Download Chapter 22 1. The electric flux of a uniform field is given by Eq. 22

Chapter 22
1. The electric flux of a uniform field is given by Eq. 22-1b.
 E  E A  EA cos    580 N C    0.13m  cos0  31N m2 C
2
(a)
 E  E A  EA cos    580 N C    0.13m  cos 45  22 N m 2 C
2
(b)
 E  E A  EA cos    580 N C   0.13m cos90  0
2
(c)
4. (a) From the diagram in the textbook, we see that the flux outward through the hemispherical surface
is the same as the flux inward through the circular surface base of the hemisphere. On that surface all
of the flux is perpendicular to the surface. Or, we say that on the circular base, E A. Thus
E  E A   r2E .
(b) E is perpendicular to the axis, then every field line would both enter through the hemispherical
surface and leave through the hemispherical surface, and so
E  0 .
5. Use Gauss’s law to determine the enclosed charge.
E 
Qencl
o



 Qencl   E o  1840 N  m 2 C 8.85  1012 C2 N  m 2  1.63  108 C
7. (a) Use Gauss’s law to determine the electric flux.
E 
Qencl
o

1.0  106 C
8.85  10
12
C Nm
2
(b) Since there is no charge enclosed by surface A2,
2
 1.1 105 N  m 2 C
E  0 .
13. The electric field can be calculated by Eq. 21-4a, and that can be solved for the magnitude of the
charge.
Ek
Q
r
2
 Q
Er 2
k
 6.25  10

2

N C 3.50  102 m
8.988  10 N  m C
9
2
2

2
 8.52  1011 C
8
This corresponds to about 5  10 electrons. Since the field points toward the ball, the charge must be
11
negative. Thus Q  8.52 10 C .
E
15. The electric field due to a long thin wire is given in Example 22-6 as
E
(a)
1 
2 0 R

1 2
4 0 R

 8.988  10 N  m C
9
2
2


2 7.2  106 C m
5.0 m 
1 
2 0 R
.
  2.6  10
4
N C
4
N C
The negative sign indicates the electric field is pointed towards the wire.
E
(b)
1 
2 0 R

1 2
4 0 R

 8.988  10 N  m C
9
2
2


2 7.2  106 C m
1.5m 
  8.6  10
The negative sign indicates the electric field is pointed towards the wire.
18. See Example 22-3 for a detailed discussion related to this problem.
(a) Inside a solid metal sphere the electric field is 0.
(b) Inside a solid metal sphere the electric field is 0.
(c) Outside a solid metal sphere the electric field is the same as if all the charge were concentrated at
the center as a point charge.
E 
1
Q
4 0 r 2

 8.988  10 N  m C
9
2
2

5.50  10 C   5140 N C
6
 3.10 m 2
The field would point towards the center of the sphere.
(d) Same reasoning as in part (c).
5.50  10 C 
1 Q
9
2
2 
E 

8.988

10
N

m
C
 772 N C


4 0 r 2
8.00 m 2
6
The field would point towards the center of the sphere.
(e) The answers would be no different for a thin metal shell.
(f) The solid sphere of charge is dealt with in Example 22-4. We see from that Example that the
E 
field inside the sphere is given by
have these results for the solid sphere.
Q
1
4 0 r03
r.
Outside the sphere the field is no different. So we

E  r  0.250 m   8.988  109 N  m 2 C2

E  r  2.90 m   8.988  109 N  m 2 C2

E  r  3.10 m   8.988  10 N  m C
9
2

2
E  r  8.00 m   8.988  109 N  m 2 C2
6
 5.50  10 C  0.250 m   458 N C
 3.00 m 3

5.50  106 C

5.50  106 C
 3.00 m 3
 3.10 m 2
 5.50  10
6
 3.10 m 
C
2
 2.90 m  
5310 N C
 5140 N C
 772 N C
All point towards the center of the sphere.
21. (a) Consider a spherical gaussian surface at a radius of 3.00 cm. It encloses all of the charge.
 E dA  E  4 r   
2
E
1 Q
4 0 r
2
Q

0

 8.988  109 N  m 2 C2

5.50  106 C
 3.00  10 m 
2
2
 5.49  107 N C, radially outward
(b) A radius of 6.00 cm is inside the conducting material, and so the field must be 0. Note that there
6
must be an induced charge of 5.50  10 C on the surface at r = 4.50 cm, and then an induced charge
6
of 5.50  10 C on the outer surface of the sphere.
(c) Consider a spherical gaussian surface at a radius of 3.00 cm. It encloses all of the charge.
 E dA  E  4 r   
2
E
1 Q
4 0 r
2

Q

0
 8.988  10 N  m C
9
2
2

5.50  106 C
 30.0  10 m 
2
2
 5.49  105 N C, radially outward
0  r  r1 ,
27. (a) In the region
a gaussian surface would enclose no charge. Thus, due to the spherical
symmetry, we have the following.
 E dA  E  4 r  
2
(b) In the region
Qencl
0
0  E  0
r1  r  r2 ,
only the charge on the inner shell will be enclosed.


