ELECTROMAGNETICS HAYT 8 EDITION
DRILL SOLUTION FROM CHAPTERS 1 TO 5
CHAPTER 1
D1.1. Given points M(–1, 2, 1), N(3, –3, 0), and P(–2, –3, –4), find: (a) RMN; (b) RMN + RMP; (c)
|rM|; (d) aMP; (e) |2rP – 3rN|.
(a) RMN = (3 – (–1))ax + (–3 – 2)ay + (0 – 1)az = 4ax – 5ay – az
(b) RMN + RMP = RMN + (–2 – (– 1))ax + (–3 – 2)ay + (–4 – 1)az
= RMN – ax – 5ay – 5az = 4ax – 5ay – az – ax – 5ay – 5az
= 3ax – 10ay – 6az
(1)2  22  12 =
(c) |rM| =
6 = 2.45
ax  5ay  5az
(d) aMP =
(1)2  (5)2  (5)2
=
ax  5ay  5az
= –0.14ax – 0.7ay – 0.7az
51
(e) 2rP – 3rN = 2(–2ax – 3ay – 4az) – 3(3ax – 3ay) = –4ax – 6ay – 8az – 9ax + 9ay
= –13ax + 3ay – 8az
|2rP – 3rN| =
(13)2  32  (8)2 = 11 2 = 15.56
D1.2. A vector field S is expressed in rectangular coordinates as S = {125/[(x – 1)2 + (y – 2)2 +
(z + 1)2]}{(x – 1)ax + (y – 2)ay + (z + 1)az}. (a) Evaluate S at P(2, 4, 3). (b) Determine a unit vector
that gives the direction of S at P. (c) Specify the surface f(x, y, z) on which |S| = 1.
125
(a) S =
{(2  1)ax  (4  2)ay  (3  1)az }
2
(2  1)  (4  2)2  (3  1)2
125
125
(ax  2ay  4az )
(ax  2ay  4az ) =
2
2
1 2 4
21
= 5.95ax + 11.9ay + 23.8az
5.95ax  11.9ay  23.8az
5.95ax  11.9ay  23.8az
(b) aS =
=
27.27
5.952  11.92  23.82
=
2
= 0.218ax + 0.436ay + 0.873az
(c) 1 = |S| =
2
2
2

 
 

125( x  1)
125( y  2)
125( z  1)



2
2
2 
2
2
2 
2
2
2 
 ( x  1)  ( y  2)  ( z  1)   ( x  1)  ( y  2)  ( z  1)   ( x  1)  ( y  2)  ( z  1) 
1 =
125 ( x  1)2  ( y  2)2  ( z  1)2
1252 [( x  1)2  ( y  2)2  ( z  1)2
=
[( x  1)2  ( y  2)2  ( z  1)2 ]2
( x  1)2  ( y  2)2  ( z  1)2
Transposing,
( x  1)2  ( y  2)2  ( z  1)2
( x  1)2  ( y  2)2  ( z  1)2
= 125
Conjugating,
( x  1)2  ( y  2)2  ( z  1)2
( x  1)2  ( y  2)2  ( z  1)2
( x  1)2  ( y  2)2  ( z  1)2
( x  1)2  ( y  2)2  ( z  1)2
= 125
( x  1)2  ( y  2)2  ( z  1)2 = 125
D1.3. The three vertices of a triangle are located at A(6, –1, 2), B(–2, 3, –4), and C(–3, 1, 5). Find:
(a) RAB; (b) RAC; (c) the angle θBAC at vertex A; (d) the (vector) projection of RAB on RAC.
(a) RAB = (–2 – 6)ax + (3 – (–1))ay + (–4 – 2)az
= –8ax + 4ay – 6az
(b) RAC = (–3 – 6)ax + (1 – (–1))ay + (5 – 2)az
= –9ax + 2ay + 3az
(c) Locating angle θBAC ,
RAB RAC = (–8)(–9) + 4(2) + (–6)3 = 72 + 8 – 18 = 62
Using dot product,
RAB RAC = |RAB||RAC| cos θBAC
1
θBAC = cos1
R AB R AC
R AB R AC
= cos1

62
(8)2  42  (6)3

(9)2  22  33

= 53.6°
(d) Analyzing the projection,
|RAB (L)| = RAB aAC = RAB
R AC
R AC
=
62
(9)2  22  32
= 6.395
 R AC 
9ax  2a y  3az
RAB (L) = |RAB (L)| aAC = |RAB (L)| 
= (6.395)
 R AC 
(9)2  22  32


