Solucionario Clayton R.Paul (Eletromagnetismo para Engenheiros)

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Solucionario 
 
Eletromagnetismo para Engenheiros 
 
Clayton R. Paul 
1-1
Chapter 1 
Problem Solutions 
111. .
(a) 25 10 250 10 2504 3× = × =Ω Ω Ωk
(b) 0 035 10 35 10 3504 2. .× = × =Ω Ω Ω
(c) 0 00045F 450 10 4506. = × =− F Fµ
(d) 0 003 10 0 3 107 9. .× = × =− −F F 0.3nF
(e) 0 005 10 50 10 502 6. × = × =− −H H Hµ
112. .
(a) 30 12in 254cm 1 1 48 28miles 5280ft
1mile 1ft 1in
m
100cm
km
1000m
km× × × × × =. .
(b) 1 12in 1000mils 12 000ft
1ft 1in
mils× × = ,
(c) 100yds 12in 2 54cm 1 9144× × × × =3ft
1yd 1ft 1in
m
100cm
m. .
(d) 5mm 100cm
254cm
1000 19685mils× × × × =1m
1000mm 1m
1in mils
1in.
.
(e) 20 100cm
254cm
1000 0 7874µ
µ
m 1m
10 m 1m
1in mils
1in
mils6× × × × =.
.
(f) 880yds 12in 2 54cm 1 804 67× × × × =3ft
1yd 1ft 1in
m
100cm
m. .
121. .
(a) λ = × =3 10 33333
8 m s
90 Hz
km = 2,071.2 miles. 
(b) λ = × =3 10 300
8 m s
10 Hz
km = 186.41 miles3
(c) λ = ×
×
=
3 10 85714
8 m s
350 10 Hz
m = 0.533 miles3 .
(d) λ = ×
×
=
3 10
2
250
8 m s
1 10 Hz
m = 820.2 ft6.
(e) λ = ×
×
=
3 10
35
8 57
8 m s
10 Hz
m = 28.12 ft6 .
1-2 
(f) λ = ×
×
=
3 10
110
2 73
8 m s
10 Hz
m = 8.95 ft6 . 
(g) λ = ×
×
=
3 10
335
89 55
8 m s
10 Hz
cm = 2.94 ft6 . 
(h) λ = ×
×
=
3 10
6
5
8 m s
10 Hz
cm = 1.97 in9 
(i) λ = ×
×
=
3 10
45
6 67
8 m s
10 Hz
mm = 262.5 mils9 . 
12 2. . 
(a) 
( )
λ
λ λ
λ
=
×
= ×
× × × ∴ = =
3 10 5 10
5 10 0 0161 1
62
8
6
6
m s
60 Hz
m 50 miles = 80.46 km
80.46 10 m = ? m3
length
1 244 344 1 24 34
? .
 
(b) 
( )
λ
λ
λ
=
×
×
=
× ∴ =
3 10 600
152 4 600 0 254
8 m s
500 10 Hz
m 500feet = 152.4 m
m = ? m
3
length
. ? .124 34 124 34
 
(c) 
( )
λ
λ
λ
=
×
×
=
× ∴ =
3 10 2 73 5
137 2 73 05
8 m s
110 10 Hz
m 4 feet = 1.37 m
m = ? m
6
length
. .
. . ? .123 124 34
 
(d) 
( )
λ
λ
λ
=
×
×
=
× ∴ =
15 10 75
508 75 0 677
8. .
. . ? .
m s
2 10 Hz
cm 2 inches = 5.08 cm
cm = ? cm
9
length
1234 124 34
 
 
 
 
1-3 
12 3. . 
(a) 
β ωβ λ
µ φ β
= × = = = × = =
= = =
−2 2 10 1 2 856 10 285 6
10 5 66
2 8. , . , .
.
rad
m
MHz, m
s
m,
s, = rad = 3781.5o
f v v
f
T d
v
d
 
(b) 
β ωβ λ
φ β
= = = = × = =
= = =
75 4 3 2 5 10 833
4 0 41 7 66
8. , . , .
. .
rad
m
GHz, m
s
mm,
inches = 0.102 m, ns, = rad = 438.92o
f v v
f
T d
v
d
 
(c) 
β ωβ λ
φ β
= = = = × = =
= = =
315 150 299 10 1995
20 20 4 19 2
8. , . , .
. .
rad
m
MHz, m
s
m,
feet = 6.1m, ns, = rad = 1100.2o
f v v
f
T d
v
d
 
(d) 
β ωβ λ
φ β
= × = = = × = =
= = =
−0126 10 3 15 10 50
50 0 54 1014
3 8. , . ,
. .
rad
m
kHz, m
s
km,
miles = 80.5 km, ms, = rad = 580.9o
f v v
f
T d
v
d
 
 2-1 
Chapter 2 
Problem Solutions 
211. . 
F m d1
2
= ω θsin , F mg2 =
sin
cos
θ
θ
, F F1 2= , ∴ =cosθ
ω
g
d2
, ω π π= × =2 2rpm
60
, 
( )cos
.
.
.θ
π
=
×
=
9 78
2 05
052 , ∴ =θ 603.
o 
 
 
 
 
 
 
 
 
 
212. . 
W 45o× = ×cos sin200 θ , sin cosθ = W
200
45o , ∴ =θ 813. o , 
GS + Wsin45o = 200cosθ , GS = 169.71 mi
hr
 
 
 
 
 
 
 
 
�
�
�
d
m
F2F1
mg
�
W
GS
200 mi/hr
45°
2-2 
213. . 
4 3sinθ = , sinθ = 3
4
, ∴ =θ 48 6. o 
 
 
 
 
 
 
214. . 
F N mv
r
= =sinθ
2
, − = =W N mgcosθ , ∴ =tanθ v
rg
2
, 
v = × × =60 5280 1 88mi
hr
ft
1mi
hr
3600s
ft
s
, ∴ =tan .θ 016 , ∴ =θ 916. o 
 
 
 
 
 
2 31. . 
C A B= + , 
( ) ( )∴ = • = + • + = • + • + • = + +C A B AB2 2 22 2C C A B A B A A B B A B cosα 
But α θ= −180o AB and ∴ = −cos cosα θ AB 
∴ = + −C A B AB AB
2 2 2 2 cosθ which is the law of cosines. 
 
 
 
 
�
3 mi/hr
4 mi/hr
N
�
�AB
C
B
A
N
W
—W
F
N
�
2-3 
2 41. . 
A B C+ + = 0 , ( )A A B C A A A B A C× + + = = × + × + × =0 0
0
123 
( )A B a a× = − = − −AB ABC n o C nsin sinα θ180 where an is a unit normal into the 
page. Also A C a a× = =AC ACB n B nsin sinα θ . ∴ =AB ACC Bsin sinθ θ giving the 
law of sines: B C
B Csin sinθ θ
= . Similarly for ( )B A B C× + + = 0 and 
( )C A B C× + + = 0 . 
 
 
 
 
 
 
2 4 2. . 
(a) ( )A B• gives a scalar which cannot be “crossed with” a vector. (c) ( )A B• gives a 
scalar which cannot be “dotted with” a vector. (d) ( )B C• gives a scalar which cannot be 
added to a vector. 
2 51. . 
(a) ( ) ( ) ( )( )A a a a a a a= − + − − + − − = − +3 0 4 2 5 4 3 6 9x y z x y z 
(b) ( ) ( ) ( )A = + − + =3 6 9 11222 2 2 . m 
(c) a A a a aA x y zA
= = − +0 27 053 0 8. . . 
2 5 2. . 
(a) A B a a a+ = + −3 4 3x y z , (b) B C a a a− = − + −2 2 3x y z , (c) 
A B C a a a+ 3 - 2 = − + −x y z8 9 , (d) A = + + =2 3 1 374
2 2 2 . , (e) 
( )a B a a a a a aB x y z x y zB= = + − = + −16 2 41 0 41 0 82. . . , (f) A B• = 7 , (g) B A• = 7 , 
(h) B C a a a× = − − −x y z7 4 , (i) C B a a a× = + +x y z7 4 , (j) A B C• × = −19
C
C
B
A
B
A
A
�B
�B
�C �C
�B�C
2-4 
2 5 3. . 
(a) A B A B• = − − = − =2 2 3 3 cosθ . Therefore B A B
A
cos .θ = • = =3
14
0 8 , 
(b) cos .θ θ= • = − ⇒ =A B
A B
3
14 6
1091o , (c) 
unit vector = ×
×
=
− − −
+ +
= − − −
A B
A B
a a a
a a ax y z x y z
7 5
1 49 25
0115 0808 0 577. . . . 
2 5 4. . 
A B• = + − =α 2 3 0 ∴ =α 1 
2 5 5. . 
( ) ( ) ( )A B a a a× = − − + + + − =18 3 3 9 2 0β α β αx y z . Hence α = −3 and β = −6 . 
2 5 6. . 
A B a a a× = − − +14 9x y z gives a vector that is perpendicular to the planes containing 
both A and B and hence is perpendicular to both A and B. The length of this vector is 
( ) ( ) ( )− + − + =14 9 1 16 672 2 2 . . Hence 
( )C a a a a a a= − − + = − − +1016 67 14 9 8 4 54 0 6. . . .x y z x y z . Check that A C• = 0 and 
B C• = 0 . 
2 5 7. . 
( ) ( ) ( )B C a a a× = − + − + −B C B C B C B C B C B Cy z z y x z x x z y x y y x z so that 
( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )
A B C a
a
a
× × = − − −
+ − − −
+ − − −
A B C B C A B C B C
A B C B C A B C B C
A B C B C A B C B C
y x y y x z z x x z x
z y z z y x x y y x y
x z x x z y y z z y z
 
( ) ( ) ( )
( )
B A C a a
a
• = + + + + +
+ + +
B A C A C A C B A C A C A C
B A C A C A C
x x x y y z z x y x x y y z z y
z x x y y z z z
 
( ) ( ) ( )
( )
C A B a a
a
• = + + + + +
+ + +
C A B A B A B C A B A B A B
C A B A B A B
x x x y y z z x y x x y y z z y
z x x y y z z z
 
Matching components we find that ( ) ( ) ( )A B C B A C C A B× × = • − • . 
2-5 
2 5 8. . 
( ) ( ) ( )A B a a a× = − + − + −A B A B A B A B A B A By z z y x z x x z y x y y x z so that 
( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )
A B C a
a
a
× × = − − −
+ − − −
+ − − −
A B A B C A B A B C
A B A B C A B A B C
A B A B C A B A B C
z x x z z x y y x y x
x y y x x y z z y z y
y z z y y z x x z x z
 
Comparing terms to the result in the previous problem we see that 
( ) ( )A B C A B C× × ≠ × × . 
2 5 9. . 
The distance is ( )( ) ( ) ( )( )D = − − + − − + − − =3 1 1 2 5 4 10 2962 2 2 . . By integration we 
integrate D dlP
P
= ∫
1
2 . A straight line between the two points is governed by y x= − +3
4
5
4
 
and z x= −9
4
7
4
. Hence dldx dx= + −

 +



 =1
3
4
9
4
2 574
2 2
. and 
D dx
x
x
= ∫ =
=−
=
2 574 10 296
1
3
. . . 
2 510. . 
The surface is drawn below and lies in the yz plane at x=1. Hence the surface area is 
( )A dydz z dz
z
z
y
y z
z
z
= ∫ ∫ = −∫ =
=
=
=
= −
=
=
1
2
1
2 1
1
2
2 2 1 . Directly it is the area of a triangle of height 1 and 
base 2 or ( )( )A = =1
2
2 1 1. 
 
