Download Chapter22Solutionstoassignedproblems

CHAPTER 22: Gauss’s Law
Solutions to Assigned Problems
2.
Use Eq. 22-1b for the electric flux of a uniform field. Note that the surface area vector points
radially outward, and the electric field vector points radially inward. Thus the angle between the two
is 180.

 E  E A  EA cos   150 N C  4 RE2 cos180  4 150 N C  6.38  106 m

2
 7.7  1016 N m2 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,
2
E A. Thus  E  E A   r E .
(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 .
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
 8.52  1011 C
2
This corresponds to about 5  10 electrons. Since the field points toward the ball, the charge must
8
be negative. Thus Q  8.52 1011 C .
15. The electric field due to a long thin wire is given in Example 22-6 as E 
(a) E 
1 

1 2

 8.988  10 N  m C
9
2
2
1 
2 0 R
.

  2.6  10

  8.6  10

2 7.2  106 C m

2 7.2  106 C m
2 0 R 4 0 R
5.0 m 
The negative sign indicates the electric field is pointed towards the wire.
(b) E 
1 

1 2

 8.988  109 N  m2 C2
2 0 R 4 0 R
1.5m 
The negative sign indicates the electric field is pointed towards the wire.
4
N C
4
N C
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.
5.50  106 C
1 Q
9
2
2
E 
 8.988  10 N  m C
 5140 N C
4 0 r 2
 3.10 m 2




© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37
Physics for Scientists & Engineers with Modern Physics, 4 th Edition
Instructor Solutions Manual
The field would point towards the center of the sphere.
(d) Same reasoning as in part (c).
5.50  106 C
1 Q
9
2
2
E 
 8.988  10 N  m C
 772 N C
4 0 r 2
8.00 m 2
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
1 Q
field inside the sphere is given by E 
r. Outside the sphere the field is no different.
4 0 r03
So we have these results for the solid sphere.
6
9
2
2 5.50  10 C
E  r  0.250 m   8.988  10 N  m C
 0.250 m   458 N C
 3.00 m 3






6

 5.50  10 C  2.90 m   5310 N C

 5.50  10
E  r  2.90 m   8.988  109 N  m 2 C2
E  r  3.10 m   8.988  109 N  m 2 C 2

E  r  8.00 m   8.988  109 N  m 2 C2
 3.00 m 3
6
 3.10 m 

C
 5140 N C
2
5.50  106 C
 772 N C
 3.10 m 2
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.
Q
2
 E dA  E 4 r  

E
1 Q
4 0 r 2

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
6
there must be an induced charge of 5.50  10 C on the surface at r = 4.50 cm, and then an
6
induced charge 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.
Q
2
 E dA  E 4 r  

E
1 Q
4 0 r
2


0
 8.988  109 N  m 2 C2

5.50  106 C
 30.0  10 m 
2
2
 5.49  105 N C, radially outward
27. (a) In the region 0  r  r1 , a gaussian surface would enclose no charge. Thus, due to the spherical
symmetry, we have the following.
Q
2
 E dA  E  4 r   encl0  0  E  0
(b) In the region r1  r  r2 , only the charge on the inner shell will be enclosed.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
38
Gauss’s Law
Chapter 22


2
 E dA  E 4 r 
Qencl
0

 1 4 r12
0
 E
 1r12
 0r 2
(c) In the region r2  r, the charge on both shells will be enclosed.


2
 E dA  E 4 r 
Qencl
0

 1 4 r12   2 4 r22
0
 E
 1r12   2 r22
 0r 2
(d) To make E  0 for r2  r, we must have  1r12   2 r22  0 . This implies that the shells are of
opposite charge.
(e) To make E  0 for r1  r  r2 , we must have  1  0 . Or, if a charge Q  4 1r12 were placed
at the center of the shells, that would also make E  0.
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.
Q
V
V
 E dA  E  2 Rl   encl0  E 0encl  E  2E 0enclRl
¬
R0
R
(a) For R  R0 , the enclosed volume of the shell is
Vencl   R02 l .
E
 EVencl
 R2
 E 0 , radially outward
2 0 Rl
2 0 R
(b) For R  R0 , the enclosed volume of the shell is Vencl   R 2 l .
E
 EVencl
 R
 E , radially outward
2 0 Rl
2 0
44. Due to the spherical symmetry of the problem, Gauss’s law using a sphere of radius r leads to the
following.
Q
Q
2
 E dA  E  4 r   encl0  E  4encl0r 2
(a) For the region 0  r  r1 , the enclosed charge is 0.
E
Qencl
4 0 r 2
 0
(b) For the region r1  r  r0 , the enclosed charge is the product of the volume charge density times
the volume of charged material enclosed. The charge density is given by    0
r1
. We must
r
integrate to find the total charge. We follow the procedure given in Example 22-5. We divide
the sphere up into concentric thin shells of thickness dr, as shown in Fig. 22-14. We then
integrate to find the charge.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
39
Physics for Scientists & Engineers with Modern Physics, 4 th Edition
Qencl
E
r
r
1
1
Instructor Solutions Manual