= –5.94ax + 1.319ay + 1.979az
D1.4. The three vertices of a triangle are located at A(6, –1, 2), B(–2, 3, –4), and C(–3, 1, 5). Find:
(a) RAB × RAC; (b) the area of the triangle; (c) a unit vector perpendicular to the plane in which the
triangle is located.
a x a y az
(a) RAB × RAC = 8
9
6 = [4(3) – (–6)(2)]ax + [(–6)(–9) – (–8)(3)]ay + [(–8)(2) – 4(–9)]az
3
4
2
= 24ax + 78ay + 20az
1
1
(b) A = base  height =
R AB  RBC
2
2
Finding the direction of vector RBC
RBC = (–3 – (–2))ax + (1 – 3)ay + (5 – (–4))az = –ax – 2ay + 9az
ax a y az
RAB × RBC = 8
4
1 2
6 = 24ax + 78ay + 20az
9
|RAB × RBC| = 242  782  202 = 84
A = (1/2)(84) = 42
24ax  78ay  20az
(c) aRAB  RAC =
= 0.286ax + 0.928ay + 0.238az
84
D1.5. (a) Give the rectangular coordinates of the point C(ρ = 4.4, ϕ = –115°, z = 2). (b) Give the
cylindrical coordinates of the point D(x = –3.1, y = 2.6, z = –3). (c) Specify the distance from C to D.
(a) Converting cylindrical to rectangular coordinates
Table 1 Cylindrical to rectangular coordinate systems
x = ρ cos ϕ
y = ρ sin ϕ
z = z
x = 4.4 cos –115° = –1.86
y = 4.4 sin –115° = –3.99
z = 2
C(x = –1.86, y = –3.99, z = 2)
(b) Converting rectangular to cylindrical coordinates
Table 2 Rectangular to cylindrical coordinate systems
ρ = x2  y2
ϕ = tan–1 y/x
z = z
ρ =
(3.1)2  2.62 = 4.05
(ϕ is added with 180° which it lies on the second quadrant.)
ϕ = tan–1 y/x + 180° = tan–1 (2.6/–3.1) + 180° = 140°
z = –3
D(ρ = 4.05, ϕ = 140°, z = –3)
(c) CD = (–3.1 – (–1.86))ax + (2.6 – (–3.99))ay + (–3 – 2)az
= –1.24ax + 6.59ay – 5az
2
|CD| =
(1.24)2  6.592  (5)2 = 8.36
D1.6. Transform to cylindrical coordinates: (a) F = 10ax – 8ay + 6az at point P(10, –8, 6); (b)
G = (2x + y)ax – (y – 4x)ay at point Q(ρ, ϕ, z). (c) Give the rectangular components of the vector
H = 20aρ – 10aϕ + 3az at P(x = 5, y = 2, z = –1).
(a) Taking note that the dot product of the unit vectors in cylindrical to rectangular coordinate
systems and vice versa
Table 3 Dot products of the unit vectors in cylindrical and rectangular coordinate systems
aρ
aϕ
az
ax
cos ϕ
– sin ϕ
0
ay
sin ϕ
cos ϕ
0
az
0
0
1
F = (10ax – 8ay + 6az) aρ + (10ax – 8ay + 6az) aϕ + (10ax – 8ay + 6az) az
= (10 cos ϕ – 8 sin ϕ + 0)aρ + [(10(– sin ϕ) + (–8)(cos ϕ) + 0]aϕ + (0 + 0 + 6)az
Since ϕ = (tan–1 (–8/10) +180°) = 141.43°
= (10 cos 141.43° – 8 sin 141.43°)aρ + [(10(– sin 141.43°) + (–8)(cos 141.43°)]aϕ + 6az
= 12.81aρ + 6az
(b) G = [(2x + y)ax – (y – 4x)ay] aρ + [(2x + y)ax – (y – 4x)ay] aϕ
Segregating the unit vectors to avoid confusion,
Gρ = [(2x + y)ax – (y – 4x)ay] aρ
Converting also the variables of rectangular to cylindrical coordinates,
= (2ρ cos ϕ + ρ sin ϕ) cos ϕ – (ρ sin ϕ – 4ρ cos ϕ) sin ϕ
= 2ρ cos2 ϕ + ρ sin ϕ cos ϕ – ρ sin2 ϕ + 4ρ sin ϕ cos ϕ
= 2ρ cos2 ϕ – ρ sin2 ϕ + 5ρ sin ϕ cos ϕ
Gϕ = [(2x + y)ax – (y – 4x)ay] aϕ
= (2ρ cos ϕ + ρ sin ϕ)(–sin ϕ) – (ρ sin ϕ – 4ρ cos ϕ) cos ϕ
= –2ρ sin ϕ cos ϕ – ρ sin2 ϕ – ρ sin ϕ cos ϕ + 4ρ cos2 ϕ
= –2ρ sin ϕ cos ϕ – ρ sin2 ϕ – ρ sin ϕ cos ϕ + 4ρ cos2 ϕ
= 4ρ cos2 ϕ – ρ sin2 ϕ – 3ρ sin ϕ cos ϕ
Adding together,
G = (2ρ cos2 ϕ – ρ sin2 ϕ + 5ρ sin ϕ cos ϕ) aρ + (4ρ cos2 ϕ – ρ sin2 ϕ – 3ρ sin ϕ cos ϕ) aρ
(c) Applying the dot product from Table 3,
H = (20aρ – 10aϕ + 3az) ax + (20aρ – 10aϕ + 3az) ay + (20aρ – 10aϕ + 3az) az
Since ϕ = (tan–1 (2/5) +180°) = 201.8°
Hx = 20 cos ϕ + 10 sin ϕ + 0 = 20 cos 201.8° + 10 sin 201.8° = 22.3
Hy = 20 sin ϕ – 10 cos ϕ + 0 = 20 sin 201.8° – 10 cos 201.8° = –1.86
Hz = 0 + 0 + 3 = 3
D1.7. Given the two points, C(–3, 2, 1) and D(r = 5, θ = 20°, ϕ = –70°), find: (a) the spherical
coordinates of C; (b) the rectangular coordinates of D; (c) the distance from C to D.
(a) Converting rectangular to spherical coordinates,
Table 4 Rectangular to spherical coordinate systems
x2  y2  z2
z
θ = cos1
r
–1
ϕ = tan y/x
r =
r =
(3)2  22  12 = 3.74
θ = cos–1 (1/3.74) = 74.5°
ϕ = tan–1 (2/–3) + 180° = 146.3°
C(r = 3.74, θ = 74.5°, ϕ = 146.3°)
(b) Converting spherical to rectangular coordinates,
Table 5 Spherical to rectangular coordinate systems
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
3
x = 5 sin 20° cos –70° = 0.585
y = 5 sin 20° sin –70° = –1.607
z = 5 cos 20° = 4.7
D(x = 0.585, y = –1.607, z = 4.7)
(c) Using rectangular coordinates to find the distance
|CD| =
(0.585  (3))2  (1.607  2)2  (4.7  1)2 = 6.29
D1.8. Transform the following vectors to spherical coordinates at the points given: (a) 10ax at
P(x = –3, y = 2, z = 4); (b) 10ay at Q(ρ = 5, ϕ = 30°, z = 4); (c) 10az at M(r = 4, θ = 110°, ϕ = 120°).
(a) Taking note that the dot product of the unit vectors in spherical to rectangular coordinate
systems and vice versa
Table 6 Dot products of the unit vectors in spherical and rectangular coordinate systems
ar
aθ
aϕ
ax
sin θ cos ϕ
cos θ cos ϕ
– sin ϕ
ay
sin θ sin ϕ
cos θ sin ϕ
cos ϕ
az
cos θ
– sin θ
0
10ax = 10 (sin θ cos ϕ ar + cos θ cos ϕ aθ – sin ϕ aϕ)
From Table 4, we calculate and have θ = 42.03° and ϕ = 146.31°. Substituting,
= 10 sin 42.03° cos 146.31° ar + 10 cos 42.03° cos 146.31° aθ – 10 sin 146.31° aϕ
= –5.57ar – 6.18aθ – 5.55aϕ
(b) 10ay = 10 (sin θ sin ϕ ar + cos θ sin ϕ aθ + cos ϕ aϕ)
We manipulate to convert in cylindrical to spherical coordinates. From Table 1, we calculate
and have x = 4.33, y = 2.5 and z = 4. Then from Table 4, we convert rectangular to spherical
coordinates. We calculate and have θ = 51.34° and ϕ = 30°. Substituting,
= 10 sin 51.34° sin 30° ar + 10 cos 51.34° sin 30° aθ + 10 cos 30° aϕ
= 3.9ar – 3.12aθ – 8.66aϕ
(c) 10az = 10 (cos θ ar + (– sin θ) aθ)
= 10 cos 110° ar – 10 sin 110° aθ = –3.42ar – 9.4aθ
CHAPTER 2
D2.1. A charge QA = –20 μC is located at A(–6, 4, 7), and a charge QB = 50 μC is at B(5, 8, –2) in
free space. If distances are given in meters, find: (a) RAB; (b) RAB. Determine the vector force
exerted on QA by QB if ϵ0 = (c) 10–9/(36π) F/m; (d) 8.854 × 10–12 F/m.
(a) RAB = (5 – (–6))ax + (8 – 4)ay + (–2 – 7)az m = 11ax + 4ay – 9az m
(b) RAB =
112  42  (9)2 m = 14.76 m
Q1Q2 R AB
Q1Q2
Q1Q2
R AB
a AB =
=
2
3
2
4πϵ 0 RAB RAB
4πϵ 0 RAB
4πϵ 0 RAB
There is an ambiguous value of FAB with regards from the answer of the textbook. Trying in
the higher decimal number of RAB = 14.7648,
(20 106 )(50 106 )
=
(11ax  4ay  9az ) = (–2.7961 × 10–3)(11ax + 4ay – 9az)
9
3
4π (10 36π) (14.7648)
(c) FAB =
= 30.76ax + 11.184ay – 25.16az mN
Q1Q2
(20 106 )(50 106 )
R
(11ax  4ay  9az )
(d) FAB =
=
AB
3
4πϵ 0 RAB
4π(8.854 1012 )(14.7648)3
= (–2.7923 × 10–3)(11ax + 4ay – 9az) = 30.72ax + 11.169ay – 25.13az mN
D2.2. A charge of –0.3 μC is located at A(25, –30, 15) (in cm), and a second charge of 0.5 μC is
at B(–10, 8, 12) cm. Find E at: (a) the origin; (b) P(15, 20, 50) cm.
(a) Solving first the vector and magnitude from charge A to the origin O,
RAO = (0 – 25)ax + (0 – (–30))ay + (0 – 15)az cm = –25ax + 30ay – 15az cm
= –0.25ax + 0.3ay – 0.15az m
|RAO| =
EA =
(0.25)2  0.32  (0.15)2 m = 0.4183 m
QA
QA R AO
QA
a
R AO
=
=
AO
2
2
3
4πϵ 0 RAO
4πϵ 0 RAO
RAO
4πϵ 0 RAO
4
0.3 106
=
(0.25ax  0.3ay  0.15az ) V/m
4πϵ0(0.4183)3
= 9.21ax 11.05ay  5.53az kV/m
For the charge B to the origin,
RBO = (0 – (–10))ax + (0 – 8)ay + (0 – 12)az cm = 10ax – 8ay – 12az cm
= 0.1ax – 0.08ay – 0.12az m
0.12  (0.08)2  (0.12)2 m = 0.1755 m
|RBO| =
EB =
QB
0.5 106
R
=
(0.1ax  0.08a y  0.12az ) V/m
BO
3
4πϵ 0 RBO
4πϵ0(0.1755)3
= 83.13ax  66.51ay  99.76az kV/m
Combining EA and EB,
E = 92.3ax – 77.6ay – 94.2az kV/m
(b) Same concept in (a)
RAP = (15 – 25)ax + (20 – (–30))ay + (50 – 15)az cm = –0.1ax + 0.5ay + 0.35az m
|RAP| = 0.6185 m
EA = 1.14ax  5.7ay  3.99az kV/m
RBP = (15 – (–10))ax + (20 – 8)ay + (50 – 12)az cm = 0.25ax + 0.12ay + 0.38az m
|RBP| = 0.4704 m
EB = 10.79ax  5.18ay  16.41az kV/m
E = 11.9ax – 0.52ay – 12.4az kV/m
1  (1) m
; (b)
2
m 0 m  1
5
D2.3. Evaluate the sums: (a) 
(a)
(0.1) m  1
.