 
 
 
 
 
(1, 3, 2)y = 2z – 1
(1, 3, 1)(1, 1, 1)
y
z
x
2-6 
2 61. . 
Drawing the rectangular and cylindrical coordinate system axes as shown below we see 
that x r= cosφ , y r= sin φ , and z z= . From this we form 
( )x y r2 2 2 2 2
1
+ = +cos sinφ φ
1 244 344
 and hence r x y= +2 2 . Similarly we form 
y
x
r
r
= =
sin
cos
tanφφ φ . 
 
 
 
 
 
 
 
2 6 2. . 
Draw the coordinate system and use the right-hand rule. 
2 6 3. . 
Drawing the vector in the xy plane as shown below shows that A A Ax r= −cos sinφ φφ 
and A A Ay r= +sin cosφ φφ . 
 
 
 
 
 
 
 
 
P(r, �, z)
�
y
z
z
x
r
r
�
�
Ay
Ax
Ar
A�
y
x
2-7 
2 6 4. . 
At point P, φ π= =
3
60o . Hence Ax = +2 3cos sinφ φ =3.598 and 
Ay = −2 3sin cosφ φ =0.232 and Bx = −4 6cos sinφ φ =-3.196 and 
By = +4 6sin cosφ φ =6.464. Directly in cylindrical coordinates 
A B• = − − = −8 18 2 12 . In rectangular coordinates 
( ) ( )A B• = × − + × − = −3598 3196 0 232 6 464 2 12. . . . . 
2 6 5. . 
From problem 2.6.1 x r= cosφ , y r= sin φ , z=z. At P1 2 2 1, ,
π


 , x y z1 1 10 2 1= = =, , 
and at P2 3 3
2= −

, ,
π , x y z2 2 215 2 598 2= = = −. , . , . Hence the distance between the two 
points is ( ) ( ) ( )D x x y y z z= − + − + − =2 1 2 2 1 2 2 1 2 3407. . 
2 6 6. . 
The surface is 1/6 of the surface of a cylinder of length 4-1=3 and radius 2. Hence the 
surface area is ( )S = × × =2 2 3 6 2π π . By direct integration we have 
( )S r d dr
dsz r
= ∫ =∫ =
= =1
4
0
3
2 2φ π
φ
π
1 24 34 . 
2 6 7. . 
The volume is 1/6 of the volume of a cylinder of radius 2 minus the volume of a cylinder 
of radius 1 or ( ) ( )( )V = × − × =16 2 1 1 1 22 2π π π . By direct integration, 
V rdrd dz
z dvr
= ∫ ∫ ∫ =
= = =0
1
0
3
1
2
2φ
π
φ π124 34 . 
2 71. . 
Drawing the rectangular and spherical coordinate system axes as shown below we see that 
x r= sin cosθ φ , y r= sin sinθ φ , and z r= cosθ . From this we form 
( )x y z r r r2 2 2 2 2 2 2 2 2 2
1
+ + = + + =sin cos sin cosθ φ φ θ
1 244 344
 and hence 
2-8 
r x y z= + +2 2 2 . Similarly we form y
x
r
r
= =
sin sin
sin cos
tanθ φ
θ φ φ and 
x y
z
r
r
2 2+
= =
sin
cos
tanθ
θ
θ . 
 
 
 
 
 
 
 
2 7 2. . 
Draw the coordinate system and use the right-hand rule. 
2 7 3. . 
Drawing the vector in the zx or zy plane as shown below shows that 
A A Az r= −cos sinθ θθ . The components parallel to the xy plane are A Ar sin cosθ θθ+ 
and the φ component, Aφ . Hence the x and y components of these are 
( )A A A Ax r= + −sin cos cos sinθ θ φ φθ φ and 
( )A A A Ay r= + +sin cos sin cosθ θ φ φθ φ . 
 
 
 
 
 
 
 
 
�
�
y
z
z
x
r sin �
r
z y
x
Ar sin �
A� sin �
Ar sin � + A� cos �
Ax
Ay
A� cos �
A�
A�
Ar cos � Ar
x, y plane
�
�
�
2-9 
2 7 4. . 
At point P, θ π= =2
3
120o and φ π= =
3
60o . Hence 
Ax = + −2 3sin cos cos cos sinθ φ θ φ φ =-0.75 and 
Ay = + +2 3sin sin cos sin cosθ φ θ φ φ =0.701 and Az = −2 3cos sinθ θ =-3.598. 
Similarly, Bx = 383. , By = 0 634. , and Bz = −3732. . Directly in spherical coordinates 
A B• = + − =8 6 3 11. In rectangular coordinates 
( ) ( ) ( )A B• = − × + × + − × − =0 75 383 0 701 0 634 3598 3732 11. . . . . . . 
2 7 5. . 
From problem 2.7.1 x r= sin cosθ φ , y r= sin sinθ φ , and z r= cosθ . At P1 2 2
2
3
, ,π π

 , 
x y z1 1 11 1732 0= − = =, . , and at P2 3 3 6
= −



, ,
π π , x y z2 2 22 25 1299 15= = − =. , . , . . 
Hence the distance between the two points is 
( ) ( ) ( )D x x y y z z= − + − + − =2 1 2 2 1 2 2 1 2 4 69. . 
2 7 6. . 
The surface is 1/8 of the surface of a sphere of radius 4. Hence the surface area is 
( )( )
S =
×
=
4 4
8 8
2π
π . By direct integration we have 
( )S r d d
dsr
= ∫ =∫ =
= =θ π
π
φ π
π
θ θ φ π
2
2
2
4 8sin1 2444 3444 . 
2 7 7. . 
The volume is ( )S r d d
dsr
= ∫ =∫ =
= =θ π
π
φ
π
θ φ θ
4
3 2
0
2
2 521sin .1 2444 3444 . 
2-10 
2 81. . 
F l•∫ = ∫ + ∫ − ∫
=− = =−
d xdx dy ydz
x y z
2 4
1
2
3
4
2
1
. Third integral with respect to z contains y. So we 
need to determine the equation of the path as y z= +1
3
11
3
. Hence the line integral 
becomes F l•∫ = ∫ + ∫ − +

∫ = −
=− = =−
d xdx dy z dz
x y z
2 4 1
3
11
3
7
21
2
3
4
2
1
. 
2 8 2. . 
W d xdx zdy dz
P
P
x y z
= •∫ = ∫ + ∫ + ∫
= = =
F l
1
2
2 3 4
0
1
0
1
0
2
. But along the path z y= 2 which when 
substituted into the second integral gives W=1+3+8=12J. 
2 8 3. . 
(a) F l•∫ = ∫ + ∫ − ∫
= = =
d xdx xydy ydz
P
P
x y z1
2
0
3
0
2
2
0
2 . Along this path x y= 3
2
 and y z= − + 2 . 
substituting these gives ( )F l•∫ = ∫ + ∫ − − +∫ =
= = =
d xdx y dy z dz
P
P
x y z1
2
0
3 2
0
2
2
0
3 2 29
2
. (b) The 
integral is the sum of the integrals along the two paths: 
( )F l•∫ = ∫ + ∫ − =∫ + ∫ + =

∫ − ∫
= = = = = =
d xdx xydy y dz xdx x y ydy ydz
P
P
x y z x y z1
2
0
0
0
0
2
0
0
3
0
2
0
0
2 0 2 3
2
=0+0+0+
9/2+8+0=25/2. Along the first segment of this path neither x nor y change and y=0. 
Hence the integral along this first path is zero. Along the second segment of the path, 
there is no change in z and we substitute the equation for the path, x y= 3
2
, into the y 
integration. 
2 8 4. . 
F l•∫ = ∫ + ∫ + ∫d rdr zrd dz
P
P
1
2
2 4φ . The two paths are sketched below. 
(a) ( ) ( )F l•∫ = =∫ + =∫ + ∫ = + − = −
= = =
d r dr z r d dz
P
P
r z1
2
2 0 0 4 0 0 12 12
0
0
0
0
3
0
φ
φ
 
(b) F l F l F l F l•∫ = •∫ + •∫ + •∫d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2
 
2-11 
( )F l•∫ = ∫ + =∫ + ∫ = + + =
=
=
=
d rdr z rd dz
P
P
r z1
3
2 3 4 8 0 0 8
0
8
4
4
3
3
φ
φ π
π
 
( ) ( )F l•∫ = =∫ + =∫ + ∫ = + − = −
=
=
=
d r dr z r d dz
P
P
r z3
4
2 8 8 4 0 0 12 12
8
8
4
4
3
0
φ
φ π
π
 
( )F l•∫ = ∫ + =∫ + ∫ = − + + = −
=
=
=
d rdr z rd dz
P
P
r z4
2
2 0 4 8 0 0 8
8
0
4
4
0
0
φ
φ π
π
 
F l F l F l F l•∫ = •∫ + •∫ + •∫ = − − = −d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2
8 12 8 12 
 
 
 
 
 
 
 
 
2 8 5. . 
F l•∫ = ∫ + ∫ − ∫d rdr rd zdz
P
P
1
2
2 φ . The two paths are sketched below. 
(a) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
2
1
0
0
2
 
( ) ( )F l•∫ = =∫ + =∫ − ∫ = + + =
= = =
d r dr r d zdz
P
P
r z1
0
0 2 0 0 0 2 2
0
0
0
0
2
0
φ
φ
 
( )F l•∫ = ∫ + ∫ − =∫ = + + =
= = =
d rdr rd z dz
P
P
r z0
2
0
3
0
0
0
0
2 0 9
2
0 0 9
2
φ
φ
 
y
z
x
P3(2, 2, 3)
P4(2, 2, 0)
P2(0, 0, 0)
P1(0, 0, 3)2-12 
F l F l F l•∫ = •∫ + •∫ = + =d d d
P
P
P
P
P
P
1
2
1
0
0
2
2 9
2
13
2
 
(b) F l F l F l F l•∫ = •∫ + •∫ + •∫d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2
 
F l•∫ = ∫ + ∫ − ∫ = + + =
=
=
=
d rdr rd dz
P
P
r z1
3
0
3
2
2
2
2
2 4 9
2
0 0 9
2
φ
φ π
π
 
( ) ( )F l•∫ = =∫ + =∫ − ∫ = + + =
=
=
=
d r dr r d dz
P
P
r z3
4
3 2 3 4 0 0 2 2
3
3
2
2
2
0
φ
φ π
π
 
( ) ( )F l•∫ = =∫ + =∫ − ∫ = − + = −
=
=
=
d r dr r d dz
P
P
r z4
2
3 2 3 4 0 3 0 3
3
3
2
0
0
0
φ π π
φ π
 
F l F l F l F l•∫ = •∫ + •∫ + •∫ = + − = −d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2 9
2
2 3 13
2
3π π 
 
 
 
 
 
 
 
 
 
2 8 6. . 
F l•∫ = ∫ + ∫ + ∫d rdr rd r d
P
P
1
2
2 3 2θ θ φsin . The two paths are sketched below. 
(a) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
2
1
0
0
2
 
y
z
x
P3(0, 3, 2)
P4(0, 3, 0)
3
3
2
P2(3, 0, 0)
P1(0, 0, 2)
2-13 
F l•∫ = ∫ + ∫ + ∫ = − + + = −
= = =
d rdr rd r d
P
P
r1
0
2 3 2 9 0 0 9
3
0
0
0
0
0
θ θ φ
θ φ
sin 
F l•∫ = ∫ + ∫ + ∫ = − + + =
= = =
d rdr rd r d
P
P
r0
2
2 3 2 9 0 0 9
0
3
2
2
0
0
θ θ φ
θ π
π
φ
sin 
F l F l F l•∫ = •∫ + •∫ = − + =d d d
P
P
P
P
P
P
1
2
1
0
0
2
9 9 0 
(b) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
2
1
3
3
2
 
