**for**

**11.6 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
11.4. This assumption results in 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, expressed as

.

If we make the additional assumption that
there are no charges within the domain D, then
Gauss' Law for electrostatic fields implies that the line integral of
the outward normal component of taken
around any small rectangle lying inside D
is identically zero. A heuristic argument similar to the
one for steady state temperatures with replaced
by will
show that the value of the line integral is

.

This quantity is zero, so we conclude that is
a harmonic function. If we 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 .

**Example
11.18.** Consider two parallel conducting planes
that pass perpendicular to the z
plane through the lines , which
are kept at the potentials
and ,
respectively. Then according to the result of Example
11.1, (see Section
11.1), the electrical potential is

.

**Example 11.19.** Find
the electrical potential in
the region between two infinite coaxial
cylinders , which
are kept at potentials , respectively.

Solution. The transformation

maps the annular region between the circles onto
the infinite strip in
the w plane, as shown in Figure 11.36. The
potential in
the infinite strip has the boundary values

and for
all v.

If we use the result of Example 11.18, the electrical
potential is

.

Because , we
can use this equation to conclude that the
potential is

.

The equipotentials are
concentric circles centered on the origin, and the lines of flux are
portions of rays emanating from the
origin. If , then
the situation is as illustrated in Figure 11.36.

The electrical field in a coaxial cylinder, where .Figure 11.36

**Example
11.20.** Find the electrical
potential produced
by two charged half-planes that are perpendicular to the z
plane and pass through the rays where
the planes are kept at the fixed potentials

Solution. The result of Example 10.13, (see Section
10.4), shows that the function

is a conformal mapping of the z plane
slit along the two rays onto
the vertical strip .
The new problem is to find the potential that
satisfies the boundary values

From Example 11.1, (see Section
11.1),

.

As in the discussion of Example 11.17, (see Section
11.5),the solution in the z plane
is

.

Several equipotential curves are shown in Figure 11.37.

The electric field produced by two charged half-planesFigure 11.37

that are perpendicular to the complex plane.

**Example 11.21.** Find
the electrical potential in
the disk that
satisfies the boundary values

Solution. The mapping is
a one-to-one conformal mapping of D
onto the upper half-plane with
the property that
is mapped onto the negative u axis
and
is mapped onto the positive u
axis. The potential
in the upper half-plane that satisfies the new boundary values

is given by

(11-29) .

A straightforward calculation shows that

We substitute the real and imaginary parts, and from
this equation, into Equation (11-29) to
obtain the desired solution:

.

The level curve in
the upper half-plane is a ray emanating from the origin, and the
preimage in
the unit disk is an arc of a circle that passes through the
points . Several
level curves are illustrated in Figure 11.38.

** Figure
11.38** The
potentials and .

**Extra Example
1.** Find the electrical
potential in
the crescent-shaped region that lies inside the disk and
outside the circle that
satisfies the following boundary values shown in Figure 11.41:

Solution. The result of Example 10.7, (see Section
10.2), shows that the function

is a conformal mapping of the crescent-shaped region that lies inside
the disk and
outside the circle onto
the horizontal strip .

The electrical potential inside and outside .Figure 11.41

Revisit and Explore Solution 10.7.

The new problem in the w plane is to find
the potential that
satisfies the boundary values

In the w plane the solution is

A straightforward calculation shows that

We substitute the imaginary part, from
this equation, into to
obtain the desired solution:

**Extra Example
2.** Find the electrical
potential in
the semi-infinite strip
that has the boundary values shown in Figure 11.42:

Solution. The result of Example 10.13, (see Section
10.4), shows that the function

is a conformal mapping of the semi-infinite
strip onto
the upper half-plane .

The new problem in the w plane is to find
the electrical potential that
satisfies the boundary values

This is a three-value Dirichlet problem in the upper half-plane
defined by . For
the w plane, the solution in Equation
(11-5) becomes

Here we have and , which
we substitute into the above equation for to
obtain

Now make the substitutions and to
get the solultion in
the z plane

The electrical potential in the semi-infinite strip .Figure 11.42

**Exercises
for Section 11.6. Two-Dimensional
Electrostatics****
**

**The Next Module
is**

**Return to the Complex
Analysis Modules **

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to the Complex Analysis Project__

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

(c) 2012 John H. Mathews, Russell W. Howell