2 1.5
m 1 (4  m )
4
1  (1)0 1  (1)1 1  (1) 2 1  (1)3 1  (1) 4 1  (1) 5
= 2.52
 2
 2
 2
 2
 2
02  1
1 1
2 1
3 1
4 1
5 1
(0.1)1  1 (0.1)2  1 (0.1)3  1 (0.1)4  1
= 0.176



(4  12 )1.5 (4  22 )1.5 (4  32 )1.5 (4  42 )1.5
D2.4. Calculate the total charge within each of the indicated volumes: (a) 0.1 ≤ |x|, |y|, |z| ≤ 0.2: ρv
= 1/(x3y3z3); (b) 0 ≤ ρ ≤ 0.1, 0 ≤ ϕ ≤ π, 2 ≤ z ≤ 4; ρv = ρ2z2sin 0.6ϕ; (c) universe: ρv = e–2r /r2.
(a) The differential volume of the rectangular coordinates is dv = dx dy dz. Using the formula,
0.2 0.2 0.2
0.2
0.2
0.2
1
3
3
3
x
dx
y
dy
dx
dy
dz
Q =  ρv dv =   
=
0.1
0.1
0.1 z dz
0.1 0.1 0.1 x 3 y 3 z 3
vol
(b)
x 2
=
2
0.2
y 2
2
0.1
0.2
0.1
z 2
2
0.2
= 112.5 C
0.1
Notice that the charge Q is a large number. Neglecting the rate of change of the cubical volume,
we have xyz = (0.1)3 and xyz = (0.2)3 which are very small; then the volume Δv is nearest to 0.
Yielding,
Q = ρvΔv = [1/(x3y3z3)] 0 = 0
(b) The differential volume of the cylindrical coordinates is dv = ρ dρ dϕ dz. Using the formula,
Q =
0.1
π
4
0
0
2
 
( ρ2 z 2 sin 0.6 ) ρ dρ d dz =

0.1
0
π
4
0
2
ρ3 dρ  sin 0.6 d  z 2 dz
ρ
  cos 0.6 π   z 
 0.14
   cos 0.6(180)  cos 0.6(0)   43 23 





 0

=
= 


  
 4 0 
0.6 0   3 2 
0.6
0.6
 3 3 
 4





= 1.018 mC
(c) The differential volume of the spherical coordinates is dv = r2 sin θ dθ dϕ dr. Assuming this
universe to be a perfect sphere, we have limits as 0 ≤ r ≤ ∞, 0 ≤ ϕ ≤ 2π, 0 ≤ θ ≤ π. Using the
formula,
4 0.1
Q =
3 4
2π
π