r

   E dV    0 1 4  r  dr   4 r1 0  r dr   2 r1 0 r 2  r12
r
r
r
Qencl
4 0 r 2


2 r1 0 r 2  r12
4 0 r 2


 0 r1  r 2  r12 
2 0 r 2
(c) For the region r  r0 , the enclosed charge is the total charge, found by integration in a similar
fashion to part (b).
r0
r0
r1
E
Qencl
4 0 r 2


1

2 r1 0 r  r
2
0
2
1
4 0 r 2

 0 r1  r  r12 
2
0
2 0 r 2
(d) See the attached graph. We
have chosen r1  12 r0 . Let
E0  E  r  r0  

r1

4  r  dr   4 r1 0  r dr   2 r1 0 r02  r12
r
r
Qencl    E dV    0
1.00
0 r1  r02  r12 
0.75
.
E /E 0
2 0 r02
The spreadsheet used for this
problem can be found on the
Media Manager, with filename
“PSE4_ISM_CH22.XLS,” on tab
“Problem 22.44d.”
0.50
0.25
0.00
0.0
0.5
1.0
1.5
2.0
r /r 0
52. (a) We use Gauss’s law for a spherically symmetric charge distribution, and assume that all the
charge is on the surface of the Earth. Note that the field is pointing radially inward, and so the
dot product introduces a negative sign.
 E  dA   E  4 r   Q
2
encl
Qencl  4 0 ER
2
Earth

0 

 150 N C  6.38  106 m
8.988  10
9
Nm C
2
2


2
 6.793  105 C  6.8  105 C
(b) Find the surface density of electrons. Let n be the total number of electrons.
Q
ne
  

A
A
n
A

Q
eA

2
4 0 EREarth

2
e 4 REarth


0E
e
8.85  10 C N m  150 N C 

1.60  10 C 
12
2
2
19
 8.3  109 electrons m 2
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
40
Gauss’s Law
Chapter 22
61. Consider this sphere as a combination of two spheres. Sphere 1 is a solid sphere of radius r0 and
charge density  E centered at A and sphere 2 is a second sphere of radius r0 / 2 and density   E
centered at C.
(a) The electric field at A will have zero contribution from sphere 1 due to its symmetry about point
A. The electric field is then calculated by creating a gaussian surface centered at point C with
radius r0 / 2.
 E  dA 
qenc
0
 E  4  r
1
2 0

2

   E  43   12 r0 3
0

E
 E r0
6 0
Since the electric field points into the gaussian surface (negative) the electric field at point A
points to the right.
(b) At point B the electric field will be the sum of the electric fields from each sphere. The electric
field from sphere 1 is calculated using a gaussian surface of radius r0 centered at A.
 r03   E 
 r
 E1  E 0
0
0
3 0
At point B the field from sphere 1 points toward the left. The electric field from sphere 2 is
calculated using a gaussian surface centered at C of radius 3r0 / 2.
 E1  dA 
qenc
 E1  4 r0 2 
4
3
   E  43   12 r0 3
 E r0
0
0
54 0
At point B, the electric field from sphere 2 points toward the right. The net electric field is the
sum of these two fields. The net field points to the left.
E
2
 dA 
qenc
E  E1  E2 
 E2  4  r
3
2 0

2

 E2  
 E r0   E r0 17  E r0


.
3 0
54 0
54 0
62. We assume the charge is uniformly distributed, and so the field of the pea is that of a point charge.
1 Q
E r  R 

4 0 R 2

 
Q  E 4 0 R 2  3  106 N C 4 8.85  10 12 C 2 N m 2
  0.00375 m 
2
 5  10 9 C
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41