CHAPTER 18
Complex Analysis and Potential Theory
This is perhaps the most important justification for teaching complex analysis to engineers,
and it also provides for nice applications of conformal mapping.
SECTION 18.1. Electrostatic Fields, page 750
Purpose.
To show how complex analysis can be used to discuss and solve twodimensional
electrostatic problems and to demonstrate the usefulness of
complex potential,
a major
concept in this chapter.
Main Content, Important Concepts
Equipotential lines
Complex potential (2)
Combination of potentials by superposition
The main reason for using complex methods in twodimensional potential problems is the
possibility of using the complex potential, whose real and imaginary parts both have a
physical meaning, as explained in the text. This fact should be emphasized in teaching
from this chapter.
SOLUTIONS TO PROBLEM SET 18.1, page 753
2.
F5
0.4
x
1
6.0 kV,
F
5
0.4
z
1
6.0
4.
F
is expected to be linear because of the boundary (two parallel straight lines). From
the boundary conditions and by inspection,
F
(
x
,
y
)
5
100(
y
2
x
)/
k
.
This is the real part of the complex potential
F
(
z
)
52
100(1
1
i
)
z
/
k
.
The real potential can also be obtained systematically, starting from
F
(
x
,
y
)
5
ax
1
by
1
c
.
By the first boundary condition, for
y
5
x
this is zero:
(1)
ax
1
bx
1
c
5
0.
By the second boundary condition, for
y
5
x
1
k
this equals 100:
(2)
ax
1
b
(
x
1
k
)
1
c
5
100.
Subtract (1) from (2) to get
bk
5
100,
b
5
100/
k
.
Substitute this into (1) to get
ax
1
x
1
c
5
(
a
1
)
x
1
c
5
0.
100
}
k
100
}
k
289
im18.qxd
9/21/05
1:09 PM
Page 289
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View Full DocumentSince this is an identity in
x
, the coefficients must vanish,
a
52
100/
k
and
c
5
0.
This gives the above result.
6.
We have
F
(
r
)
5
a
ln
r
1
b
and get from this and the two boundary conditions
F
(1)
5
b
5
100
F
(10)
5
a
ln 10
1
b
5
a
ln 10
1
100
5
1000.
Hence
a
5
900/ln 10, so that the
answer
is
F
(
r
)
5
(900 ln
r
)/ln 10
1
100.
8.
F
(
r
)
5
100
2
(50/ln 10) ln
r
,
F
(
z
)
5
100
2
(50/ln 10) Ln
z
10.
w
5
.
w
5
0 gives
z
5
z
0
1
1
i
, Arg
z
0
5
3
p
/4. Hence at
z
0
the
potential is (1/
)3
/4
5
3/4.
By considering the three given points and their images we see that the potential on
the unit circle in the
w
plane is 0 for the quartercircle in the first quadrant and 1 for
the other portion of the circle. This corresponds to a conductor consisting of two
portions of a cylinder separated by small slits of insulation at
w
5
1 and
w
5
i
, where
the potential jumps.
12. CAS Experiment.
(a)
x
2
2
y
2
5
c
,
xy
5
k
(b)
xy
5
c
,
x
2
2
y
2
5
k
; the rotation caused by the multiplication by
i
leads to the
interchange of the roles of the two families of curves.
(c)
x
/(
x
2
1
y
2
)
5
c
gives (
x
2
1/(2
c
))
2
1
y
2
5
1/(4
c
2
). Also,
2
y
/(
x
2
1
y
2
)
5
k
gives the circles
x
2
1
(
y
1
1/(2
k
))
2
5
1/(4
k
2
). All circles of both families pass through
the origin.
(d) Another interchange of the families, compared to (c), (
y
2
1/(2
c
))
2
1
x
2
5
1/(4
c
2
),
(
x
2
1/(2
k
))
2
1
y
2
5
1/(4
k
2
).
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 Partial Differential Equations, Boundary value problem, Boundary conditions, Green's function

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