0
0
2
 
(e2 r r 2 )(r 2 sin θ dθ d dr ) =
   cos θ    e
= 0
2π
π
0
2 r
2

0

2π
0
π

0
2
d  sin θ dθ  e2 r dr
 = (2π – 0)(–cos 180° – (–cos 0°))(–e
–∞
2 – (–e 0 2))
5
= 6.28 C
D2.5. Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and y axes in
free space. Find E at: (a) PA(0, 0, 4); (b) PB(0, 3, 4).
ρL
(a) E =
aρ where ρ is just a radical distance R, then
2πϵ0 ρ
ρL R
ρL
R
=
2πϵ0 R2
2π ϵ0 R R
Finding Ex and Ey at Point A since infinite uniform line charges lie along x and y axes,
Rx = 4az
5 109
Ex =
(4az ) = 22.469az V/m
2π ϵ0 42
Ry = 4az
5 109
Ey =
(4az ) = 22.469az V/m
2πϵ0 42
Adding together,
E = 22.469az + 22.469az V/m = 45az V/m
(b) Finding R and E at Point B along x and y axes,
Rx = 3ay + 4az
Ry = 4az
5 109
(3a y  4az ) = 10.785ay + 14.38az V/m
Ex =
2
2
2
2πϵ0 3  4
=


5 109
Ey =
2πϵ0
 
42
2
(4az ) = 22.469 az V/m
E = 10.785ay + (14.38 + 22.469)az V/m = 10.8ay + 36.9az V/m
D2.6. Three infinite uniform sheets of charge are located in free space as follows: 3 nC/m 2 at
z = –4, 6 nC/m2 at z = 1, and –8 nC/m2 at z = 4. Find E at the point: (a) PA(2, 5, –5); (b)
PB(4, 2, –3); (c) PC(–1, –5, 2); (d) PD(–2, 4, 5).
(a) Obtaining first the electric field due to charges,
ρs
3 109
aN =
Ez = –4 =
az = 169.41az V/m
2ϵ 0
2ϵ 0
Ez = 1 =
6 109
az = 338.82az V/m
2ϵ 0
8 109
az = –451.76az V/m
2ϵ 0
Since the point A is located below all the sheets of charge, all directions of electric field E are
negative az direction with relative to normal, which is aN, of PA. Combining together,
E = –E1 – E2 – E3 = –169.41az – 338.82az – (–451.76az) V/m = –56.5az V/m
(b) The location of point B suggests that the electric field E contributed by the surface charge at
z = –4 will be in positive az direction while others are negative. Combining together,
E = E1 – E2 – E3 = 282.3az V/m
(c) Same arguments in (b), only z = 4 will be in negative az direction. Combining together,
E = E1 + E2 – E3 = 960az V/m
(d) We know that the point D is located above all the sheets of charge and all electric fields E are
positive az direction. Combining together,
E = E1 + E2 + E3 = 56.5az V/m
D2.7. Find the equation of that streamline that passes through the point P(1, 4, –2) in the field
Ez = 4 =
4x 2
8x
E = (a)
ax + 2 ay; (b) 2e5x[y(5x + 1)ax + xay].
y
y
(a)
x
4 x2 y2
Ey
dy
=
=
= 
2y
8 x y
Ex
dx
6
2ydy = –xdx
 2 ydy
  xdx
=
x2 C 2

2
2
For P(x = 1, y = 4, z = –2),
12 C 2
42 =  
; C 2 = 33
2 2
x2 + 2y2 = 33
y2 = 
(b)
x
dy
=
y (5 x  1)
dx
 ydy
=
x
 5x  1 dx
Letting u = 5x + 1 and du = 5 dx, and then x = (u – 1)/5,
y2
u  1  1  du 
1
1
u 1
1
= 
(u  ln u )
du =  du  
du =
   = 
25
25
25u
25u
2
5  u  5 
1
C2
y2
=
[(5 x  1)  ln 5 x  1] 
25
2
2
For P(x = 1, y = 4, z = –2),
1
C2
42
=
; C 2 = 15.66
[(5(1)  1)  ln 5(1)  1] 
25
2
2
y2
15.66
= 0.04(5 x  1)  0.04ln 5 x  1 
2
2
y2 = 15.7 + 0.4x – 0.08 ln |5x + 1|
CHAPTER 3
D3.1. Given a 60-μC point charge located at the origin, find the total electric flux passing through:
(a) that portion of the sphere r = 26 cm bounded by 0 < θ < π/2 and 0 < ϕ < π/2; (b) the closed
surface defined by ρ = 26 cm and z = ±26 cm; (c) the plane z = 26 cm.
(a) D =
Q
ar
4πr 2
Q
Q
a (r2 sin θ dθ d )ar =
sin θ dθ d
2 r
4πr
4π
π2
Q π2
Q π2 π2
Q
π/2
π/2
Ψ =
sin θ dθ  d =
( cos θ 0 ) ( 0 )
sin θ dθ d =



0
0
0
0
4π
4π
4π
6
60 10
( cos90  ( cos 0))((π 2)  0) = 7.5 µC
=
4π
(b) Deriving Gauss’s law of the cylinder,
dΨ =
Q =
 DS dS = DS
DS = Dρ =
Q
2πρL

π
L
ρ d dz = DS ρ  d  dz = DS ρ  0 z 0 = DS 2πρL
or Dρ =
0
2π
L
0
Q
aρ
2πρL
Solving Ψ,
2π
L
Q
Q
Q
Q
dΨ =
aρ (ρ dϕ dz)aρ =
ρ dϕ dz =
ρ  d  dz =
(2πρL)
0
0
2πρL
2πρL
2πρL
2πρL
Ψ = Q = 60 µC
(c) Getting the equation in (b) with changing the limits of ϕ as 0 ≤ ϕ ≤ π because only the flux lines
pass in half through the plane,
π
L
Q
Q
dΨ =
ρ  d  dz =
(πρL)
0
2πρL 0
2πρL
Ψ = Q 2 = 30 µC
7
D3.2. Calculate D in rectangular coordinates at point P(2, –3, 6) produced by: (a) a point charge
QA = 55 mC at Q(–2, 3, –6); (b) a uniform line charge ρLB = 20 mC/m on the axis; (c) a uniform
surface charge density ρSC =120 μC/m2 on the plane z = –5 m.
Q
Q
(a) D = ϵ0E =
a =
R
2 R
4πR
4πR3
RQP = (2 – (–2))ax + (–3 – 3)ay + (6 – (–6))az = 4ax – 6ay + 12az
RQP = 14
55 103
Q
DQP =
=
(4ax – 6ay + 12az) = 6.38ax – 9.57ay + 19.14az µC/m2
R
QP
3
3
4πRQP
4π14
ρL
ρL
aR =
R
2πR2
2πR
Rx = (2 – x)ax – 3ay + 6az
Since the infinite line charge density is along x axis, the electric field E at point P is having
only y and z components present. Canceling x component due to symmetry,
Rx = –3ay + 6az
(b) D = ϵ0E =
Rx =
Dx =
45
ρL
20 103
R
=
(–3ay + 6az) = –212ay + 424az µC/m2
x
2
2πRx2
2π 45


