Chapter 27 Solutions
(solutions marked with * are graded problems)
27.3. Visualize:
Figure 27.6 shows the electric field for an infinite plane of charge. For two parallel planes, this is the only shape of the
electric field vectors that matches the symmetry of the charge distribution.
27.5.
Model:
The electric flux “flows”
out
of a closed surface around a region of space containing a net positive charge
and
into
a closed surface surrounding a net negative charge.
Visualize:
Please refer to Figure Ex27.5. Let
A
be the area of each of the six faces of the cube.
Solve:
The electric flux is defined as
e
cos
E A
EA
!
"
=
#
=
r
r
, where
θ
is the angle between the electric field and a line
perpendicular
to the plane of the surface. The electric flux out of the closed cube surface is
(
)
(
)
2
out
10 N/C 10 N/C 10 N/C
5 N/C
cos0
35
N m /C
A
A
!
=
+
+
+
° =
Similarly, the electric flux into the closed cube surface is
(
)
(
)
(
)
2
in
15 N/C
20 N/C
cos 180
35
N m /C
A
A
!
=
+
° =
"
Hence,
Φ
out
+
Φ
in
=
0 N m
2
/C. Since the net electric flux is zero, the closed box contains no charge.
27.9.
Model:
The electric field is uniform over the entire surface.
Visualize:
Please refer to Figure Ex27.9. The electric field vectors make an angle of 30
°
below the surface. Because the
normal
ˆ
n
to the planar surface is at an angle of 90
°
relative to the surface, the angle between
ˆ
n
and
E
r
is
=
120
°
.
Solve:
The electric flux is
(
)
(
)
2
2
2
e
cos
200 N/C 1.0 10 m
cos120
1.0 N m /C
E A
EA
"
#
=
$
=
=
%
° =
"
r
r
27.11.
Model:
The electric field is uniform over the rectangle in the
xy
plane.
Solve:
(a)
The area vector is perpendicular to the
xy
plane. Thus
(
)
(
)
4
2
ˆ
ˆ
2.0 cm 3.0 cm
6.0 10 m
A
k
k
!
=
"
=
"
r
The electric flux through the rectangle is
(
)(
)
4
2
2
2
e
ˆ
ˆ
ˆ
50
100
6.0 10
N m /C
6.0 10 N m /C
E A
i
k
k
!
!
"
=
#
=
+
#
$
=
$
r
r
(b)
The electric flux is
(
)
(
)
4
2
2
e
ˆ
ˆ
ˆ
50
100
6.0 10
N m /C
0 N m /C
E A
i
j
k
!
"
=
#
=
+
#
$
=
r
r
Assess:
In (b),
E
r
is in the plane of the rectangle. That is why the flux is zero.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document27.17.
Solve:
For
any
closed surface that encloses a total charge
Q
in
, the net electric flux through the surface is
(
)(
)
6
19
2
in
e
12
2
2
0
55.3 10
1.60 10
C
1.0 N m /C
8.85 10
C /N m
Q
!
"
"
#
"
#
$
=
=
=
"
#
27.19.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 ALL
 Charge, Electrostatics, Electric charge, Surface, Qin

Click to edit the document details