( ) ( ) ( )F l•∫ = =∫ + =∫ + =∫ = + + =
= = =
d r dr r d r d
P
P
r1
3
2 3 3 3 2 3 0 9
2
0 9
23
3
0
2
2
2
θ θ φ π π
θ
π
φ π
π
sin 
( ) ( ) ( )F l•∫ = =∫ + =∫ + = 

∫ = + − = −
= = =
d r dr r d r d
P
P
r3
2
2 3 3 3 2 3
2
0 0 3 3
3
3
2
2
2
0
θ π φ π π
θ π
π
φ π
sin 
F l F l F l•∫ = •∫ + •∫ = − =d d d
P
P
P
P
P
P
1
2
1
3
3
2 9
2
3 3
2
π
π
π 
 
 
 
 
 
 
 
 
 
 
y
z
x
P3(0, 3, 0)
3
3
3
P2(3, 0, 0)
P(0, 0, 3)
P0(0, 0, 0)
2-14 
2 8 7. . 
F l•∫ = ∫ + ∫ + ∫d rdr r d r d
P
P
1
2
2 32 θ θ φsin . The two paths are sketched below. 
(a) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
0
1
3
3
0
 
( ) ( ) ( )F l•∫ = =∫ + =∫ + =∫ = + + =
= = =
d r dr r d r d
P
P
r1
3
2 2 2 3 2 0 2 0 2
2
2 2
0
4
2
2
θ θ φ π π
θ
π
φ π
π
sin 
F l•∫ = ∫ + ∫ + ∫ = − + + = −
= = =
d rdr r d r d
P
P
r3
0
2
0 2
4
4
2
2
2 3 2 0 0 2θ θ φ
θ π
π
φ π
π
sin 
F l F l F l•∫ = •∫ + •∫ = −d d d
P
P
P
P
P
P
1
0
1
3
3
0
2 2π 
(b) F l•∫ = ∫ + ∫ + ∫ = − + + = −
= = =
d rdr r d r d
P
P
r1
0
2
0 2
0
0
0
0
2 3 2 0 0 2θ θ φ
θ φ
sin 
 
 
 
 
 
 
 
2 91. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ + ∫∫d xdydz ydxdz zdxdy . 
Top: ( )
y x
z dxdy
=− =−
∫ =∫ =
1
1
1
1
1 4 . Bottom: ( )− ∫ = −∫ =
=− =−y x
z dxdy
1
1
1
1
1 4 . 
Right: ( )
z x
y dxdz
=− =−
∫ =∫ =
1
1
1
1
1 4 . Left: ( )− ∫ = −∫ =
=− =−z x
y dxdz
1
1
1
1
1 4 . 
Front: ( )
z y
x dydz
=− =−
∫ =∫ =
1
1
1
1
1 4 . Back: ( )− ∫ = −∫ =
=− =−z y
x dydz
1
1
1
1
1 4 . 
Total=4+4+4+4+4+4=24. 
y
z
x
P0(0, 0, 0)
P1(0, 0, 2) 2
45°
P3(0, 2 , )2
2-15 
 
 
 
 
 
 
 
 
2 9 2. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d xydydz yzdxdz xzdxdy . 
Top: ( )− ∫ =∫ = −
= =y x
x z dxdy
0
2
0
1
3 3 . Bottom: ( )
y x
x z dxdy
= =
∫ =∫ =
0
2
0
1
0 0 . 
Right: ( )
z x
y zdxdz
= =
∫ =∫ =
0
3
0
1
2 9 . Left: ( )− ∫ =∫ =
= =z x
y zdxdz
0
3
0
1
0 0 . 
Front: ( )
z y
x ydydz
= =
∫ =∫ =
0
3
0
2
1 6 . Back: ( )− ∫ =∫ =
= =z y
x ydydz
0
3
0
2
0 0 . 
Total=-3+0+9+0+6+0=12. 
 
 
 
 
 
 
 
 
 
y
z
x
(–1, 1, 1)
(–1, 1, –1)
y
z
x
2
1
3
2-16 
2 9 3. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d dz drdz rdrd2 3 22 φ φ φ . 
Top: − ∫ ∫ = −
= =φ
π
φ π
0
2
0
2
2 2rdrd
r
. Bottom: 
φ
π
φ π
= =
∫ ∫ =
0
2
0
2
2 2rdrd
r
. 
Right: 
z r
drdz
= =
∫ =

∫ =0
3
0
2
3
2
9φ π π . Left: ( )− ∫ =∫ =
= =z r
drdz
0
3
0
2
3 0 0φ . 
Front: ( )
z
r d dz
= =
∫ =∫ =
0
3 2
0
2
2 2 12φ π
φ
π
. 
Total=-2π+2π+9π+0+12π=21π. 
 
 
 
 
 
 
 
 
 
2 9 4. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d dz drdz zrdrd3 2 φ φ φ . 
Top: ( )− ∫ =∫ =
=− =φ π
π
φ π
2
2
0
2
4 8z rdrd
r
. Bottom: ( )
φ π
π
φ
=− =
∫ =∫ =
2
2
0
2
0 0z rdrd
r
. 
Right: 
z r
drdz
= =
∫ =

∫ =0
4
0
2
2
4φ π π . Left: − ∫ = −

∫ =
= =z r
drdz
0
4
0
2
2
4φ π π . 
Front: ( )
z
z r d dz
= =−
∫ =∫ =
0
4 2
2
2
2 48φ π
φ π
π
. 
Total=8π+0+4π+4π+48π=64π. 
z
3
2
y
x
2-17 
 
 
 
 
 
 
 
 
2 9 5. . 
A sketch of the surface is given below. 
F s•∫ = ∫∫ + ∫∫ − ∫∫d r d d r drd rdrd3 2sin sinθ φ θ θ φ φ θ . 
Front: ( )
θ
π
φ
π
θ φ θ π
= =
∫ =∫ =
0
2 3
0
2
2 4r d dsin . Bottom: 
φ
π
π φ π
= =
∫ ∫ =
0
2
0
2
2
2
2r drd
r
sin . 
Right: − ∫ =

∫ = −
= =θ
π
φ π θ π
0
2
0
2 2
2 2
rdrd
r
. Left: ( )
θ
π
φ θ
= =
∫ =∫ =
0
2
0
2
0 0rdrd
r
. 
Total=4 2 2
0 6
2
2 2
π π π π π+ − + = − .
z
4
2
y
x
2
y
x
z
2-18 
2 9 6. . 
A sketch of the surface is given below. 
F s•∫ = ∫∫ − ∫∫ + ∫∫d r d d r drd rdrd3 2 3sin sinθ φ θ θ φ φ θ . 
Front: ( )
θ
π
φ π
π
θ φ θ π
= =−
∫ =∫ =
0
2 3
2
2
3 27r d dsin . Bottom: − ∫ ∫ = −
=− =φ π
π
π φ π
2
2
0
3
2
2
9r drd
r
sin . 
Right: 
θ
π
φ π θ π
= =
∫ =

∫ =0
2
0
3 2
3
2
27
8
rdrd
r
. Left: − ∫ = −

∫ =
= =θ
π
φ π θ π
0
2
0
3 2
3
2
27
8
rdrd
r
. 
Total= 27 9 27
8
27
8
18 27
4
2 2 2
π π
π π
π
π
− + + = + . 
 
 
 
 
 
 
 
3
y
x
z
 3-1 
Chapter 3 
Problem Solutions 
311. . 
Q dv zrdzd drv
r z
= ∫∫∫ = ∫ ∫ ∫ =
=
=
=
ρ φ π
φ π
π
0
1
4
2
0
2
2
2
C 
312. . 
The surface is shown below. ( )Q ds xy z dxdys
x y
x
= ∫∫ = ∫ =∫ =
= =
− +
ρ
1
2
1
2 5
3 2 13C . 
 
 
 
 
 
 
313. . 
The problem is sketched below. A vector from the second charge to the first is 
( )( ) ( ) ( )( )R a a a a a a21 1 1 1 0 1 2 2 3= − − + − + − − = + +x y z x y z whose length is 
R21 14= . A unit vector pointing from the second charge to the first is a
R
21
21
21
=
R
. 
Coulomb’s law yields 
( )( )
F a a a a21
9
6 6
21
2 219 10
100 10 50 10
1718 0 859 2577= ×
× ×
= + +
− −
R
x y z. . . N . 
 
 
 
 
 
y
z
x
y = –2x + 5
(1, 1, 2) (1, 3, 2)
(2, 1, 2)
y
z
x
(–1, 0, 2)
Q2 = 50 �C
Q1 = 100 �C
(1,1, 1)
3-2 
314. . 
(a) F=0, (b) 
( )
( )F a= × ×
×
×
−
4 9 10
100 10
2
459
6 2
2 cos
o
z , 
(c) 
( )
( )F a a1
9
6 2
29 10
100 10
1
90= ×
×
=
−
x x , 
( )
( )F a a2
9
6 2
29 10
100 10
1
90= ×
×
=
−
y y , 
( )F a a3 905= +cos sinθ θx y , ( )F a a4 905= +sin cosθ θx y . But cosθ = 25 and 
sinθ = 1
5
. Hence F F F F F a a= + + + = +1 2 3 4 11415 11415. .x x . 
 
 
 
 
 
 
 
 
315. . 
The problem is sketched below. For the forces exerted on Q3 by the other two charges to 
be equal and oppositely directed, we must have 
( )( ) ( )( )
( )9 10
18 10 8 10
9 10
72 10 8 10
0 03
9
6 6
2
9
6 6
2×
× ×
= ×
× ×
−
− − − −
d d.
. Solving for d gives 
d=1cm. 
 
 
 
 
 
F2
F4
F3
F1
–1
�
�
�
�
1
y
x
5
5
1
2
4
3
d
3 cm
Q1 = 18 �C Q3 = –8 �C Q2 = 72 �C
–1 1
3-3 
316. . 
The charge of an electron is e = − × −16 10 19. C. Placing the positive charge on the left 
and the negative charge on the right, the force exerted on the electron by the positive 
charge is directed to the left and is 
( )( )
( )F1
9
6 19
2 2
129 10
35 10 16 10
10 10
504 10= ×
× ×
×
= ×
− −
−
−
.
. N . The force exerted on the 
electron by the negative charge is of the same magnitude and in the same direction so that 
the net force is 1 10 11× − N directed toward the positive charge. 
317. . 
The problem is sketched below. In order that the forces balance, the Coulomb force 
acting in the horizontal direction is ( )F
Q
lo
=
2
24 2π ε θsin
. The component of the 
restraining force along the string that is horizontally directed is T sinθ and the force of 
gravity acting downward on the charges is T mgcosθ = . Hence F mg= sin
cos
θ
θ
. 
Equating the two and solving gives ( )Q l mgo2 2 31 16= cos sinθ πε θ . 
 
 
 
 
 
3 21. . 
The problem is sketched below. The electric field due to the positive charge is 
( )E a a1
9 1
29 10 4
2 8125= × =Q y y, . . The electric field due to the negative charge is 
( )E a a2
9 2
29 10 2
22 500= − × = −Q z z, . Hence the total electric field is 
E E E a a= + = −1 2 2 8125 22 500, . ,y z
V
m
.
�
2l sin �
QF
mg
Q
T l
3-4 
 
 
 
 
 
 
 
3 2 2. . 
The problem is sketched below. The angle θ is θ = 60o . The distances from the triangle 
vertices to the center is, according to the law of cosines, ( )5 2 22 2 2 2= + −d d d cos θ 
giving d=2.887 m. The vector contributions are 
( )E Q
d
Q
do o
o
= + =
4
2
4
60 1082 2π ε π ε
cos kV
m
. 
 