ρS
aR
2
The infinite surface change density is an infinite x-y plane located at z = –5 and the charge is
spread on that plane. Going back the formula,
D = ( ρS 2)az = (120 2) μaz = 60az µC/m2
(c) D = ϵ0E =
D3.3. Given the electric flux density, D = 0.3r2ar nC/m2 in free space: (a) find E at point P(r = 2,
θ = 25°, ϕ = 90°); (b) find the total charge within the sphere r = 3; (c) find the total electric flux
leaving the sphere r = 4.
(a) E = D ϵ 0 = (0.3  10–9 r2ar ) ϵ 0 = (0.3  10–9 (22 )ar ) ϵ0 = 135.5ar V/m
(b) Q =
D
dS = 0.3 109 r2ar 
2π
0

π
0
r2 sin θ dθ d ar = 0.3 109 r2 (4πr2 )
= 0.3 109 (4πr4 ) = 0.3 109 (4π(34 )) = 305 nC
(c) Same procedure in (b),
Ψ = Q = 1.2 109 πr4 = 1.2 109 π(4)4 = 965 nC
D3.4. Calculate the total electric flux leaving the cubical surface formed by the six planes x, y,
z = ±5 if the charge distribution is: (a) two point charges, 0.1 μC at (1, –2, 3) and 1/7 μC at
(–1, 2, –2); (b) a uniform line charge of π μC/m at x = –2, y = 3; (c) a uniform surface charge of
0.1 μC/m2 on the plane y = 3x.
(a) Both given charges are enclosed by the cubical volume according to Gauss’s law. Adding,
Ψ = Q1 + Q2 = 0.1 µC + (1 7) μC = 0.243 µC
(b) The line charge distribution passes through x = –2 and y = 3; it is parallel to z axis. The total
length of that charge distribution enclosed by the given cubical volume is 10 units as z = ±5.
Applying Gauss’s law of line charge,
Ψ = ρL L = π(10) µC = 31.4 µC
(c) A straight line equation, y = 3x, passes through the origin. We need to find the length of that
line which is enclosed by the given volume. The length is moving up and down along z axis as
z = ±5 to a form a plane. Putting y = 5,
5 = 3; x = 5 3
Finding the length of that line on the plane formed by positive x and y axes,
52  (5 3)2 = 5.27
The same length we get on the plane formed by negative x and y axes. Adding two lengths,
5.27 + 5.27 = 10.54
That straight line moves between z = ±5 to form a plane. Knowing the area,
8
10(10.54) = 105.4
Applying Gauss’s law of surface charge,
Ψ = ρS S = (0.1 µC)105.4 = 10.54 µC
D3.5. A point charge of 0.25 μC is located at r = 0, and uniform surface charge densities are located
as follows: 2 mC/m2 at r = 1 cm, and –0.6 mC/m2 at r = 1.8 cm. Calculate D at: (a) r = 0.5 cm; (b)
r = 1.5 cm; (c) r = 2.5 cm. (d) What uniform charge density should be established at r = 3 cm to
cause D = 0 at r = 3.5 cm?
Note: Always remember the radius r within the charge Q and charge density D are different.
Assume them as r(Q) and r(D).
Q1
0.25 106
(a) D =
=
a
ar = 796ar µC
r
4πr( D )2
4π(0.5 102 )2
(b) Finding the second charge Q2,
Q2 = ρS S = ρS (4πr(Q)2) = 2 × 10–3(4π(1 × 10–2)2) = 2.513 × 10–6 C
Q1  Q2
0.25 106  2.513 106
D =
ar =
ar = 977ar µC
4πr( D )2
4π(1.5 102 )2
(c) Finding the third charge Q3,
Q3 = ρS (4πr(Q)2) = –0.6 × 10–3(4π(1.8 × 10–2)2) = –2.443 × 10–6 C
Q1  Q2  Q3
0.25 106  2.513 106  2.443 106
a
D =
=
ar = 40.8ar µC
r
4πr( D )2
4π(2.5 102 )2
Q1  Q2  Q3  ρS S
Q1  Q2  Q3  Q4
0.32 106  ρS 4πr(Q )2
a
a
=
=
ar
r
r
4πr( D )2
4πr( D )2
4πr( D )2
(d) 0 = D =
0 = (0.32 106  4 ρS πr(Q)2 )ar
–0.32 × 10–6 = 4ρSπr(Q)2
0.32 106
0.32 106
ρS = 
=
= –28.3 µC/m2

2 2
2
4πr(Q )
4π(3 10 )
D3.6. In free space, let D = 8xyz4ax + 4x2z4ay + 16x2yz3az pC/m2. (a) Find the total electric flux
passing through the rectangular surface z = 2, 0 < x < 2, 1 < y < 3, in the az direction. (b) Find E at
P(2, –1, 3). (c) Find an approximate value for the total charge contained in an incremental sphere
located at P(2, –1, 3) and having a volume of 10–12 m3.
(a) Ψ =
=
 D dS where dS = dx dy az
  (8xyz a  4 x z a  16 x yz a )
3
1
2
4
0
3
x
2 4
y
2
2
3
x
dxdyaz =
3
2
  16x yz
1
2
0
3
dxdy
= z3  y dy  16 x2 dx = z3 ( y2 2) 1 (16 x3 3) 0 = [16z3(4) (8 3) ]z = 2 = 1365 pC
1
0
3
2
(b) E = D ϵ0 = [((8xyz4ax + 4x2z4ay + 16x2yz3az)  10–12 ) ϵ0 ]x = 2, y = –1, z = 3
= –146.4ax + 146.4ay – 195.2az V/m
 Dx Dy Dz 



  Δv
y
z 
 x
Finding the partial derivatives of each term,
Dx x = (8xyz4 ) x = (8yz4)x = 2, y = –1, z = 3 = –648
(c) Q = Charge enclosed in volume Δv
Dy y = (4 x2 z4 ) y = 0
Dz z = (16 x2 yz3 ) z = (48x2yz2)x = 2, y = –1, z = 3 = –1728
Q = (–648 – 1728) × 10–12 Δv = [(–648 – 1728) × 10–12 ](10–12) = –2.38 × 10–21 C
D3.7. In each of the following parts, find a numerical value for div D at the point specified: (a)
D = (2xyz – y2)ax + (x2z – 2xy)ay + x2yazC/m2 at PA(2, 3, –1); (b) D = 2ρz2 sin2 ϕ aρ + ρz2 sin 2ϕ aϕ +
2ρ2z sin2 ϕ azC/m2 at PB(ρ = 2, ϕ = 110°, z = –1); (c) D = 2r sin θ cos ϕ ar + r cos θ cos ϕ aθ –
r sin ϕ aϕC/m2 at PC(r = 1.5, θ = 30°, ϕ = 50°).
(a) div D = Dx x + Dy y + Dz z
Finding the partial derivatives of each term,
Dx x = (2 xyz  y2 ) x = 2yz
9
Dy y = ( x2 z  2 xy) y = –2x
Dz z = ( x2 y) z = 0
div D = (2yz – 2x)x = 2, y = 3, z = –1 = –10
1 ( ρDρ ) 1 D Dz
(b) div D =


