COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpub.com, http://www.jbpub.com/

CHAPTER 10 APPLICATIONS of HARMONIC FUNCTION S

Section 10.6 Two Dimensional Electrostatics

A two-dimensional electrostatic field is produced by a system of charged wires, plates, and cylindrical conductors that are perpendicular to the z-plane. The wires, plates, and cylinders are assumed to be so long that the effects at the ends can be neglected as mentioned in Section 10.4. This sets up an electric field that can be interpreted as the force acting on a unit positive charge placed at the point . In the study of electrostatics the vector field is shown to be conservative and is derivable from a function , called the electrostatic potential , as expressed by the equation

= = - .

The function is harmonic. Let be the harmonic conjugate, then

is the complex potential (not to be confused with the electrostatic potential). The curves are called the equipotential curves , and the curves are called the lines of flux . If a small test charge is allowed to move under the influence of the field , then it will travel along a line of flux. Boundary value problems for the potential function are mathematically the same as those for steady state heat flow, and they are realizations of the Dirichlet problem where the harmonic function is .

Load Maple's "contourplot" procedure.
Make sure this is done only ONCE during a Maple session.

> with(plots):

Example 9.18, Page 425. Consider two parallel conducting planes that pass perpendicular to
the -plane through the lines and , which are kept at the potentials and ,
respectively. Then according to the result of Example 10.1, the electrical potential is
.

Example 9.19, Page 425. Find the electrical potential in the region between the two
infinite coaxial cylinders and , which are kept at potentials and , respectively.

For an illustration we will use the boundary values:
when ,
when .

> a:='a': b:='b': u:='u': U1:='U1': U2:='U2': x:='x': y:='y':
a := 1:
b := 2:
U1 := 0:
U2 := 40:
`a ` = a, ` b ` = b;
`When |z| = a, f(x,y) ` = U1;
`When |z| = b, f(x,y) ` = U2;
u := proc(x,y)
U1 + (U2-U1)/(ln(b)-ln(a)) *
(ln(x^2+y^2)/2 - ln(a))
end:

> `u(x,y) ` = u(x,y);
contourplot(u(x,y), x=-2..2, y=-2..2,
title=`u(x,y) = 20 ln(x^2+y^2)/ln(2)`,
axes=boxed, grid=[50,50]);

Example 9.20, Page 425. Find the electrical potential produced by two
charged half-planes that are perpendicular to the -plane and pass trought the rays
, and , where the planes are kept at the fixed potentials:
for ,
for .

> u:='u': U:='U': x:='x': y:='y':
u := proc(x,y)
600/Pi*Re(arcsin(x + I*y))
end:
U := proc(x,y)
600/Pi*arcsin((sqrt((x+1)^2 + y^2) -
sqrt((x-1)^2 + y^2))/2)
end:
`u(x,y) ` = u(x,y);
`U(x,y) ` = U(x,y);

> contourplot(u(x,y), x=-4..4, y=-4..4,
title=`u(x,y) = 600/Pi Re(arcsin(x + I y))`,
axes=boxed, grid=[50,50],
scaling=constrained);

Example 10.21, Page 426. Find the electrical potential in the
disk : that satisfies the boundary values:
for and < < ,
for and < < .

> f:='f': u:='u': U:='U': v:='v': V:='V':
w:='w': W:='W': x:='x': y:='y':
w := (1-I)*(z - I)/(z - 1): `w ` = w;
W := subs(z=x+I*y,w): `w ` = W;
W := simplify(evalc(W)): `w ` = W;
U := ((x-1)^2-2*y)/((x-1)^2+y^2):
V := (1-x^2-y^2)/((x-1)^2+y^2): `w ` = U + I*V;
`log(w) ` = log(W);
`log(w) ` = evalc(log(W));

Use the imaginary part.

> f := proc(x,y)
80/Pi*argument((1-I)*(x+I*y-I)/(x+I*y-1))
end:
v := proc(x,y)
80/Pi*arctan(1-x^2-y^2,(x-1)^2-2*y)
end:
`f(x,y) ` = f(x,y);
`v(x,y) ` = v(x,y);
contourplot(v(x,y), x=-1..1, y=-1..1,
title=`v(x,y) = 80 Arg((1-I)(z-I)/(z-1))/Pi`,
axes=boxed, grid=[100,100],
scaling=constrained);

>

>

End of Section 10.6.