460 CHAPTER 11 • APPLICATIONS OF HARMONIC FUNCTIONS- EXAMPLE 11 .18 Consider two parallel conducting planes that pass per-
pendicular to the z plane through the lines x = a and x = b, which are kept at
the potentials U 1 and U2, respectively. Then, according to the result of Example
11.1, the electrical potential is
U2 - U1
-a
- EXAMPLE 11.19 Find the electrical potential
(x, y) in the region between
two infinite coaxial cylinders r = a and 1· = b, which are kept at the potentials
U1 and U2, respectively.Solution The function w = log z = In lzl + i arg z maps the annular region
between the circles r = a and r = b onto the infinite strip In a < u < In b in the
w plane, as shown in Figure ll.36. The potential <I> (u, v) in the infinite strip
has the boundary valuesIf we use the result of Example 11.18, the electrical potential q> ( u, v) is
U2-U1(u, v) = Ui + lnb - In a (u - Ina).
Because u = In lzl, we can use this equation to conclude t hat the potential
<f>(x,y)isU2-U 1
(x, y) = U1 + lnb- Ina (ln lzl - Ina).
The equipotentials tf>(x,y) = constant are concentric circles centered on the
origin, and the lines of flux are portions of rays emanating from the origin. If
U2 < U1, then the situation is as illustrated in Figure 11.36.- EXAMPLE 11 .2 0 Find the electrical potential </> (x, y) produced by two
charged half-planes that are perpendicular to the z plane and pass through the
rays x < -1, y = 0 and x > 1, y = 0, where the planes are kept at the fixed
potentials</> (x , O) = - 300, for x < - 1, and </> (x, 0) = 300, for x > 1.
Solution The result of Example 10.13 shows that the function w = Arcsinz is
a conformal mapping of the z plane slit along the two rays x < -1, y = 0 and