 
 
 
 
 
 
3 2 3. . 
The problem is sketched below. (a) First we determine the electric field along the z axis. 
Superimposing the fields due to the two charges gives 
E a a a a=
−




−
+




=
−



 +




=
−






1
4
2
1
4
2
2
2 2
2
4
2 2 2 2
2
2 2π ε π ε π ε π εo
z
o
z
o
z
o
z
Q
z l
Q
z l
Ql z
z l z l
Ql z
z l
z
y
(0, 4, 2)
Q1 = 5 �C
Q2 = –10 �C
z = 2
y = 4
E1
E2 E
5 m 5 m
5 m
Q
—Q —Q
d d
d
� �
3-5 
(b) Now we determine the electric field along the y axis. Superposing the fields as shown 
gives E a a= −
+





 +




= −
+






2
4
1
4
2
4
4
4
2
2
2
2
1
2
2
2
3
2
Q
y l
l
y l
Ql
y l
o
z
o
zπ ε
π ε
θsin
1 244 344
. 
 
 
 
 
 
 
 
3.2.4 
The problem is sketched below. Divide the charge into chunks of charge, {dQ adl
dl
= ρ φ . 
At a distance d from the center and on a line perpendicular to the ring, the horizontal 
components cancel out leaving only the vertical components so that 
 ( )E a= ∫
=
1
4 20
2
π ε
ρ φ
α
φ
π
o
l
z
ad
R
cos 
where R d a= +2 2 and ( )cos α = d
R
. Substituting these gives 
 
( )
( )
E a
a
=
+
∫
=
+
>
=
1
4
1
2
0
2 2
3
20
2
2 2
3
2
π ε
ρ φ
ε
ρ
φ
π
o
l
z
o
l
z
ad
d a
d
ad
d a
z
 
At a large distance from the center, d a>> , this result reduces to 
 
E a
a
= >>
= >>
1
2
2
4
2
2
ε
ρ
π ρ
π ε
o
l
z
l
o
z
a
d
d a
a
d
d a
 
y
l
2 �
� �
E+
E+
E–
E–
E
z
z
y
l
2
3-6 
 
 
 
 
 
 
 
 
3.2.5 The problem is sketched below. The chunks of charge are dQ dr rds
ds
= ρ φ123 and 
again, by symmetry, the horizontal components cancel leaving only the vertical (z-
directed) components. Summing these contributions gives 
 
( )
E a
a
a
= ∫ ∫
+ +
= −
+






>
= >>
= =r
a
s
o
R
z
s
o
z
s
o
z
rd dr
d r
d
d r
d
d a
z
a
d
d a
0 0
2
2 2
1
2 2
2 2
2
2
4
1
2
1 0
4
2
ρ φ
π ε
ρ
ε
π ρ
π ε
φ
π
α
124 34 1 24 34
cos
 
 
 
 
 
 
 
 
 
 
E
R
z = d
a
y
z
x
�
�
�l ad�
E
R
z = d
�s rdrd�
a
r
y
z
x
�
�
3-7 
3.2.6 
The problem is sketched in the xy plane below. Using the results of Example 3.3 and 
superpositioning the fields gives (a) 
E a a a=
−




−
+




=
−






ρ
π ε
ρ
π ε
ρ
π ε
l
o
y
l
o
y
l
o
y
y l y l
l
y l
2
1
2
2
1
2
2
4
2
2
. 
Similarly the fields along the x axis become E a= −
+ +
2
2
4
2
4
2
2
2
2
ρ
π ε
θ
l
o
y
x l
l
x l
cos
1 24 34
. 
 
 
 
 
 
 
 
 
 
3.2.7 
The problem is sketched below. Place the strip in the xz plane centered on the origin. 
Divide the strip into infinite line charges with distribution ρ ρl sdz=
C
m
. Use the result 
in Example 3.3, equation (3.10). Accounting for symmetry, E a= ∫
=
2
20
2 ρ
πε
αs
oz
W
y
dz
R
cos
`
 
where R z d= +2 2 and cosα = d
R
. Hence, using the integral 
1 1
2 2
1
d z
dz
d
z
d+
∫ = 


−tan , ( )E a a= +∫ =




=
−
ρ
πε
ρ
πε
s
o z
W
y
s
o
y
d
d z
dz
d
W
d
1
22 20
2 1
`
tan . 
Er–
Er+
Er–
Er+l
2
l
2
R
R
y
y
x
x
—�
�
�
�
3-8 
 
 
 
 
 
 
 
3 31. . 
The electric field intensity vector is E V
d
= = 105 V
m
. The polarization vector is 
P D Eo= − ε . Substituting D Er o= ε ε gives 
( ) ( )P Eo r= − = × − × =−ε ε π
µ1 1
36
10 54 1 10 3899 5. . C
m2
. 
3.4.1 
The problem is sketched below. (a) the total charge enclosed is 
Q r drd d kr drd d ka
r
a
v
dv r
a
enc = ∫ ∫ =∫ ∫ ∫ =∫
= = = = = =0 0
2 2
0 0 0
2 3
0
4
φ
π
θ
π
φ
π
θ
π
ρ θ φ θ θ φ θ πsin sin1 244 344 . (b) Using 
Gauss’ law, ε ε π πo od E r Q kaE s•∫ = = =4 2 4enc giving E a= ka
ro
r
4
24ε
. (c) The charge 
enclosed is Q krenc = π
4 . Hence the electric field is E a= kr
o
r
2
4ε
. 
 
 
 
 
 
 
 
 
�
�
E–
E+
z
d
R
R
W
—z
z
y
W
2
—
W
2
�
v
 = kr
a
r dv
y
z
x
3-9 
3 4 2. . 
No. No closed surface can be found for which the electric field is perpendicularto all 
sides and hence no simplification of D s•∫ d can be obtained. 
3 4 3. . 
The problem is sketched below. Since D is directed in the z direction, there is no flux 
through the sides. Hence Gauss’ law gives 
( ) ( )Q d d z rrdrd z rrdrd a
r
a
ds r
a
ds
enc
top bottom
C= •∫ + •∫ = ∫ =∫ − ∫ =∫ = =
= = = =
D s D s123 123 123 123φ
π
φ
π
φ φ π π
0
2
0 0
2
0
3
4 0 8
3
64
3
. 
 
 
 
 
 
 
 
 
3 4 4. . 
For r b≥ ( )Q dv kr r drd d k b av r a
b
dv
enc = ∫ = ∫ ∫ ∫ = −
= = =
ρ θ φ θ π
φ
π
θ
π
0
2 2
0
2 22sin1 244 344 . By symmetry, 
the field is radially directed. Hence D s•∫ =d Qenc so that ( )ε π πo rE r k b a4 22 2 2= − . 
Thus 
( )
E
k b a
r
r
o
=
−
2 2
22ε
. For r a≤ , Er = 0 since no charge is enclosed. For a r b≤ ≤ , 
( )Q k r aenc = −2 2 2π . Hence ( )ε π πo rE r k r a4 22 2 2= − . Thus E k a
r
r
o
= −





2 1
2
2ε
. 
3.5.1 
The problem is sketched below. The work required to move q around the paths is 
W q d q xdx q ydy q zdz= − •∫ = − ∫ − ∫ − ∫E l . (a) 
z
4
Dz
a
y
x Dz
3-10 
W q zdz q ydy q zdz q ydy
z y z y
= − ∫ − ∫ − ∫ − ∫ =
= = = =0
1
0
2
1
0
2
0
0 . (b) 
W q zdz q ydy q xdx q xdx
z z x x
= − ∫ − ∫ − ∫ − ∫ =
= = = =0
1
1
0
0
1
1
0
0 . 
 
 
 
 
 
 
 
3.5.2 
The problem is sketched below. Applying superposition and equation (3.37) yields 
V Q
o
= −



 =4
1
2
1
3
15
π ε
kV . 
 
 
 
 
 
 
3.5.3 
The problem is sketched below. Applying superposition and equation (3.37) yields 
( ) ( ) ( ) ( )
V Q Q
o o
=
+ −
−





 + + −
−





 = − − = −
1
2 2
2
2 24
1
5 5 3
1
3 4
1
5 5 2
1
2
13 287 42 1478256 28
π ε π ε
, . . ,070V
y
z
x
(0, 2 m, 1 m)
(0, 2 m, 0)
(0, 0, 1 m)
x = 1
y = 2
z = 1
(0, 0, 0)
(0, 3, 2)
–
+
y = 3
Q = 10 �C
V
z
y
3-11 
 
 
 
 
 
 
 
 
3.5.4 
The problem is sketched below. Applying superposition and equation (3.37) yields 
V Q Q
o o
=
+
−





 +
+
−





 = − − = −1 2 2
2
2 24
1
5 2
1
2 4
1
5 3
1
3
28 287 42 7 28256 35 570
π ε π ε
, . , . , V 
 
 
 
 
 
 
 
 
3.5.5 
The problem is sketched below. Applying superposition and equation (3.37) yields 
V Q Q
o o
=
+
−





 +
+
−





 = − + =1 2 2
2
2 24
1
5 3
1
3 4
1
5 2
1
2
7 28256 28 287 42 21
π ε π ε
, . , . ,005V . 
 
 
(0, 5, 5)
–
+
y = 2
z = 3
V
Q2 = 5 �C
Q1 = 10 �C
z
y
(5)2 +
 (5 – 3
)2
29
=
(5 – 2)2 + (5)2 34=
(0, 0, 5)
–
+
y = 3y = –2
V
Q2 = 5 �CQ1 = 10 �C
z
y
52 + 22 52 + 32
3-12 
 
 
 
 
 
 
 
 
3.5.6 
The problem is sketched below. Applying superposition and equation (3.38) yields 
V
o o
=
+





 +
+





 = − + =
ρ
π ε
ρ
π ε
1
2 2
2
2 22
2
5 2 2
3
5 3
89145 119 30 477Vln ln , ,622 , . 
 
 
 
 
 
 
 
 
3.5.7 
The problem is sketched below. By the law of cosines, the distance from each charge to 
the center of the triangle is related to the side length as ( )l d d d o2 2 2 22 120= + − cos so 
that d l=
1732.
. Applying superposition and equation (3.42) yields 
V Q
l
Q
l
o
= 



= × =3
4
1732
4 677 10 31177V10
π ε
.
. , .
y
z
x
x = 3
Q1 = 5 �C
Q2 = –10 �C
(0, 5, 0)
z = 2
52 + 22
52 + 32
–
+
y = 2
z = 5
y = –3
V
�l = –10 �C/m �1 = 5 �C/m
z
y
52 + 32 52 + 22
3-13 
 
 
 
 
 
 
 
 
3.5.8 
The problem is sketched below. Using (3.44) gives 
V
ad
z a
a
z a
l
o
l
o
=
+
∫ =
+=
ρ φ
π ε
ρ
εφ
π
4 22 20
2
2 2
. The potential is only a function of z and hence 
the gradient is 
( ) ( )E a a a= − = − = − + = +
−
gradientV V
z
a
z
z a
a z
z a
z
l
o
z
l
o
z
∂
∂
ρ
ε
∂
∂
ρ
ε2 2
2 2
1
2
2 2
3
2
 which 
agrees with the results of Problem 3.2.4. 
 