ρ ρ
ρ 
z
Finding the partial derivatives of each term,
1 ( ρDρ )
1 (2 ρ2 z2 sin2  )
=
= (4 z2 sin2  )ρ 2, 110, z 1 = 3.532
ρ ρ
ρ
ρ
1 D
1 ( ρz2 sin 2 )
=
= (2 z2 cos 2 )ρ 2, 110, z 1 = –1.532
ρ 
ρ

Dz
(2 ρ2 z sin2  )
=
= (2ρ2 sin2  )ρ 2, 110, z 1 = 7.064
z
ρ
div D = 3.532 + (–1.532) + 7.064 = 9.06
1 (r2 Dr )
1 (sin θDθ )
1 D
(c) div D = 2


r
r
r sin θ
θ
r sin θ 
1 (r2 Dr )
1 (2r3 sin θ cos  )
=
= (6sin θ cos  )r 1.5,θ 30, 50 = 1.928
r2 r
r2
r
1 (sin θDθ )
1 (r sin θ cos θ cos  Dθ )
=
r sin θ
θ
r sin θ
θ
Applying product rule of the derivative of sin θ cos θ,
cos 
=
(cos2 θ  sin2 θ )r 1.5,θ 30, 50 = 0.643
sin θ
1 D
1 (r sin  )
=
= ( cos  sin θ )r 1.5,θ 30, 50 = –1.286
r sin θ 
r sin θ

div D = 1.928 + 0.643 – 1.286 = 1.29
D3.8. Determine an expression for the volume charge density associated with each D field: (a)
2x 2 y
2x 2
4xy
ax +
ay – 2 az; (b) D = z sin ϕ aρ + z cos ϕ aϕ + ρ sin ϕ az; (c) D = sin θ sin ϕ ar +
z
z
z
cos θ sin ϕ aθ + cos ϕ aϕ.
Note: The volume charge density ρv is equal to div D.
(a) Dx x = (4 xy z) x = 4y z
D=
Dy y = (2 x2 z ) y = 0
Dz z = ( 2 x2 y z2 ) z = 4x2 y z3
ρv = 4y z + 4x2 y z3 = (4 y z3 )( x2  z2 )
(b) (1 ρ) ( ρDρ ) ρ = (1 ρ) ( ρz sin ) ρ = ( z sin ) ρ
(1 ρ) D  = (1 ρ) ( z cos  )  = ( z sin  ) ρ
Dz z = ( ρ sin  ) z = 0
ρv = ( z sin  ) ρ + ( z sin  ) ρ = 0
(c) (1 r2 ) (r2 Dr ) r = (1 r 2 ) ( r 2 sin θ sin ) r = (2sin θ sin ) r
(1 r sin θ ) (sin θDθ ) θ = (1 r sin θ )  (sin θ cos θ sin ) θ
= (sin  r sin θ )(cos2 θ  sin2 θ )
(1 r sin θ ) D  = (1 r sin θ ) (cos  )  =  sin  (r sin θ )
ρv = (2sin  sin  ) r + (sin  r sin  )(cos2   sin2  ) + ( sin  (r sin  ))
2sin2  sin   cos2  sin   sin2  sin   sin 
sin2  sin   cos2  sin   sin 
=
r sin 
r sin 
Since sin2   cos2  = 1,
sin  (sin2   cos2  )  sin 
0
=
=
= 0
r sin 
r sin 
=
10
D3.9. Given the field D = 6ρ sin (1/2)ϕ aρ + 1.5ρ cos (1/2)ϕ aϕ C/m2, evaluate both sides of the
divergence theorem for the region bounded by ρ = 2, ϕ = 0, ϕ = π, z = 0, and z = 5.
Using the divergence theorem with the formula:  D dS ,
S

S


D dS =
=
S
S
(6 ρ sin ( 2) aρ  1.5 ρ cos ( 2) a
6 ρ sin ( 2) ρ d dz 
π
5
0
0

=
( ρ d dzaρ  dρ dza )
 1.5 ρ cos (
6 ρ sin ( 2) ρ d dz 
2) dρ dz
S
2
5
0
0
  1.5 ρ cos (
π
5
2
0
0
0
2) dρ dz
5
= 6 ρ2  sin ( 2) d  dz  (1.5 cos ( 2))  ρ dρ  dz

π
= 24 2cos  2 0

 z 
5
0
2
 (1.5cos  2) ρ2 2 0
0
 z 
5
0
Since the second surface lies at ϕ = 0°,
= 24(2)(5)  (1.5cos  2)(2)(5) = 240 – 15 = 225
Using another divergence theorem with the formula:

vol
 D dv but solving first the  D,
 D = div D = (1 ρ) ( ρDρ ) ρ + (1 ρ) D  + Dz z
= (1 ρ) (6 ρ2 sin  2) ρ + (1 ρ) (1.5 ρ cos  2)  + 0
= 12sin  2 + (1.5 2)sin  2 = 11.25sin  2

