ECE329
Spring 2009
Homework 7 — Solution
March 10, 2009
1. Continuity equation.
a) Taking the divergence of Ampere’s law we have
∇ ·
(
∇ ×
H
) =
∇ ·
J
+
∇ ·
∂
D
∂t
.
Since
∇ ·
(
∇ ×
H
) = 0
,
we can write
∇ ·
J
+
∇ ·
∂
D
∂t
=
∇ ·
J
+
∂
∇ ·
D
∂t
= 0
.
Finally, using Gauss’ law
∇ ·
D
=
ρ,
we show
∇ ·
J
+
∂ρ
∂t
= 0
.
b) Taking the divergence of
J
=
(
4
z
2
ˆ
x
+ 3
x
3
y
ˆ
y
+ 2
z
(
y

y
o
)
2
ˆ
z
)
A
/
m
2
we get
∇ ·
J
=
∂
∂x
4
z
2
+
∂
∂y
3
x
3
y
+
∂
∂z
2
z
(
y

y
o
)
2
= 3
x
3
+ 2(
y

y
o
)
2
.
Since
∇ ·
J
is time independent, we have that
∂ρ
∂t
=
∇ ·
J
→
ρ
(
r
, t
) =

3
x
3
+ 2(
y

y
o
)
2
t
+
ρ
o
C
m
3
.
Evaluating at
r
=
0
and considering that
ρ
o
= 0
and
y
o
= 1
,
we find
ρ
(
0
, t
) =

2
t
C
m
3
.
c) Since the units of
J
x
, J
y
,
and
J
z
are A
/
m
2
,
we get
[
J
x
] =
h
4
z
2
i
=
A
m
2
→
[4] =
A
m
4
,
[
J
y
] =
h
3
x
3
y
i
=
A
m
2
→
[3] =
A
m
6
,
[
J
z
] =
h
2
z
(
y

y
o
)
2
i
=
A
m
2
→
[2] =
A
m
5
.
2. In HW6, we found that the charge density in a conductor with conductivity
σ
and permittivity
o
is
ρ
(
r
, t
) = cos(100
x
)
e

σ
o
t
C
/
m
3
.
1
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ECE329
Spring 2009
a) Using the continuity equation
∇ ·
J
+
∂ρ
∂t
= 0
and noting that
J
has only a component along
x
direction (i.e.,
J
=
J
x
ˆ
x
), we can write
∂J
x
∂x
=

∂ρ
∂t
.
Solving this equation we get
J
=
σ
100
o
sin(100
x
)
e

σ
o
t
ˆ
x
A
m
2
.
Then, the associated electric field is
E
=
1
σ
J
=
1
100
o
sin(100
x
)
e

σ
o
t
ˆ
x
V
m
.
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 Spring '08
 FRANKE
 Electromagnet, Electric charge, charge density, surface charge density, Jx, Surface charge

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