 
 
 
 
 
 
 
 
 
120°
120°120°
l
ll
Q
Q Q
dd
d
R
a
y
z
x
�l ad�C
3-14 
3.5.9 
The problem is sketched below. Using (3.44) gives 
V rdrd
z r
z a z
r
a
s
o
s
o
= ∫
+
∫ = + −


= =0 2 20
2 2 2
4 2
ρ φ
π ε
ρ
εφ
π
. The potential is only a function of z 
and hence the gradient is 
E a a a= − = − = − + −

 = − +






gradientV V
z z
z a z z
z a
z
s
o
z
s
o
z
∂
∂
ρ
ε
∂
∂
ρ
ε2 2
12 2
2 2
 which 
agrees with the results of Problem 3.2.5. 
 
 
 
 
 
 
 
 
3.5.10 
(a) 
( ) ( ) ( )
E a a a
a a a
= − = − − −
=
+ +
+
+ +
+
+ +
gradientV V
x
V
y
V
z
x
x y z
y
x y z
z
x y z
x y z
x y z
∂
∂
∂
∂
∂
∂
2 2 2
3
2 2 2 2
3
2 2 2 2
3
2
 
(b) 
E a a a
a a a
= − = − − −
= − + +− − −
gradientV V
r r
V V
z
e e re
r z
z
r
z z
z
∂
∂
∂
∂ φ
∂
∂
φ φ φ
φ
φ
1
cos sin cos
 
R
�s rdrd�
�s
a
r y
z
x
3-15 
E a a a
a a a
= − = − − −
= − +
gradientV V
r r
V
r
V
r r r
r
r
∂
∂
∂
∂ θ θ
∂
∂ φ
θ φ θ φ φ
θ φ
θ φ
1 1
2 3 3 3
sin
sin cos cos cos sin
 
3 61. . 
E V
d
= = 10 kV
m
, D Eo r= =ε ε
µ0 478. C
m2
, P D Eo= − =ε
µ0 389. C
m2
, 
C A
d
pFr o= =ε ε 4775. . 
3 6 2. . 
( )C A
do o
= = =
−
ε ε
π 01
10
277 8
2
3
.
. pF . 
3 6 3. . 
There are two capacitors in series: C A
dr o1 1 1
= ε ε and C A
dr o2 2 2
= ε ε . Capacitors in 
series add like resistors in parallel so that C C C
C C
A d d
d d
o
r r
r r
=
+
=
+
1 2
1 2
1 2
1 2
1
1
2
2
ε
ε ε
ε ε
. The total free 
charge on the upper (and lower) plate is Q CVf = . Using Gauss’ law and surrounding 
the upper plate with a closed surface yields DA Q f= . Therefore D
Q
A
CV
A
f
= = . In the 
upper dielectric, E D CV
Ar o r o
1
1 1
= =
ε ε ε ε
 and in the lower dielectric, 
E D CV
Ar o r o
2
2 2
= =
ε ε ε ε
. Evaluating these gives C = 70 74. pF , E1 4000=
V
m
, 
E2 2000=
V
m
. 
3 6 4. . 
There are two capacitors in parallel: C A
dr o1 1
1
= ε ε and C A
dr o2 2
2
= ε ε . Capacitors in 
parallel add like resistors in series so that ( )C C C A A
do
r r
= + =
+
1 2
1 1 2 2ε
ε ε
. Using 
Gauss’ law and surrounding each portion of the upper plate with a closed surface yields 
D A Q f1 1= and D A Q f2 2= . Therefore E
D C V
Ar o r o
1
1
1
1
1 1
= =
ε ε ε ε
 and 
 
(c) 
3-16 
E D C V
Ar o r o
2
2
2
2
2 2
= =
ε ε ε ε
. Evaluating these gives C1 88 42= . pF , C2 707 4= . pF , 
C = 7958. pF , E1 5000=
V
m
, E2 5000=
V
m
. Observe that V E d E d= = =1 2 10V and 
Q Q Q CVf f f= + = =1 2 7 96. nC . 
3 6 5. . 
We observe that there are essentially two spherical capacitors in series. First we obtain 
the capacitance of a spherical capacitor with a homogeneous dielectric filling the interior. 
Using Gauss’ law and surroundingthe inner sphere with a sphere of radius r gives 
D
Q
r
f
=
4 2π
. Hence the electric field is radially directed and is E
Q
r
f
=
4 2π ε
. The voltage 
between the spheres is V
Q
r
dr
Q
a b
f
b
a f
= − ∫ = −

4 4
1 1
2π ε π ε
. Hence the capacitance is 
( )C
Q
V
ab
b a
f
= =
−
4π . This result was derived in Exercise Problem 3.8. Hence the 
capacitances are ( )C
ar
r a1
1
1
4
=
−
π and ( )C
br
b r2
1
1
4
=
−
π . Since these are in series and capacitors 
in series add like resistors in parallel we obtain 
C C C
C C
r b a r
o
r r
=
+
=
−



 + −




1 2
1 2
2 1 1 1
4 1
1 1 1 1 1 1
π ε
ε ε
. Substituting the values gives 
0.303pF. 
3 6 6. . 
The per-unit-length capacitance for a coaxial cable filled with a homogeneous dielectric 
was obtained in Example 3.15 as c
b
a
= 



2π ε
ln
F
m
. For this problem we observe that there 
ae two such capacitors in series: c
r
a
o r
1
1
1
2
= 



π ε ε
ln
 and c
b
r
o r
2
2
1
2
= 



π ε ε
ln
. Since capacitors in 
3-17 
series add like resistors in parallel, c c c
c c b
r
r
a
o r r
r r
=
+
= 


 +




1 2
1 2
1 2
1
1
2
1
2π ε ε ε
ε εln ln
. Evaluating this 
for the given dimensions yields 82 06. pF m . 
3.7.1 
Converting the radius from mils to meters gives 
16 254cm 1 4 064 10 4mils 1inch
1000mils 1inch
meter
100cm
m× × × = × −. . . Hence the resistance is 
( )R A= × × = × =−
1000
4 064 10
3323
4 2
m
= 5.8 107σ π .
. Ω . 
3.7.2 
The electric field is E V
d
= . Hence the current density is J E V
d
= =σ σ . The total 
current is I JA A V
d
= = σ . Hence the resistance is R V
I
d
A
= =
σ
Ω . Evaluating this 
gives 50mΩ. 
3.7.3 
The electric field intensity at a radius r was determined in Problem 3.6.5 as E
Q
r
f
=
4 2π ε
. 
The voltage between the spheres was also determined as 
V
Q
r
dr
Q
a b
f
b
a f
= − ∫ = −

4 4
1 1
2π ε π ε
. The current flowing between the two spheres is 
I JA EA
Q
r
r
Qf f
= = = =σ σ
π ε
π σ
ε4
42
2 . Hence the resistance is 
R V
I a b
= = −




1
4
1 1
π σ
. Evaluating this for the given dimensions yields 106.1Ω. 
3.7.4 
The voltage between the inner wire and the shield was determined in Example 3.15 in 
terms of the charge distribution on the inner wire of ρ l
C
m
 as V b
a
l
=




ρ
π ε2
ln . 
Similarly, the electric field in this region was determined in Example 3.7 to be 
3-18 
E
r
l
=
ρ
π ε2
. Hence the current flowing from one cylinder to the other (per-unit-of line 
length) is I JA
r
rl l= = =σ
ρ
π ε
π σ
ρ
ε2
2 . Thus the resistance per unit length is 
r V
I
b
a
= =




1
2π σ
ln /Ω m . Evaluating this for the given dimensions yields 01. /Ω m 
3.7.5 
From the results of Example 3.16, the magnetic flux density vector at a perpendicular 
distance from the midpoint of the current element is B a=
+
µ
π φ
o I
r
L
r L
2
2
4
2
2
. (a) At 
(0,3m,0) the field is 0.2858 Taφ µ . (b) Off the ends of the curent element d Rl a× is 
zero and hence the field is zero off the ends. 
3.7.6 
Utilizing the result obtained in Example 3.16 for an infinite current, the magnetic flux 
density at a distance r is B I
r
o
=
µ
π2
Wb
m2
. Hence the total magnetic flux penetrating the loop 
is ψ = •∫ B sd . By the right-hand rule the B field is directed into the page and hence the 
dot product can be removed. However, the B field depends on distance away from the 
current and cannot be removed from the integral. Hence 
ψ µ
π
µ
π
= ∫ ∫ = 


= =z
l
o
r r
r
oI
r
drdz Il r
r0
2
12 21
2
ln Wb . For the given parameters we obtain 
ψ µ= 0139. Wb . 
3.7.7 
The problem is sketched below. Using the results of Example 3.16 we may superimpose 
the contributions from the four identical sides. Observe that the contributions from 
opposite sides cause the net B from those two to be directed in the +z direction. similarly 
the other two sides cause a similar result. Hence B a=
+
4
2
2
4
2 2
µ
π
αo z
I
R
l
R l
cos where 
3-19 
R z l= +2
2
4 and cosα =
l
R
2 . Combining these gives 
B a=
+

 +
2 4
4 2
2
2 2 2 2
µ
π
o
z
I l
z l z l
. For l = 2m and I = 10A , at the center of the 
loop, z = 0 , we obtain B a= 566. µ T z . 
 
 
 
 
 
 
 
 
 
 
 
 
3.7.8 
The problem is sketched below. Using the results of Example 3.16 we may superimpose 
the contributions from the three identical sides. The radial distance from the center of 
each side to the center of the triangle is r l= 0 577
2
. . Using (3.60) of Example 3.16 we 
obtain B I
r
l
r l
I
l
o o
=
+
=3
2
2
4
9
22 2
µ
π
µ
π
 into the page. For l = 5cm and I = 4A we obtain 
B = 144µ T . 
 
R
�
��
�
R
I
I
I
I
l
2
( ,l
2
, 0)l
2
(– ,l
2
, 0)l
2
l
2
y
z
x
3-20 
 
 
 
 
 
 
 
3.7.9 
The problem is sketched below. According to the Biot Savart law, the magnetic field off 
the ends of a current is zero since d Rl a× is zero. Hence there are no contributions to the 
magnetic field at point P due to sides DA and BC. For side CD d rdRl a× = θ and is out 
of the page. Hence the Biot Savart law gives for this contribution 
( )
µ
π
θ µ
π
θ
θ
θ
o oI
r
r d I
r4 42
2 2
0 2=
∫ = . Similarly the contribution from the segment AB is µ
π
θo I
r4 1
 
which is into the page. Hence the total is B I
r r
o
= −




µ θ
π4
1 1
1 2
 which is directed into the 
page since r r2 1> . 
 
 
 
 
 
 
 
 
I
30°
l
l l
r
rr
I
I
�
I
AD
C
B
P
I
I
I
r1
r2
3-21 
3.9.1 
The problem is sketched below. Use the result for the magnetic field from an infinitely 
long current filament obtained in Example 3.19: H I
r
=
2π
. (a) At the center, the H fields 
add giving H I
d
I
d
= 



=2
2
2
2
π
π
. (b) At a distance D from the center and in a plane 
containing the currents we superimpose the fields to give 
( ) ( )H
I
D d
I
D d
I d
D d
=
−
−
+
=
−



2 2 2 2
2
4
2 2π π π
. 
 