vol
 D dv =
2
π
5
0
0
0


2
π
5
0
0
0
(11.25sin  2)ρ dρ d dz = 11.25 ρ dρ  (sin  2) d  dz
2

π
= 11.25 ρ2 2 0 2cos  2 0
 z 
5
0
= 11.25(2)(2)(5) = 225
CHAPTER 4
D4.1. Given the electric field E = (1/z2)(8xyzax + 4x2zay – 4x2yaz) V/m, find the differential amount
of work done in moving a 6-nC charge a distance of 2 μm, starting at P(2, –2, 3) and proceeding
in the direction aL = (a) –6/7ax + 3/7ay + 2/7az; (b) 6/7ax – 3/7ay – 2/7az; (c) 3/7ax + 6/7ay.
(a) dW = –QE dL
Finding first the differential length dL,
dL = aL dL = (–6/7ax + 3/7ay + 2/7az)(2 × 10–6) = (–12/7ax + 6/7ay + 4/7az)(1 × 10–6)
dW = – (6 × 10–9)(1/z2)(8xyzax + 4x2zay – 4x2yaz) (–12/7ax + 6/7ay + 4/7az)(1 × 10–6)
= –6 × 10–15[(1/z2)((–96/7)xyz + (24/7)x2z – (16/7)x2y)]x = 2, y = –2, z = 3 = –6 × 10–15(224/9)
= –149.3 fJ
(b) Same procedure in (a),
dL = (12/7ax – 6/7ay + 4/7az)(1 × 10–6)
dW = – (6 × 10–9)(1/z2)(8xyzax + 4x2zay – 4x2yaz) (12/7ax – 6/7ay + 4/7az)(1 × 10–6)
= –6 × 10–15[(1/z2)((96/7)xyz – (24/7)x2z – (16/7)x2y)]x = 2, y = –2, z = 3 = –6 × 10–15(–224/9)
= 149.3 fJ
(c) Same procedure in (a) and (b),
dL = (6/7ax + 12/7ay + 0az)(1 × 10–6)
dW = – (6 × 10–9)(1/z2)(8xyzax + 4x2zay – 4x2yaz) (6/7ax + 12/7ay + 0az)(1 × 10–6)
= –6 × 10–15[(1/z2)((48/7)xyz + (48/7)x2z]x = 2, y = –2, z = 3 = –6 × 10–15(0)
= 0
D4.2. Calculate the work done in moving a 4-C charge from B(1, 0, 0) to A(0, 2, 0) along the path
y = 2 – 2x, z = 0 in the field E = (a) 5ax V/m; (b) 5xax V/m; (c) 5xax + 5yay V/m.
A
(a) W = Q  E dL where dL = dx ax + dy ay + dz az
B
0
0
= 4 (5ax  0ay  0az ) (dx ax  dy ay  dz az ) = 4 5 dx = 20 x 1 = –20(–1) = 20 J
1
0
1
(b) Same procedure in (a),
0
W = 4 5 x dx = 20( x2 2) 1 = –20(–0.5) = 10 J
0
1
(c) Same procedure in (a) and (b),
11
 x2 y2 (0,2) 
 = –20(1.5) = –30 J
W = 4 (5 x dx  5 y dy) = 20  
(1,0)
 2 2 (1,0) 


D4.3. We will see later that a time-varying E field need not be conservative. (If it is not
final
conservative, the work expressed by equation: W = –Q ∫initial E dL may be a function of the path
used.) Let E = yax V/m at a certain instant of time, and calculate the work required to move a 3-C
charge from (1, 3, 5) to (2, 0, 3) along the straight-line segments joining: (a) (1, 3, 5) to (2, 3, 5)
to (2, 0, 5) to (2, 0, 3); (b) (1, 2, 5) to (1, 3, 3) to (1, 0, 3) to (2, 0, 3).
(a) Use the formula in D4.2 (a). One differential unit vector of each line segments is used. For
segment (1, 3, 5) to (2, 3, 5),
(0,2)
2
 
3
W1 = 3 ( y ax  0ay  0az ) dx ax = 3y  dx = 3 y x 1
1
2
1
y 3
= –9 J
For segment (2, 3, 5) to (2, 0, 5),
2
W2 = 3 ( y ax  0ay  0az ) dy ay = 0
1
For segment (2, 0, 5) to (2, 0, 3),
2
W3 = 3 ( y ax  0ay  0az ) dz az = 0
1
Adding the total work, W = W1 + W2 + W3 = –9 J
(b) Same procedure in (a). For segment (1, 2, 5) to (1, 3, 3),
1
 
W1 = 3y  dx = 3 y x 1
1
2
= 0
y 3
Both W2 and W3 are zeroes no matter what the line segments are. So adding the total work, W = 0.
D4.4. An electric field is expressed in rectangular coordinates by E = 6x2ax + 6yay + 4az V/m.
Find: (a) VMN if points M and N are specified by M(2, 6, –1) and N(–3, –3, 2); (b) VM if V = 0 at
Q(4, –2, –35); (c) VN if V = 2 at P(1, 2, –4).
M
(2,6, 1)
N
( 3, 3,2)
(a) VMN =   E dL = 
(2,6, 1)
(6 x2 dx  6 y dy  4dz ) = (2 x2  3 y2  4 z ) ( 3,3,2)
= (70  81  12) = –139 V
(2,6, 1)
M
(b) VMQ =   E dL = (2 x2  3 y2  4 z ) (4,2,35) = (112  96  136) = –120 V
Q
VMQ = VM – VQ
VM = VMQ + VQ = –120 V + 0 = –120 V
( 3, 3,2)
N
(c) VNP =   E dL = (2 x2  3 y2  4 z ) (1,2,4)
P
= (56  15  24) = 17 V
VNP = VN – VQ
VN = VNP + VP = 17 V + 2 V = 19 V
D4.5. A 15-nC point charge is at the origin in free space. Calculate V1 if point P1 is located at
P1(–2, 3, –1) and (a) V = 0 at (6, 5, 4); (b) V = 0 at infinity; (c) V = 5 V at (2, 0, 4).
(a) Assume that second point is M. Applying the potential difference VAB, or VPM in the problem,
Q  1
1 

VPM =


4πϵ 0  RP RM 
RP =
(2) 2  32  (1) 2 =
RM =
62  52  42 =
14
77
15 109  1
1 


 = 20.67 V
4πϵ 0  14
77 
VPM = VP – VM
VP = VPM + VM = 20.67 V + 0 = 20.67 V
(b) Same procedure in (a),
VPM =
RP = 14
RM = 
VPM =
15 109  1
1
  = 36 V

4πϵ 0  14  
12
VP = VPM + VM = 36 V + 0 = 36 V
(c) Same procedure in (a) and (b),
RP =
14
RM =
22  42 = 20
15 109  1
1 

VPM =

 = 5.89 V
4πϵ 0  14
20 
VP = VPM + VM = 5.89 V + 5 V = 10.89 V
D4.6. If we take the zero reference for potential at infinity, find the potential at (0, 0, 2) caused by
this charge configuration in free space (a) 12 nC/m on the line ρ = 2.5 m, z = 0; (b) point charge of
18 nC at (1, 2, –1); (c) 12 nC/m on the line y = 2.5, z = 0, –1.0 < x <1.0.
ρL (r' ) dL'
(a) V (r) = 
where dLʹ = ρ dϕ
4πϵ0 r  r'
Since the distance is converted in rectangular coordinates, we can mentally solve it where ϕ = 0°.
|r – rʹ| =
V (r) =
(b) V (r) =
(0  2.5)2  (2  0)2 = 3.20

2π
0
 
12 109 ρ 2π
12 109 ρ d
12 109 ρ 2π
=
=

d

4πϵ0 (3.20) 0
4πϵ0 (3.20) 0
4πϵ0(3.20)
ρ  2.5
= 529 V
Q1
4πϵ0 r  r1
|r – rʹ| =
(0  1)2  (0  2)2  (2  (1))2 =
14
18 109
= 43.2 V
4πϵ0 14
(c) Same formula in (a) but now dLʹ = dx. Solving the distance |r – rʹ| where x is symmetric,
V (r) =
|r – rʹ| =
V (r) =
x2  10.25
(0  x)2  (0  2.5)2  (2  0)2 =
1
12 109
1
4πϵ0 x  10.25

