Mathematical Methods for Physics and Engineering : A Comprehensive Guide

APPLICATIONS OF COMPLEX VARIABLES

deduce that

F(w)=f(z)=V+ikz=V+ikw^1 /^2 (25.9)

is the required potential. Expressed in terms ofr,sandρ=(r^2 +s^2 )^1 /^2 ,w^1 /^2 is given by

w^1 /^2 =ρ^1 /^2

[(

ρ+r

2 ρ

) 1 / 2

+i

(

ρ−r

2 ρ

) 1 / 2 ]

, (25.10)

and, in particular, the electrostatic potential is given by

Φ(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)|=|^12 ikw−^1 /^2 |=^12 k(r^2 +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 thez-plane into the first quadrant of the
w-plane and the conducting planey=0intothewedger>0,s=0andr=0,s>0.
The complex potential now becomes
F(w)=V+ikw^2
=V+ik[(r^2 −s^2 )+2irs], (25.12)

showing that the electrostatic potential isV− 2 krsand that the electric field has components

E=(2ks, 2 kr). (25.13)

Figure 25.3(c) indicates the approximate equipotentials and field lines. (Note that, in both
transformations,g′(z)iseither0or∞at the origin, and so neither transformation is
conformal there. Consequently there is no violation of result (ii), given at the start of
section 24.7, concerning the angles between intersecting lines.)

Themethod of images, discussed in section 21.5, can be used in conjunction with

conformal transformations to solve some problems involving Laplace’s equation

in two dimensions.

A wedge of angleπ/αwith its vertex atz=0is formed by two semi-infinite conducting

plates, as shown in figure 25.4(a). A line charge of strengthqper unit length is positioned

atz=z 0 , perpendicular to thez-plane. By considering the transformationw=zα,findthe

complex electrostatic potential for this situation.

Let us consider the action of the transformationw=zαon the lines defining the positions
of the conducting plates. The plate that lies along the positivex-axis is mapped onto the
positiver-axis in thew-plane, whereas the plate that lies along the direction exp(iπ/α)is
mapped into the negativer-axis, as shown in figure 25.4(b). Similarly the line charge atz 0
is mapped onto the pointw 0 =zα 0.
From figure 25.4(b), we see that in thew-plane the problem can be solved by introducing
a second line charge of opposite sign at the pointw∗ 0 , so that the potential Φ = 0 along
ther-axis. The complex potential for such an arrangement is simply

F(w)=−

q

2 π 0

ln(w−w 0 )+

q

2 π 0

ln(w−w∗ 0 ).