 
 
 
 
 
 
 
3.9.2 
The problem is sketched below. Place the strip on the z axis centered about the origin. 
Treat the linear current density as filaments of current Kdz A . Superimpose the magnetic 
fields using equation (3.81) of Example 3.19. The magnetic field intensity vector is 
H a= ∫
=
2
20
2 Kdz
Rz
w
zπ
αcos where R d z= +2 2 and cosα = d
R
. Hence 
( )H
Kd
d z
dz K z
d
K w
dz
w
z
w
=
+
∫ = 

 =
=
−
=
−
π π π
1
22 20
2 1
0
2 1tan tan . For an infinite strip as 
w → ∞ , H K→
2
 which agrees with the result for an infinite strip in Example 3.21. 
 
d
2
d
2
II
D
3-22 
 
 
 
 
 
 
 
3.9.3 
For (b) a r b< < , H rd NIφ
φ
π
φ
=
∫ =
0
2
. Hence H NI
rφ π= 2 . For (a) r a< and (c) r b> the 
net current penetrating this loop of radius r is zero and hence H=0. 
3.9.4The problem is sketched below. Construct a rectangular contour as shown. By symmetry, 
the magnetic field is directed along the solenoid and is directed right to left. Applying 
Ampere’s law to this yields H l•∫ = = =d HL I InLenclosed . Hence H nI= and 
B H nIr o r o= =µ µ µ µ . 
 
 
 
 
 
3.11.1 
The magnetic flux density along the axis of the solenoid was obtained in Problem 3.9.4 as 
B nIr o= µ µ . Since this is uniform over the cross section of the core and is axially 
directed, the flux is ψ π= •∫ =B sd B a
core
cross section
2 . The flux linkages per unit length are 
�
��
z
d
R
R
Kdz
y
z =
w
2
z = _
w
2
L
C
H H
II
3-23 
Λ = nψ . Hence the per-unit-length self inductance is l
I
n ar o= =
Λ µ µ π2 2 H
m
. For 
the given dimensions, ( )l = × × × 

 =
−1000 4 10 2000 0 01 1587
2
2π π
turns
m
H
m
. . . 
3.11.2 
Treating this as a long parallel wire line of length l and separation w and neglecting the 
contribution from the end segments gives, using the result of Example 3.25 for an infinite 
line, L l w
a
o≅ 


µ
π
ln . 
3.11.3 
The problem is sketched below. Assuming the plate width is much greater than the plate 
separation we can use the result in Example 3.18 for the magnetic flux density from an 
infinite plate carrying a linear current density of K A
m
. The magnetic flux density is 
parallel to the plate and opposite to the direction of the current: B Ko= µ
2
. 
Superimposing the fields due to both plates gives B Ko= µ which is constant across the 
cross section between the plates. The flux penetrating the surface between the plates is 
ψ µ= •∫ = =B sd Bs l Ks lo∆ ∆ . The total current on each plate is I Kw= . Hence the 
flux is ψ µ= o I l
s
w
∆ . Thus the inductance per unit length is l I
l
s
wo
= =
ψ
µ
∆
. 
 
 
 
 
 
 
 
 
 
s
w
B
3-24 
3.11.4 
First we solve the basic problem shown below. The magnetic field intensity due to the 
infinitely long current is B I
r
o
=
µ
π2
. The flux through the loop is obtained by integration 
as ψ µ
π
µ
π
= ∫ = 


=
l I
r
dr Il b
a
o
r a
b
o
2 2
ln . Applying this to the original problem and 
superimposing the fluxes due to each current gives 
( )
( )
( )
( )ψ
µ
π
µ
π2 2
2 2
2 2
2
2 2
2 2
=
− +
− −





 −
+ +
+ −





o o
Il D s w
D s w
I D s w
D s w
ln ln . Hence the mutual inductance 
is 
( )( )
( )( )M I
l D s w D s w
D s w D s w
o
= =
− + + −
− − + +






ψ µ
π
2
2
2 2 2 2
2 2 2 2
ln . 
 
 
 
 
 
 
3.12.1 
The particle is traveling with constant velocity so that F ma= = 0 . In order for the 
vertical forces to balance so that the particle passes through the hole we must have the 
electric force, F qEe = , equal the magnetic force, F qvBm = . Solving gives the critical 
velocity of the particle as v E
B
= . Evaluating this for the given conditions yields 
v =
×
= ×
−
2000
1 10
2 103
6 m
s
. 
3.12.2 
The magnetic flux density vector is radially directed about each wire and is given by 
B I
r
o
=
µ
π2
. The magnetic flux density at wire 2 due to the current of wire 1 is therefore 
b
I
a l
3-25 
B I
s
o
21
1
2
=
µ
π
 and is perpendicular (into the page) to current I2 . The force exerted on a ∆l 
section of wire 2 is, according to the Lorentz force equation, F I lB I l I
s
o
21 2 21 2
1
2
= =∆ ∆ µ
π
. 
Hence the force per unit length exerted on wire 2 is f F
l
I I
s
o
21
21 1 2
2
= =
∆
µ
π
N
m
. 
3.12.3 
From the previous problem, the force exerted on the left side of the loop is 
F I I
a
wo1 1 22
=
µ
π
, and the force exerted on the right side of the loop is F I I
b
wo3 1 22
=
µ
π
. 
Along the upper and lower segments of the loop, the B field varies along the wire. Hence 
the force along the upper segment is F I I
r
dr I I b
a
Fo
r a
b
o
2
1 2 1 2
42 2
= ∫ = 

 =
=
µ
π
µ
π
ln . 
3.12.4 
The sliding bar cuts the magnetic field resulting in a voltage source , Bvw , inserted in it 
as shown below. Hence the current is I Bvw
R
= − . 
 
 
 
 
 
3.12.5 
The vertical side of the loop cuts the magnetic field so that a voltage is induced in it. The 
horizontal sides have no inserted source since v B× is perpendicular to the wire. Shown 
below is a view in the xy plane. The tangential velocity is v l= ω and 
( )v B× = l B tω ωsin assuming that the loop starts at the x axis at t=0. The voltage 
source is inserted as shown below so that the current through the resistor is 
( )I wl B t
R
= −
ω ωsin
. 
I
R
Bvw+
–
3-26 
 
 
 
 
 
 
 
 
 
 
B
B
l
v
�t
�
x
y
I
R
l� B sin �t+
–
4-1 
Chapter 4 
Problem Solutions 
411. . 
The flux in the left loop is ψ1 301 1 0 2 10= × × = × −B t. . Wb , and the flux in the right 
loop is ψ 2 301 0 5 01 10= × × = × −B t. . . Wb . The induced sources are shown below: 
V d
dt1
1 0 2= =ψ . mV and V d
dt2
2 01= =ψ . mV . Solving the resulting circuit gives 
V = − −
+
= −01 100
100 50
0 2 0 233. . .mV mV mV . 
 
 
 
 
 
 
412. . 
The flux in the inner loop is ψ1 301 0 5 01 10= × × = × −B t. . . Wb , and the flux in the 
outer loop is ψ 2 30 3 1 0 6 10= × × = × −B t. . Wb . The induced sources are shown below: 
V d
dt1
1 01= =ψ . mV and V d
dt2
2 0 6= =ψ . mV . Solving the resulting circuit gives 
V = −
+
=0 6 100
100 50
01 0533. . .mV mV mV . 
 
 
 
 
 
 
 
–
V
+
50 100
0.1 mV
+ –
0.2 mV
+ –
–
V
+
50 100
0.1 mV
0.6 mV
+ –
+ –
4-2 
413. . 
The flux in the left loop is ( )ψ π1 305 0 2 10 2 60= × × = −B t. . sin Wb , and the flux in the 
right loop is ( )ψ π2 303 0 2 0 6 10 2 60= × × = × −B t. . . sin Wb . The induced sources are 
shown below: ( )V d
dt
t1 1
377
2 60 2 60= = ×ψ π πcos1 2444 3444 mV and 
( )V d
dt
t2 2
266
2 60 0 6 2 60= = × ×ψ π π. cos1 24444 34444 mV . Solving the resulting circuit gives 
( ) ( ) ( )V t t t= × +
+
× = ×226 2 60 50
200 50
377 2 60 3016 2 60cos cos . cosπ π πmV mV mV . 
 
 
 
 
 
 
 
414. . 
(b) and (d) are not correct. In (b) a current must be established that will induce a 
secondary current I that will oppose the change in the original magnetic field. For this 
case the induced current must be counterclockwise to keep the field from decreasing. 
Similarly, in (d) a current must be established that will induce a secondary current I that 
will oppose the change in the original magnetic field. For this case the induced current 
must be clockwise to keep the field from increasing. 
415. . 
The flux in the loop is 
( ) ( )ψ π π= × = × × × ×B t t t tArea mWb = mWb2 2 60 05 10 10 2 60cos . cos . The induced 
–
V
+
200 50
V1 V2
+– +–
4-3 
voltage is ( ) ( )[ ]V ddt t t t= = × − ×
ψ
π π10 2 60 3770 2 60cos sin mV . Hence the current is 
( ) ( )[ ]I V t t t= = × − ×100 01 2 60 37 7 2 60. cos . sinπ π mA . 
 
 
 
 
 
 
416. . 
The position of the bar is ( )L d tt= ∫ =100 10 10 10
0
cos sin( )τ τ m . The flux in the loop is 
( ) ( )ψ = × = × ×B t tArea mWb = 0.05 Wb10 05 10 10 10. sin sin . Hence the induced voltage 
is ( )V d
dt
t= =ψ 05 10. cos V . Hence the currentis ( ) ( )I t t= =05 10
100
5 10
. cos
cos mA . 
 
 
 
 
 
 
 
417. . 
A view in the xy plane is shown below. The flux through the loop is 
( ) ( )ψ ω ω= •∫ = × × =B sd B t tArea mWbcos cos1 . Hence an induced voltage in the 
loop is ( ) ( )V d
dt
t t= = −ψ ω ω ωsin sinmV = -5 mV with polarity shown below. Hence 
the current is ( )I V t= = −
2
2 5. cos ω mA . 
I
V
100
50 cm
v
+ –
I
100
50 cm
v
+–
4-4 
 
 
 
 
 
 
 
 
418. . 
The magnetic flux density a distance r from an infinitely long current was obtained in the 
previous chapter as B t I t
r
o( ) ( )= µ
π2
 and is circumferentially directed about the wire. 
Hence the flux through the loop formed by the household power wiring is 
ψ µ
π
= •∫ = ∫ ∫ = × 

 = ×
=
− −B sd I t
r
drdz I t I t
z 0
3 7 8
2
6 10 109 517 10
m
o
r=1km
1.09km km
1km
( ) ln . ( ) . ( ) . 
The induced voltage is V d
dt
dI t
dt
= = × −
ψ 517 10 8. ( ) and is sketched below. Hence 
V = 2 585, V for 0 1< <t µ s and is V = −287V for 1 10µ µs s< <t . 
 
 
 
 
 
 
 
419. . 
The magnetic flux density threading the loop is B I
r
o
=
µ
π2
. Assuming that the loop starts 
at t=0 barely touching the wire, at some time t the flux through the loop is (downward) 
I
V
2
+ –
B
B
�t
x
y
2585 V
–287 V
V (t)
1 �s
10 �s
t
4-5 
ψ µ
π
µ
π
= •∫ = ∫ ∫ = +


= =
+
B sd I
r
drdz Il vt w
vtz
l
o
r vt
vt w
o
0 2 2
ln . The induced voltage in the loop 
(tending to push current counter clockwise) is 
( )V
d
dt
d
dt
Il vt w
vt
Il vt
vt w
w
vt
Ilw
vt w t
o o o
= =
+


 = +



 −



 = − +
ψ µ
π
µ
π
µ
π2 2 22
ln . Hence 
( )I
V
R
Ilw
R vt w t
o
= − =
+
µ
π2
. 
4110. . 
A sketch of the problem looking down on the plane of rotation is shown below. The 
further we go out along the bar length, the greater the velocity cutting the magnetic field 
lines. The linear velocity of a section of the bar at a radius r is v r= ω . The voltage 
induced in the bar is ( )v B l× •∫ = ∫ =
=
d B rdr B l
r
l
ω
ω
0
2
2
. By the right-hand rule, the 
rotating end is the positive end. 
 
