charge of opposite sign at the point w0 ∗ , so that the potential Φ = 0 alongthe r-axis. The complex potential for such an arrangement is simplyF(w) =−q ln(w − w 0 )+q ln(w − w2πɛ 0 2πɛ0).∗0878