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Chapter 24
47
P24.66
The electric field throughout the region is directed along
x
; therefore,
E
will be
perpendicular to
dA
over the four faces of the surface which are perpendicular
to the
yz
plane, and
E
will be parallel to
dA
over the two faces which are parallel
to the
yz
plane. Therefore,
Φ
Ex
xa
x
xac
EA
E
A
a
a
b
a
c
a
b
a
b
c
a
c
=−
+
=− +
+
+
+
=
+
==
+
ej
e
j
e
j
af
32
2 2
2
2
.
Substituting the given values for
a
,
b
, and
c
, we find
Φ
E
=⋅
0269
.
N
m
C
2
.
Q
E
=∈
=
×
=
−
0
12
238 10
238
Φ
..
C
p
C
FIG. P24.66
P24.67
⋅=
=
∈
z
dEr
q
4
2
0
π
in
(a)
For
rR
>
,
qA
rr
d
r
AR
R
in
z
22
0
5
44
5
ππ
and
E
AR
r
=
∈
5
0
2
5
.
(b)
For
<
,
d
r
Ar
r
in
z
0
5
4
4
5
and
E
Ar
=
∈
3
0
5
.
P24.68
The total flux through a surface enclosing the charge
Q
is
Q
∈
0
. The flux through the
disk is
Φ
disk
z
d
where the integration covers the area of the disk. We must evaluate this integral
and set it equal to
1
4
0
Q
∈
to find how
b
and
R
are related. In the figure, take
d
A
to be
the area of an annular ring of radius
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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