 
 
 
 
 
 
 
4111. . 
Forming ∇ × = −

 + −



 + −



E a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
E
y
E
z
E
z
E
x
E
x
E
y
z y
x
x z
y
y x
z . From the 
form of E this reduces to 
( ) ( )∇ × = − + = − + −E a a a a∂∂
∂
∂ β α ω β α α ω β
E
z
E
x
E x t z E x t zy x
y
z m x m zsin cos cos sin
l
B
�
4-6 
. From Faraday’s law: ∇ × = −E Hµ ∂∂o t we obtain 
( ) ( )H a a= − − + −E x t z E x t zm
o
x
m
o
z
β
ω µ
α ω β α
ω µ
α ω βsin sin cos cos . 
4112. . 
Forming ∇ × = −

 + −



 + −



E a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
E
y
E
z
E
z
E
x
E
x
E
y
z y
x
x z
y
y x
z . From the 
form of E this reduces to ( )∇ × = =E a a∂∂ β β ω
E
z
E z tx y m ycos cos . From Faraday’s 
law: ∇ × = −E Hµ ∂∂o t we obtain ( )H a= −
E z tm
o
y
β
ω µ
β ωcos sin . 
4 21. . 
Above a frequency for which σ
ω ε εr o
= 1 the displacement current dominates the 
conduction current. This occurs for f > 45kHz . 
4 2 2. . 
Forming ∇ × = −

 + −



 + −



H a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
H
y
H
z
H
z
H
x
H
x
H
y
z y
x
x z
y
y x
z . From 
the form of H this reduces to 
( ) ( )∇ × = −

 = − + −H a a
∂
∂
∂
∂ β α α ω β
H
z
H
x
H H x t zx z y x z ysin cos . From Ampere’s 
law: ∇ × =H Eε ∂∂o t we obtain 
( ) ( )E a= − + −β α
ωε
α ω βH H x t zx z
o
ysin sin 
4 2 3. . 
Forming ∇ × = −

 + −



 + −



H a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
H
y
H
z
H
z
H
x
H
x
H
y
z y
x
x z
y
y x
z . From 
the form of H this reduces to ( )∇ × = − =H a a∂∂ β β ω
H
z
H z ty x m xsin sin . From 
Ampere’s law: ∇ × =H Eε ∂∂o t we obtain ( )E a= −
β
ωε
β ωH z tm
o
xsin cos 
4-7 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +ε ∂∂
∂
∂
∂
∂o
x y zE
x
E
y
E
z
E 0 . 
But for the given electric field, ( )E a= −E x t zm ysin sinα ω β , it has only a y component 
which is independent of y. Hence, the divergence is zero. 
4 3 2. . 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +ε ∂∂
∂
∂
∂
∂o
x y zE
x
E
y
E
z
E 0 . 
But for the given electric field, ( )E a= E z tm xsin cosβ ω , it has only an x component 
which is independent of x. Hence, the divergence is zero. 
4 3 3. . 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +µ ∂∂
∂
∂
∂
∂o
x y zH
x
H
y
H
z
H 0 . 
For the given magnetic field, 
( ) ( )H a a= − + −H x t z H x t zx x z zsin sin cos cosα ω β α ω β , so that 
( ) ( )∂∂
∂
∂ α α ω β β α ω β
H
x
H
z
H x t z H x t zx z x x z z+ = − + −cos sin cos sina a . Hence the 
divergence is zero and Gauss’ law is satisfied only if α βH Hx y+ = 0 . 
4 3 4. . 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +µ ∂∂
∂
∂
∂
∂o
x y zH
x
H
y
H
z
H 0 . 
But for the given magnetic field, ( )H a= H z tm ycos sinβ ω , the y component is 
independent of y. Hence, the divergence is zero. 
4 61. . 
The Poynting vector is S E H a= × = − −

 +








−715 1
2
2 8
4
1
2 4
8. cos cose t zz zω
π π W
m2
. 
The Poynting vector averaged over one cycle is S aaverage 2
W
m
=




−
715
2 4
8. cose z z
π . This 
is perpendicular only to the sides at z=0 and z=1m. Hence the average power exiting the 
cube is 
4 31. .
4-8 
( ) ( )
P S S
e e
z z
z z
average average average m
m
Area+ Area
= - = -25.27W
= − × ×



 +




= =
− = − =
0 1
8 0 8 1715
2 4
715
2 4
. cos . cosπ π
 
4 71. . 
The components of E that are tangential to the boundary must be continuous. Hence 
E a a2, tan = +β γy z . The components of D that are normal to the boundary must be 
continuous. Hence D E a D E1 1 1 1 2 2 2, , , , norm norm norm norm= = = =ε ε α εx so that 
E a2 1
2
, norm =
ε
ε
α x . Therefore, E E E a a a2 0 2 2
1
2
x x y z= = + = + +, , norm tan
ε
ε
α β γ . 
4 7 2. . 
The components of B that are normal to the boundary must be continuous. Hence 
B a2, norm = α x . The components of H that are tangential to the boundary must be 
continuous. Hence H B a a H B1
1
1
1 1
2
2
2
1 1
, , , , tan tan tan tan= = + = =µ
β
µ
γ
µ µy z
 so that 
B a a2 2
1
2
1
, tan = +
µ
µ
β µ
µ
γy z . Therefore, 
B B B a a a2 0 2 2
2
1
2
1
x x y z= = + = + +, , norm tan α
µ
µ
β µ
µ
γ . 
4 7 3. . 
The components of D that are normal to the boundary must be continuous. Hence 
D a2, norm = α x . The components of E that are tangential to the boundary must be 
continuous. Hence E D a a E D1
1
1
1 1
2
2
2
1 1 1 1
, , , , tan tan tan tan= = + = =ε ε
β
ε
γ
εy z
 so that 
D a a2 2
1
2
1
, tan = +
ε
ε
β ε
ε
γy z . Therefore, 
D D D a a a2 0 2 2
2
1
2
1
x x y z= = + = + +,tan ,norm α β
ε
ε
γ ε
ε
. 
4 7 4. . 
The componentsof H that are tangential to the boundary must be continuous. Hence 
H a a2, tan = +β γy z . The components of B that are normal to the boundary must be 
continuous. Hence B H a B H1 1 1 1 2 2 2, , , , norm norm norm norm= = = =µ µ α µx so that 
4-9 
B a a2 2
1
2
1
, tan = +
µ
µ
β µ
µ
γy z . Therefore, 
B B B a a a2 0 2 2
2
1
2
1
x x y z= = + = + +, , norm tan α
µ
µ
β µ
µ
γ . 
4 7 5. . 
The tangential components of E must be zero at the surface of a perfect conductor. Hence 
γ = 0 . Also, the x and y components of E must be equal in order that there be no 
tangential component from these two components and hence α β= . the normal 
components of B must be zero at the surface of a perfect conductor. Hence the x and y 
components must form a resultant that is tangent to the surface so that σ δ= − . 
4 81. . 
The components tangent to the plane are 2 3a ax y− . Reversing these gives − +2 3a ax y . 
Similarly reversing the z component gives −4a z . Hence the image current is 
I a a aimage A= − + −2 3 4x y z at (0,0,-2). 
4 91. . 
To convert a phasor quantity to the time domain we simply multiply by e ej t j tω π= ×2 10
6
 
and take the real part of the result. Hence, 
(a) 
( ) ( )E E a a
a a
= = − + −




= − × +



 + × +




Re $ Re Re
cos cos
e j e
j
e
t t
j t j t
x
j t
y
x y
ω ω ω
π
π
π
π
30 10
30 2 10
2
10 2 10
2
6 6
, 
(b) 
( )H H a a= =   = × + Re $ Re cose e e tj t j t j z zω ω π π π10 10 2 10 434 3 6 , 
(c) 
( ) ( ) ( )B B a a= = = × −− − −Re $ Re cose e e e e t zj t z j t j z x z xω ω π π π4 4 2 10 42 4 2 6 , 
(d) $E a= − −10 3 3e ex j z yπ , (e) ( )$ sinB a= −5 3 6z e j z x . 
4-10 
The phasor form of the field is $ sinE a= −E xem j z yα β . Faraday’s law in phasor form is 
∇ × = −$ $E Hj oω µ . Writing out the curl gives 
∇ × = −





 + −



 + −





$
$ $ $ $ $ $
E a a a∂∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
E
y
E
z
E
z
E
x
E
x
E
y
z y
x
x z
y
y x
z . From the form of 
the field this reduces to 
∇ × = − + = +− −$
$ $
sin cosE a a a a
∂
∂
∂
∂ β α α α
β βE
z
E
x
j E xe E xey x
y
z m
j z
x m
j z
z . Hence 
( )$ $ $ sin cos
sin cos
H E E a a
a a
= − ∇ × = ∇ × = +
= − +
− −
− −
1 1 1
j
j j j E xe E xe
E xe j E xe
o o o
m
j z
x m
j z
z
m
o
j z
x
m
o
j z
z
ω µ ω µ ω µ
β α α α
β
ω µ
α
α
ω µ
α
β β
β β
The time-domain field is 
( )
( )
( ) ( )
H H
a a
a a
=
= − − + − +


= − − − −
Re $
sin cos cos cos
sin cos cos sin
e
E x t z E x t z
E x t z E x t z
j t
m
o
x
m
o
z
m
o
x
m
o
z
ω
β
ω µ
α ω β α
ω µ
α ω β π
β
ω µ
α ω β α
ω µ
α ω β
2
 
4 9 3. . 
Consider two field vectors, ( )E Ee= Re $ j tω and ( )H He= Re $ j tω . The instantaneous 
power density or Poynting vector is S E H= × . Rewriting the field vectors as 
( ) ( )E Ee Ee E e= = + ∗ −Re $ $ $j t j t j tω ω ω12 and substituting into S E H= × yields 
( ) ( )S E H E H E H E H E H= × = × + × + × + ×∗ ∗ ∗ ∗ −14 14 2 2$ $ $ $ $ $ $ $e ej t j tω ω . But since 
( ) ( )$ $ $ $E H E H× = ×∗ ∗ ∗ and ( )$ $ Re $A A A+ =∗ 2 , this may be written as 
( ) ( )S E H E H= × + ×∗12 12 2Re $ $ Re $ $ e j tω . The time average of this is S Sav = ∫1 0T dt
T
 and T 
is the period of the sinusoid. The first term of our result is a constant and the second term 
is sinusoidal and hence averages to zero giving the desired result. 
4 9 2. . 
4-11 
The magnetic flux density in the core due to a current I is, from Chapter 3, B I
r
r oφ
µ µ
π
=
2
 
where the flux is circumferentially directed and r is approximately the mean radius of the 
core. The total flux through the windings is approximately ψ = BA where A is the area of 
the core cross section. According to Faraday’s law a voltage will be generated in the turns 
of wire of V N d
dt
NA
r
dI
dt
r o
= =
ψ µ µ
π2
. But dI
dt
j I⇔ ω $ . Hence $
$
$Z
V
I
NA
rT
r o
= =
ω µ µ
π2
. 
Substituting numerical values gives 
( )( ) ( )( )
( )$ . .Z
f N A
rT
r o
=
= = = = ×
=
=
−ω π µ µ
π
2 200 10 4 10
2 0 02
50 3
4
Ω Ω= 34dB 
 
