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GAUSS'S
LAW
27.1. Vialie:
As
discussed in Section 27.1, the symmetry of the electric field must match the symmetry of the charge distribution. In
particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the
cylinder axis. Also, the electric field of a cylindrically symmetric charge distribution cannot have a component tangent
to the circular cross section. The only shape for the electric field that matches the symmetry of the charge distribution
with respect to (i) translation parallel to the cylinder axis, (ii) rotation by an angle about the cylinder
axis,
and (iii)
reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure.
27.2. Visualize:
t
+++++++++++++++++
1!11!1111111111
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.
27.4.
positive charge and
inro
a closed surface surrounding
a
net negative charge.
Visualize:
Solve:
line
perpendiculur
to the plane of the surface. The electric flux
our
of the closed cube surface is
Model:
The electric flux "flows"
our
of a closed surface around a region of space containing a net
Please refer to Figure Ex27.4. Let A be the area in
m2
of each of the
six
faces of the cube.
The electric flux is defined as
=
€.A
=
EAcose, where
8
is the angle between the electric field and a
(Pa", =
(20 N /C+ 20 N
/
C+
10
N
/C)AcosO"
=
(50A) N m'
/C
Similarly. the electric tlux
inro
the closed cube surface is
=
(15 N
/
C
+
15 N
/
C
+
/
C)Acos180°
=
(UA)
N m'
/
C
The net electric flux
is (50A)
N m'
/
C

(45A) N m'
/
C
=
(5A) N m'
/
C.
Since the net electric tlux
is
positive
(i.e.
. outward). the closed box contains a positive charge.
271
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Chapter 27
27.5.
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:
line
perpendicular
to the plane of the surface. The electric flux out of the closed cube surface is
Qoul
=
(10 N
/
C
+10 N
/
C+lON
/
C
+
5 N
/
C)AcosO"
=
(35A) N m2
/
C
Model:
The electric flux "flows"
out
of a closed surface around a region of space containing a net
The electric flux is defined as
Qe
=
E.
2
=
EAcose, where
8
is the angle between the electric field and a
Similarly, the electric flux into the closed cube surface is
Qin
=(15
N/C+20N/C)Acos(l8O0)=(35A)Nm2 /C
Hence,
Qoul
+
Qin
=
0 N
m2/C.
Since the net electric flux is zero, the closed box contains no charge.
27.6.
positive charge and
into
a closed surface surrounding a net negative charge.
Visualize:
Please refer to Figure Ex27.6. Let A be the area of each of the six faces of the cube.
Solve:
The electric flux is defined as
=
~?.i
=
EAcose, where 8is the angle between the electric field and a
line
to the plane of the surface. The electric flux out of the closed cube surface
is
Similarly, the electric flux into the closed cube surface is
Because the cube contains negative charge,
Qom
+
Qm
must be negative. This means
Qom
+
Q,"
+
Therefore,
Model:
The electric flux "flows"
out
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This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.
 Spring '08
 Medvedev

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