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ECE3025A
Problem Set 3 Solutions
4.24. The surface deFned by the equation
x
3
+
y
2
+
z
= 1000, where
x
,
y
, and
z
are positive, is an
equipotential surface on which the potential is 200 V. If

E

= 50 V/m at the point
P
(7
,
25
,
32)
on the surface, Fnd
E
there:
±irst, the potential function will be of the form
V
(
x, y, z
)=
C
1
(
x
3
+
y
2
+
z
)+
C
2
, where
C
1
and
C
2
are constants to be determined (
C
2
is in fact irrelevant for our purposes). The
electric Feld is now
E
=
−∇
V
=
−
C
1
(3
x
2
a
x
+2
y
a
y
+
a
z
)
And the magnitude of
E
is

E

=
C
1
!
9
x
4
+4
y
2
+ 1, which at the given point will be

E

P
=
C
1
!
9(7)
4
+ 4(25)
2
+ 1 = 155
.
27
C
1
=50
⇒
C
1
=0
.
322
Now substitute
C
1
and the given point into the expression for
E
to obtain
E
P
=
−
(47
.
34
a
x
+16
.
10
a
y
+0
.
32
a
z
)
The other constant,
C
2
, is needed to assure a potential of 200 V at the given point.
4.32. Using Eq. (36), a) Fnd the energy stored in the dipole Feld in the region
r>a
:
We start with
E
(
r, θ
)=
qd
4
π±
0
r
3
[2 cos
θ
a
r
+ sin
θ
a
θ
]
Then the energy will be
W
e
=
"
vol
1
2
±
0
E
·
E
dv
=
"
2
π
0
"
π
0
"
∞
a
(
qd
)
2
32
π
2
±
0
r
6
#
4 cos
2
θ
+ sin
2
θ
$
%
(
3 cos
2
θ
+1
r
2
sin
θdrdθdφ
=
−
2
π
(
qd
)
2
32
π
2
±
0
1
3
r
3
±
±
±
∞
a
"
π
0
#
3 cos
2
θ
+1
$
sin
θdθ
=
(
qd
)
2
48
π
2
±
0
a
3
#
−
cos
3
θ
−
cos
θ
$
π
0
%
(
4
=
(
qd
)
2
12
π±
0
a
3
J
b) Why can we not let
a
approach zero as a limit? ±rom the above result, a singularity in the
energy occurs as
a
→
0. More importantly,
a
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This note was uploaded on 10/28/2008 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 CITRIN

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