Question 7.5.2: An Electrostatics Application. Determine the electrostatic p......
An Electrostatics Application
Determine the electrostatic potential \boldsymbol{\phi} in the domain D between the circles |z| = 1 and \left|z-{\textstyle\frac{1}{2}}\right|={\textstyle\frac{1}{2}}, shown in color in Figure 7.50(a), that satisfies the indicated boundar yconditions.
The electrostatic potential \boldsymbol{\phi} is a solution of Laplace’s equation (1) in D that satisfies the boundary conditions
{\frac{\partial^{2}\phi}{\partial x^{2}}}+{\frac{\partial^{2}\phi}{\partial y^{2}}}=0, (1)
\phi(x,y)=-10\quad\mathrm{on}\ \ x^{2}+y^{2}=1,\phi(x,y)=20\quad\mathrm{~on}\ \ (x-{\textstyle\frac{1}{2}})^{2}+y^{2}={\textstyle\frac{1}{4}}.
We proceed as in Example 1.
Step 1 The given domain D can be mapped onto the infinite horizontal strip 0 < v < 1, shown in gray in Figure 7.50(b), by a linear fractional transformation. One way to do this is to require that the points 1, i, and −1 on the circle |z| = 1 map onto the points \infty , 0, and 1, respectively.
By Theorem 7.4 in Section 7.2 the desired linear fractional transformation w = T(z) must satisfy
After solving for w = T(z), we obtain
T(z)=(1-i){\frac{z-i}{z-1}}. (5)
By construction, the circle |z| = 1 is mapped onto the line v = 0 by w = T(z). Furthermore, because the pole z = 1 of (5) is on the circle \left|z-{\textstyle\frac{1}{2}}\right|\,=\,{\textstyle\frac{1}{2}}, it follows that this circle is also mapped onto a line. The image line can be determined by finding the image of two points on the circle \left|z-{\textstyle\frac{1}{2}}\right|\,=\,{\textstyle\frac{1}{2}}. For the points z = 0 and z={\frac{1}{2}}+{\frac{1}{2}}i on the circle \left|z-{\frac{1}{z}}\right|={\frac{1}{2}}, we have T(0) = 1+i and T\left({\textstyle{\frac{1}{2}}}+{\textstyle{\frac{1}{2}}}i\right)=-1+i. Therefore, the image of the circle \left|z-{\frac{1}{2}}\right|={\frac{1}{2}} must be the horizontal line v = 1. Using the test point z = −\frac{1}{2} , we find that T\,\left(-{\textstyle{\frac{1}{2}}}\right)=1+{\textstyle{\frac{1}{3}}}i, and so we conclude that the domain shown in color between the circles in Figure 7.50(a) is mapped by w = T(z) onto the domain shown in gray between the horizontal lines in Figure 7.50(b).
Step 2 From Step 1 we have w = T(z) maps the circle |z| = 1 onto the horizontal line v = 0, and it maps the circle \left|z-{\frac{1}{2}}\right|={\frac{1}{2}} onto the horizontal line v = 1. Thus, the transformed boundary conditions are Φ = −10 on the line v = 0 and Φ = 20 on the line v = 1. See Figure 7.50(b).
Step 3 Modeled after Example 2 in Section 3.4 and Problem 12 in Exercises 3.4, a solution of the transformed Dirichlet problem is given by
Step 4 A solution of the original Dirichlet problem is now obtained by substituting the real and imaginary parts of T(z) defined in (5) for the variables u and v in Φ(u, v). By replacing the symbol z with x + iy in T(z) and simplifying we obtain:
T(x+i y)=(1-i){\frac{x+i y-i}{x+i y-1}}=(1-i){\frac{x+i(y-1)}{x-1+y i}}{\frac{x-1-i y}{x-1-i y}}\quad \quad\quad\quad\quad =\frac{x^{2}+y^{2}-2x-2y+1}{(x-1)^{2}+y^{2}}+\frac{1-x^{2}-y^{2}}{(x-1)^{2}+y^{2}}i.
Therefore,
\phi(x,y)=30\frac{1-x^{2}-y^{2}}{(x-1)^{2}+y^{2}}-10 (6)
is the desired electrostatic potential function.
