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RECITATION 4
5.
The area of each face is
A
= (1.40 m)
2
.
(a) Since the field has only a
y
component, there will be no flux through the faces pointing
in the
x
and
z
directions.
Then
(
29
(
29
(
29
(
29
(
29
net
=0
1.40
2
2
ˆ
ˆ
ˆ
ˆ
(
)j
j
(
)j
j
0.00 N C
3.00 N C
1.40m
1.40m
8.23 N m
C.
j
j
y
y
E
A
E
A
=
Φ
=
⋅ 
+
⋅
=
+
=
⋅
(b) The electric field can be rewritten as
(
29
0
ˆ
[ 3.00 N C m ] j
E
y
E
=
⋅
+
r
r
, where
(
29
(
29
0
ˆ
ˆ
4.00 N C i
6.00 N C j
E
= 
+
r
is a constant field which does not contribute to the net flux
through the cube (since any element of positive flux will be balanced by an element of
negative flux on the opposite face). Thus
net
Φ
is
2
8.23 N m
C
⋅
just as before.
(c) The charge is given by
(
29
2
enc
net
12
2
11
0
2
C
8.85 10
8.23 N m
C
7.29 10
C
N m
q
ε


=
Φ
=
×
⋅
=
×
⋅
in each case.
19.
For spherically symmetric charge distributions, Eq. 2411 is
(
29
2
0
1
,
4
q r
E
r
πε
′
=
r
where
(
29
q r
′
is the function describing how the enclosed charge varies with distance from
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This note was uploaded on 06/01/2008 for the course PHYS 232 taught by Professor Goksu during the Spring '08 term at Millersville.
 Spring '08
 Goksu
 Physics

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