Chapter 23
1. The vector area
A
and the electric field
E
are shown on the diagram below. The angle
θ
between them is 180° – 35° = 145°, so the electric flux through the area is
(
29
(
29
2
3
2
2
cos
1800 N C
3.2
10
m
cos145
1.5
10
N m
C.
E A
EA
θ


Φ =
⋅
=
=
×
° = 
×
⋅
2. We use
Φ =
⋅
E
A
, where
A
A
=
=
.
j
m
j
2
140
b
g
.
(a)
(
29
(
29
2
ˆ
ˆ
6.00 N C
i
1.40 m
j
0.
Φ =
=
�
(b)
(
29
(
29
2
2
ˆ
ˆ
2.00 N C
j
1.40 m
j
3.92 N
m
C.
Φ = 
= 
�
�
(c)
(
29
(
29
(
29
2
ˆ
ˆ
ˆ
3.00 N C
i
400 N
C
k
1.40 m
j
0
�
�
Φ =

+
=
�
�
�
.
(d) The total flux of a uniform field through a closed surface is always zero.
3. We use
Φ =
z
⋅
E dA
and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m.
(a) On the top face of the cube
y
= 2.0 m and
(
29
ˆ
j
dA
dA
=
. Therefore, we have
(
29
(
29
2
ˆ
ˆ
ˆ
ˆ
4i
3
2.0
2 j
4i
18j
E
=

+
=

. Thus the flux is
(
29
(
29
(
29
(
29
2
2
2
top
top
top
ˆ
ˆ
ˆ
4i
18j
j
18
18
2.0
N m
C
72 N m
C.
E dA
dA
dA
Φ =
=

= 
= 
= 
�
�
�
�
�
�
�
(b) On the bottom face of the cube
y
= 0 and
dA
dA
=

bg
ej
j
. Therefore, we have
E
=

+
=

4
3 0
2
4
6
2
i
j
i
j
c
h
. Thus, the flux is
963
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CHAPTER 23
(
29
(
29
(
29
(
29
2
2
2
bottom
bottom
bottom
ˆ
ˆ
ˆ
4i
6j
j
6
6
2.0
N
m
C
24 N
m
C.
E dA
dA
dA
Φ =
=


=
=
= +
�
�
�
�
�
�
�
(c) On the left face of the cube
(
29
(
29
ˆ
i
dA
dA
=

. So
(
29
(
29
(
29
(
29
2
2
2
left
left
bottom
ˆ
ˆ
ˆ
ˆ
4i
j
i
4
4 2.0
N m
C
16 N m
C.
y
E dA
E
dA
dA
Φ =
=
+

= 
= 
= 
�
�
�
�
�
�
�
(d) On the back face of the cube
(
29
(
29
ˆ
k
dA
dA
=

. But since
E
has no
z
component
0
E dA
⋅
=
. Thus,
Φ
= 0.
(e) We now have to add the flux through all six faces. One can easily verify that the flux
through the front face is zero, while that through the right face is the opposite of that
through the left one, or +16 N·m
2
/C. Thus the net flux through the cube is
Φ
= (–72 + 24 – 16 + 0 + 0 + 16) N·m
2
/C = – 48 N·m
2
/C.
4. The flux through the flat surface encircled by the rim is given by
2
.
a E
Φ = π
Thus, the
flux through the netting is
2
3
4
2
(0.11 m)
(3.0
10
N/C)
1.1
10
N m /C
a E
π
π
2


Φ = Φ = 
= 
= 
.
5. We use Gauss’ law:
0
q
ε
Φ =
, where
Φ
is the total flux through the cube surface and
q
is the net charge inside the cube. Thus,
6
5
2
12
2
2
0
1.8
10
C
2.0
10 N m
C.
8.85
10
C
N m
q
ε


×
Φ =
=
=
×
⋅
×
⋅
6. There is no flux through the sides, so we have two “inward” contributions to the flux,
one from the top (of magnitude (34)(3.0)
2
) and one from the bottom (of magnitude (20)
(3.0)
2
). With “inward” flux being negative, the result is
Φ
= – 486 N
⋅
m
2
/C. Gauss’ law
then leads to
q
enc
=
ε
0
Φ
= –4.3
×
10
–9
C.
7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the
shape of a cube, of edge length
d
, with a proton of charge
19
1.6
10
C
q

= +
situated at
the inside center of the cube. The cube has six faces, and we expect an equal amount of
flux through each face. The total amount of flux is
Φ
net
=
q
/
ε
0
, and we conclude that the
flux through the square is onesixth of that. Thus,
Φ
=
q
/6
ε
0
= 3.01
×
10
–9
N
⋅
m
2
/C.
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 Fall '08
 Many
 Physics, Charge, Electrostatics, Electric charge, R. Gauss

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