2
 E dA  E 4 r 
(c) In the region
r2  r,
Qencl
0

 1 4 r12
0
 E
 1r12
 0r 2
the charge on both shells will be enclosed.
 E dA  E  4 r  
2
Qencl
0
 1 4 r12   2 4 r22

0
 1r12   2 r22
 E
 0r 2
 r 2   2 r22  0 .
r  r,
(d) To make E  0 for 2
we must have 1 1
This implies that the shells are of opposite
charge.
 0.
Q  4 1r12
r  r  r2 ,
(e) To make E  0 for 1
we must have 1
Or, if a charge
were placed at the
center of the shells, that would also make E  0.
33. We follow the development of Example 22-6. Because of the
symmetry, we expect the field to be directed radially outward (no fringing
effects near the ends of the cylinder) and to depend only on the
perpendicular distance, R, from the symmetry axis of the shell. Because of
the cylindrical symmetry, the field will be the same at all points on a
gaussian surface that is a cylinder whose axis coincides with the axis of the
¬
R
R0
shell. The gaussian surface is of radius r and length l. E is perpendicular to this surface at all points. In
order to apply Gauss’s law, we need a closed surface, so we include the flat ends of the cylinder. Since
E is parallel to the flat ends, there is no flux through the ends. There is only flux through the curved
wall of the gaussian cylinder.
 E dA  E  2 Rl  
(a) For
R  R0 ,
Qencl
0

 Aencl
0
 E
 Aencl
2 0 Rl
the enclosed surface area of the shell is
Aencl  2 R0 l .
E
(b) For
R  R0 ,
(c) The field for
substitute
 Aencl  2 R0 l  R0


, radially outward
2 0 Rl
2 0 R l
0R
the enclosed surface area of the shell is
R  R0
Aencl  0,
and so E  0 .
due to the shell is the same as the field due to the long line of charge, if we
  2 R0 .
34. The geometry of this problem is similar to Problem 33, and so
we use the same development, following Example 22-6. See the
solution of Problem 33 for details.
 E dA  E  2 Rl  
(a) For
R  R0 ,
Qencl
0

 EVencl
0
 E
¬
R0
R
 EVencl
2 0 Rl
the enclosed volume of the shell is
Vencl   R02 l .
 EVencl
 E R02
E

, radially outward
2 0 Rl
2 0 R
(b) For
R  R0 ,
the enclosed volume of the shell is
E
Vencl   R 2 l .
 EVencl
 R
 E , radially outward
2 0 Rl
2 0
36. The geometry of this problem is similar to Problem 33, and so we use the same development,
following Example 22-6. See the solution of Problem 33 for details. We choose the gaussian cylinder to
be the same length as the cylindrical shells.
 E dA  E  2 Rl  
Qencl
0
 E
Qencl
2 0 Rl
E
(a) At a distance of R  3.0cm, no charge is enclosed, and so
Qencl
2 0 Rl
 0.
(b) At a distance of R  7.0cm, the charge on the inner cylinder is enclosed.
E
Qencl
2 0 Rl

2 Qencl
4 0 Rl
 0.88  10 C
6

 2 8.988  10 N  m C
9
2
2
  0.070 m5.0 m  4.5  10
4
N C
The negative sign indicates that the field points radially inward.
(c) At a distance of R  12.0cm, the charge on both cylinders is enclosed.
E
Qencl
2 0 Rl

2 Qencl
4 0 Rl

 2 8.988  109 N  m 2 C2
   0.120 m 5.0 m 
1.56  0.88  106 C
 2.0  104 N C
The field points radially outward.
46. Because the slab is very large, and we are considering only distances from the slab much less than its
height or breadth, the symmetry of the slab results in the field being perpendicular to the slab, with a
constant magnitude for a constant distance from the center. We assume that
field points away from the center of the slab.
E  0
and so the electric
(a) To determine the field inside the slab, choose a cylindrical
gaussian surface, of length 2x  d and cross-sectional area A. Place it
so that it is centered in the slab. There will be no flux through the
curved wall of the cylinder. The electric field is parallel to the surface
area vector on both ends, and is the same magnitude on both ends.
Apply Gauss’s law to find the electric field at a distance
the center of the slab. See the first diagram.
x d
1
2
 E dA   E dA   E dA   E dA  0 
ends
Einside 
x
; x  12 d
0
side
ends
E
E
x
1
2
x
d
from
Qencl
0
 2EA 
  2 xA

0
1
2
d
(b) Use a similar arrangement to
determine the field outside the
slab. Now let 2 x  d . See the
second diagram.
E
E
x
x
1
2
 E dA   E dA 
ends
2EA 
  dA

0
Qencl
0
d
1
2
d

Eoutside 
d
; x  12 d
2 0
Notice that electric field is continuous at the boundary of the slab.
56.
(a) The flux through any closed surface containing the total charge must be the same, so the flux
through the larger sphere is the same as the flux through the smaller sphere,
235 N m2 /C .
(b) Use Gauss’s law to determine the enclosed charge.

Qencl
0

 Qencl   0  8.85  1012 C2 N m 2
 235 N m /C  
2
2.08  109 C