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ECE 3030
Electromagnetic Fields and Waves
Fall 2009 Lecture 7
2009/9/11
Boundary Valued Problems
Spherical Solutions
1
Swartz and Rana 06/5/25
Electromagnetic Fields and Waves
– Fall 2009
Lecture 7:
Instructor:
Dr. Wesley E. Swartz
Capacitors with Dielectric Insulators
Conductors Immersed in Electric Fields
θ
z
a
x
+
+
+
+
+
++



E
o
V
+
ε
+
+
+
+
+
+
+
+








Solutions of Laplace’s Equation in
Spherical Coordinates
•
The following are solutions to
.
–
Spherically Symmetric Solution
–
A Constant Uniform Electric Field Solution
()
B
r
A
r
+
=
φ
z
A
r
A
r
−
=
−
=
cos
z
0
2
=
∇
r
+
q
3
Swartz and Rana 06/5/25
Electromagnetic Fields and Waves
– Fall 2009
Lecture 7:
–
A Dipole Oriented Along zAxis Solution
–
A Constant Potential Solution (a trivial
solution):
( )
( )
z
A
r
r
E
ˆ
=
−∇
=
2
cos
r
A
r
=
z
+
q

q
A
r
=
z
x
Solutions of Laplace’s Equation in
Cylindrical Coordinates (1)
•
The following are solutions to
.
–
Cylindrically Symmetric Line Charge
Solution:
r
A
r
ln
=
0
2
=
∇
r
+
λ
x
y
4
Swartz and Rana 06/5/25
Electromagnetic Fields and Waves
– Fall 2009
Lecture 7:
–
A Constant Uniform Electric Field Solution
x
E
r
E
r
o
o
−
=
−
=
cos
y
x
E
r
r
E
o
ˆ
=
−∇
=
x
Solutions of Laplace’s Equation in
Cylindrical Coordinates (2)
•
More solutions to
.
–
A Line Dipole Oriented Along xAxis
Solution:
0
2
=
∇
r
r
A
r
cos
=
y

+
x
r
ln
2
r
π
=
−
d
5
Swartz and Rana 06/5/25
Electromagnetic Fields and Waves
– Fall 2009
Lecture 7:
A
r
=
r
cos
2
d
cos
2
d
r
cos
2
d
r
ln
2
r
o
o
o
≈
−
+
≈
+
–
A constant Potential Solution (a trivial solution):
z
A NonUniformly Charged
Spherical Shell (1)
•
Consider a charged spherical shell where the
surface charge density is fixed and is given
by:
–
Looking from outside the potential looks like that
of a dipole. So try a
dipolelike solution
for
r
>
a
:
σ
cos
o
=
( )
C
cos
A
r
+
z
+
+
+
+
+
a
0
6
Swartz and Rana 06/5/25
Electromagnetic Fields and Waves
– Fall 2009
Lecture 7:
–
For
r
<
a
, try something that does not go to infinity
at
r
=
0
:
•
How can you tell if this is the right solution??
–
( )
r
2
out
=
D
cos
r
B
r
in
+
=
0
Why?
Why?
A NonUniformly Charged
Spherical Shell (2)
•
Our two guessed solutions are:
•
Boundary conditions:
–
(1) Potential inside and outside must be continuous
at the surface of the shell:
cos
o
=
2
cos
r
A
r
out
=
cos
r
B
r
in
=
( )
cos
z
+
+
+
+
+
a
7
Swartz and Rana 06/5/25
Electromagnetic Fields and Waves
– Fall 2009
Lecture 7:
( )
( )
( )
cos
2
a
B
a
A
r
r
a
r
out
a
r
in
=
=
=
=
o
o
o
a
r
out
o
a
r
out
o
a
r
r
in
a
r
r
out
o
B
a
A
r
r
r
r
E
E
cos
cos
cos
2
cos
3
,
,
=
+
=
∂
∂
+
∂
∂
−
=
−
=
=
=
=
–
(2) Discontinuity in the radial electric field at the surface must be
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