 
4101. . 
 5-1 
Chapter 5 
Problem Solutions 
511. . 
The time derivative is related to the phasor form by j
t
ω
∂
∂⇔ . Hence 
− ⇔ −j H z
H z t
ty
y
ω µ µ
∂
∂
$ ( )
( , )
 and − ⇔ −j E z E z t
tx
xω ε ε
∂
∂
$ ( ) ( , ) . 
51 2. . 
The time-domain equations are ( ) ( )E z t E t z E t zx m m( , ) cos cos= − + ++ −ω β ω β and 
( ) ( )H z t E t z E t zy m m( , ) cos cos= − − +
+ −
η
ω β
η
ω β . Substituting into the first equation 
gives ( ) ( )∂ ∂ β ω β β ω β
E z t
z
E t z E t zx m m
( , ) sin sin= − − ++ − and 
( ) ( )− = − − ++ −µ ∂ ∂ µω η ω β µω η ω β
H z t
t
E t z E t zy m m
( , )
sin sin . Matching the 
corresponding coefficients of the sin terms gives the requirement that β µω
η
= . But 
substituting β ω µ ε= and η µ
ε
= shows this to be true. Similar results are shown for 
the second equation of Problem 5.1.1. 
51 3. . 
(a) pvc, ( )ε r = 35. β ω µ ε= =vo r r 0392.
rad
m
, η η µ
ε
= =o
r
r
202 Ω , 
v vo
r r
= = ×
µ ε
16 108. m
s
, λ = =v
f
16 m (b) Teflon, ( )ε r = 21. 
β ω µ ε= =
vo
r r 0304.
rad
m
, η η µ
ε
= =o
r
r
260 Ω , v vo
r r
= = ×
µ ε
2 07 108. m
s
, 
λ = =v
f
20 7. m . (c) Mylar, ( )ε r = 5 β ω µ ε= =vo r r 0 468.
rad
m
, η η µ
ε
= =o
r
r
169 Ω , 
v vo
r r
= = ×
µ ε
134 108. m
s
, λ = =v
f
134. m . (d) Polyurethane ( )ε r = 7 
5-2 
β ω µ ε= =
vo
r r 0554.
rad
m
, η η µ
ε
= =o
r
r
142 Ω , v vo
r r
= = ×
µ ε
113 108. m
s
, 
λ = =v
f
113. m . 
51 4. . 
A sketch is shown below. The phase constant is β ω= =
vo
0105. rad
m
 and the intrinsic 
impedance is η η= =o 377 Ω . The electric field intensity vector is given by 
$ .E a= 10 0105e j y z or ( )E a= × +10 10 10 01056cos .π t y z . In order for the power flow 
E H× to be in the -y direction, the magnetic field intensity vector must be in the -x 
direction so that $ . .H a= −0 0265 0105e j y x or ( )H a= − × +0 0265 10 10 01056. cos .π t y x . 
 
 
 
 
 
 
 
 
515. . 
A sketch is shown below. Since the wavelength is λ = v
f
, the frequency of the wave is 
800MHz. Also the velocity of propagation is v vo
r
=
ε
. Hence ε r = 2 25. . The phase 
constant is β ω πλ= = =v
2 251. rad
m
 . The electric field is 
( )E a= × −100 16 10 2518cos .π t x z . The intrinsic impedance is η η
ε
= =
o
r
251Ω . In 
order that E H× be in the +x direction, the magnetic field must be directed in the -y 
direction. Hence ( )H a= − × −0 398 16 10 2518. cos .π t x y 
Ez
Hx
y
z
x
5-3 
 
 
 
 
 
 
 
 
51 6. . 
The frequency of the wave is 40MHz. The phase constant is β ω ε= =
vo
r 29.
rad
m
. The 
intrinsic impedance is η η
ε
= =
o
r
109Ω . The problem is sketched below. The wave is 
traveling in the +y direction. From this E must be in the z direction in order that E H× be 
in the +y direction. The magnitude of the electric field is the product of the magnitude of 
the magnetic field and the intrinsic impedance. Hence, the electric field is 
( )E a= × −10 9 8 10 2 97. cos .π t y z . 
 
 
 
 
 
 
 
 
Ez
Hy
x
y
zEz
Hx
y
z
x
5-4 
51 7. . 
The problem is sketched below. The phase constant is β ω ε µ= =
vo
r r 251
rad
m
. The 
intrinsic impedance is η η µ
ε
= =o
r
r
565Ω . The wave is traveling in the -z direction and 
the magnetic field is in the y direction. Hence the magnetic field intensity vector is given 
by ( )H a= × +0 02 4 10 2519. cos π t z y From this E must be in the -x direction in order that 
E H× be in the -z direction. The magnitude of the electric field is the product of the 
magnitude of the magnetic field and the intrinsic impedance. Hence, the electric field is 
( )E a= − × +113 4 10 2519. cos π t z x . 
 
 
 
 
 
 
 
 
51 8. . 
Since λ
ε
=
v
f
o
r
 we have that ε λλr
o
= = 2 . Hence ε r = 4 . 
5 21. . 
The frequency is 500MHz. The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 10 10 4 10 4 1 10 10
1
36
10 368 7 8 9
This evaluates to $ . .γ = ∠ = +149 67 5 57 2 138o j . Hence we identify α = 57 2. and 
β = 138 rad
m
. The intrinsic impedance is 
Hy
Ex
z
x
y
5-5 
( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
10 10 4 10 4
1 10 10 1
36
10 36
8 7
8 9
. This evaluates to 
$ .η = ∠106 22 5o . The problem is sketched below. The wave is traveling in the +x 
direction and the electric field is in the z direction. From this H must be in the -y direction 
in order that E H× be in the +x direction. Hence the magnetic field vector is 
( )H a= − × − −−0 946 10 10 138 22 557 2 8. cos ..e t xx o yπ . 
 
 
 
 
 
 
 
 
5.2.2 
(a) 60Hz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × + × × ×




− −j j j jo r o r 120 4 10 0 01 120
1
36
10 157 9 
This evaluates to $ . . .γ = × ∠ = × + ×− − −218 10 45 154 10 154 103 3 3o j . Hence we identify 
α = × −154 10 3. and β = × −154 10 3. rad
m
. The velocity of propagation is 
v = = ×ωβ 2 45 10
5. m
s
. The intrinsic impedance is 
( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× ×
+ × × ×




−
−
j
j
j
j
o r
o r
120 4 10
0 01 120 1
36
10 15
7
9
 which evaluates to 
$ .η = ∠0 22 45o . 
Ez
Hy
x
y
z
5-6 
(b) 1MHz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × × + × × × ×




− −j j j jo r o r 2 10 4 10 0 01 2 10
1
36
10 156 7 6 9
This evaluates to $ . . . .γ = ∠ = +0 281 47 4 019 0 21o j . Hence we identify α = 019. and 
β = 0 21. rad
m
. The velocity of propagation is v = = ×ωβ 303 10
7. m
s
. The intrinsic 
impedance is ( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
0 01 2 10 1
36
10 15
6 7
6 9
.which evaluates to 
$ . .η = ∠28 05 42 62o . 
(c) 100MHz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × × + × × × ×




− −j j j jo r o r 2 10 4 10 0 01 2 10
1
36
10 158 7 8 9
This evaluates to $ . . . .γ = ∠ = +814 86 58 0 49 813o j . Hence we identify α = 0 49. and 
β = 813. rad
m
. The velocity of propagation is v = = ×ωβ 7 73 10
7. m
s
. The intrinsic 
impedance is ( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
0 01 2 10 1
36
10 15
8 7
8 9
.which evaluates to 
$ . .η = ∠96 99 342o . 
(d) 10GHz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × × + × × × ×




− −j j j jo r o r 2 10 4 10 0 01 2 10
1
36
10 1510 7 10 9
This evaluates to $ . . . .γ = ∠ = +81116 89 97 0 49 81116o j . Hence we identify α = 0 49. and 
β = 8112. rad
m
. The velocity of propagation is v = = ×ωβ 7 75 10
7. m
s
. The intrinsic 
impedance is ( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
0 01 2 10 1
36
10 15
10 7
10 9
.which evaluates 
to $ . .η = ∠97 34 0 03o . 
5-7 
The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 2 10 4 10 16 2 2 10
1
36
10 99 7 9 9
This evaluates to $ . . . .γ = ∠ = +510 33 52 02 314 06 402 25o j . Hence we identify α = 314 06. 
and β = 402 25. rad
m
. The velocity of propagation is v = = ×ωβ 156 10
7. m
s
. The intrinsic 
impedance is ( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10 16
2 2 10 1
36
10 9
9 7
9 9
.which evaluates to 
$ . .η = ∠247 55 37 98o . 
5 2 4. . 
The frequency is 10GHz. Forming the propagation constant as 
$γ π π σ π
π
ε= + = × × × + × × × ×


− −200 300 2 10 4 10 2 10 1
36
1010 7 10 9j j j r . Squaring this 
gives $γ π ε π π σ2 4 4 2 4 10 75 10 12 10 16
36
10 2 10 4 10= − × + × = − × + × × × −j jr . Solving gives 
ε r = 114. and σ = 152.
S
m
. The intrinsic impedance can now be computed from 
( )$ . .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
152 2 10 1
36
10 114
10 7
10 9
.which evaluates to 
$ . .η = ∠218 98 3369o . The problem is sketched below. The wave is traveling in the +y 
direction and the magnetic field is in the x direction. From this E must be in the +z 
direction in order that E H× be in the +y direction. Hence the electric field vector is 
( )E a= × − +−219 2 10 300 3369200 10. cos .e t yy o zπ . 
 
 
 
 
 
j
5 2 3. . 
Ez
Hx
y
z
x
5-8 
5 31. . 
The surface is sketched below and has a surface area of ( ) ( )3 1 2 1 12− − × − − =( ) ( ) m2 . 
The intrinsic impedance of the medium is η η= =o
5
168 6. Ω . Hence the average power 
density is ( )1
2
10
168 6
0 3
2
.
.= W
m2
. The wave is propagating perpendicular to the surface and is 
uniform over it so that the total average power crossing the surface is 
P d
s
AV AV= •∫ =S s 356W. . 
 
 
 
 
 
 
 
 
 
 
5 3 2. . 
The surface is sketched below and has a surface area of 3 5 15m× = 2 . The intrinsic 
impedance of the medium is η η= =o
9
125 66. Ω . The magnitude of the electric field 
intensity is ( )E = × =0 2 125 66 2513. . . V
m
. Hence the average power density is 
( )1
2
2513
12566
2 51
2.
.
.= W
m2
. The wave is propagating perpendicular to the surface and is 
uniform over it so that the total average power crossing the surface is 
P d
s
AV AV W= •∫ =S s 37 7. . 
 
z
x
y
(3, –1, 2) m
(–1, –1, 2) m
(–1, 2, 2) m
(3, 2, 2) m
5-9 
 
 
 
 
 
 
 
 
 
 
5 3 3. . 
The frequency is 500MHz. The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 10 10 4 10 4 1 10 10
1
36
10 368 7 8 9
This evaluates to $ . .γ = ∠ = +149 67 5 57 2 138o j . Hence we identify α = 57 2. and 
β = 138 rad
m
. The intrinsic impedance is 
( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
10 10 4 10 4
1 10 10 1
36
10 36
8 7
8 9
. This evaluates to 
$ .η = ∠106 22 5o . The problem is sketched below. The wave is propagating in the +x 
direction and is perpendicular to the side of area 2m