Integrating 1
2
dx =
12 109
4πϵ0

1
1
1
x  10.25
2
dx
x2  10.25 is solved by trigonometric substitution and yielding,
1


12 109  x  x2  10.25 
12 109
=
=
ln
 0.615 = 66.3 V

4πϵ0 
4πϵ0
10.25
1 

D4.7. A portion of a two-dimensional (Ez = 0) potential field is shown the figure below. The grid
lines are 1mm apart in the actual field. Determine approximate values for E in rectangular
coordinates at: (a) a; (b) b; (c) c.
Requires graphical solution.
D4.8. Given the potential field in cylindrical coordinates, V = [100/(z2 + 1)]ρ cos ϕ V, and point P
at ρ = 3 m, ϕ = 60°, z = 2 m, find values at P for (a) V; (b) E; (c) E; (d) dV/dN; (e) aN ; (f) ρv in free
space.
(a) V = [100/(z2 + 1)]ρ cos ϕ = [100/(22 + 1)]3 cos 60° = 30 V
13
 V
1 V
V 
aρ 
a 
az 
(b) E = V =  
ρ 
ρ 
 ρ
 100

100
200 z
cos  aρ  2 sin  a  2
ρ
cos

a
=  2
z
 = –(10aρ – 17.32aϕ – 24az)
z 1
( z  1)2
 z 1

= –10aρ + 17.32aϕ + 24az V/m
(c) E =
(10)2  17.322  242 = 31.2 V/m
(d) dV dN = E = 31.2 V/m
(e) E = (dV dN )aN
aN = 
E
 –10aρ  17.32a  24az 
= 
 = 0.32aρ – 0.55aϕ – 0.77az
dV dN
31.2


 100

100
200 z
cos  aρ  2 sin  a  2
ρ cos  az 
(f) D = ϵ0E = ϵ 0   2
2
z 1
( z  1)
 z 1

ρv =  D = div D = (1 ρ) ( ρDρ ) ρ + (1 ρ) D  + Dz z
[ z ( z2  1)2 ]
= 0 + 0 + 200ϵ0 ρcos
z
Applying product rule of the derivative of z ( z2  1)2 and yielding,
 1
4 z2 
= 200ϵ0 ρcos  2
= –234 pC/m3

2
2
3
 ( z  1) ( z  1)  ρ 3,  60, z 2
D4.9. An electric dipole located at the origin in free space has a moment p = 3ax – 2ay + az nC ∙ m.
(a) Find V at PA(2, 3, 4). (b) Find V at r = 2.5, θ = 30°, ϕ = 40°.
1
r  r'
1
p
p r  r'
(a) V =
=
2
3
r  r'
4πϵ0 r  r'
4πϵ0 r  r '
r – rʹ = (2 – 0)ax + (3 – 0)ay + (4 – 0)az = 2ax + 3ay + 4az
|r – rʹ| =
22  32  42 =
1
V =
4πϵ0

29
29
(3ax – 2a y  az )(1109 ) (2ax  3a y  4az ) =
3

4 109
4πϵ0

29

3
= 0.23 V
(b) Using Table 5 from Chapter 1 to convert the point PA to rectangular coordinates,
x = 0.958, y = 0.803, z = 2.165
r – rʹ = (0.958 – 0)ax + (0.803 – 0)ay + (2.165 – 0)az = 0.958ax + 0.803ay + 2.165az
|r – rʹ| =
0.9582  0.8032  2.1652 = 2.5
(3ax – 2ay  az )(1109 )
3.43 109
=
= 1.97 V
(0.958
a

0.803
a

2.165
a
)
x
y
z
4πϵ02.5 3
4πϵ02.5 3
D4.10. A dipole of moment p = 6az nC ∙ m is located at the origin in free space. (a) Find V at
P(r = 4, θ = 20°, ϕ = 0°). (b) Find E at P.
(a) Same formula in D4.9(a). Converting the point P to rectangular coordinates,
x = 1.368, y = 0, z = 3.759
Since the procedure is very similar with D4.9(a) and (b) where p is at the origin,
r – rʹ = 1.386ax + 3.759az and |r – rʹ| = 4
22.55 109
(1.368ax  3.759az )(1109 )
6
a
V =
=
= 3.167 V = 3.17 V
z
4πϵ0 43
44 πϵ 0
Qd
(2cos  ar  sin  a )
(b) E =
4πϵ0 r3
Since there is no charge given, we have to derive with the help of the voltage V since we have
the answer in (a). Deriving,
Qd cos 
4πϵ0 r 2V
V =
; Qd =
cos 
4πϵ0 r 2
Substituting,
V =
14
(4πϵ0 r2 ) cos 
V
(2cos  ar  sin  a ) =
(2ar  tan  a )
3
4πϵ0 r
r
All variables are given in (a). Solving,
3.167
=
(2ar  tan 20 a ) = 1.58ar + 0.29aθ V/m
4
D4.11. Find the energy stored in free space for the region 2 mm < r < 3 mm, 0 < θ < 90°, 0 < ϕ < 90°,
given the potential field V = : (a) 200/r V; (b) (300 cos θ)/r2 V.
(a) E = V =  (V r )ar  (1 r )(V  )a  (1 r sin  )(V  )a 
E =
= [((200 r ) r )ar  0a  0a ] = (200 r2 ) ar
2
(200 r2 )
E =
= 200 r2
0.5 ϵ0 E2 dv = 0.5 ϵ0  200 r2  (r2 sin θ dθ d dr )
2
vol
vol
= 20 000ϵ0 
0.003
0.002

(1 r2 ) dr 
0.003
π2
0

sin θ dθ 
π2
0
π2
= 20 000ϵ0 1 r 0.002  cos θ 0
d
 
π2
0
= 20 000ϵ0 (500 3)(π 2) = 46.4 μJ
(b) Same procedure in (a),
E = {[(300cos θ ) r2 ] r ar  (1 r )[(300cos θ ) r2 ] θ aθ  0a }
= (600cos  ) r3 ar + (300sin  ) r3 a
2
[(600cos  ) r3 ]  [(300sin  ) r3 ]2 =
E =
(360 000cos2  ) r6  (90 000sin2  ) r6
0.5 ϵ0 E2 dv = 0.5 ϵ0{(360 000cos2 θ ) r6  (90 000sin2 θ ) r6 }(r2 sin θ dθ d dr )
vol
vol
= 45 000ϵ0  (1 r4 ) (4sin θ cos2 θ  sin3 θ ) dθ d dr
vol
= 45 000ϵ0 
0.003
0.002
(1 r4 ) dr 
π2
(4sin θ cos2 θ  sin3 θ )dθ 
0
π2
0
d
Integrating (4sin θ cos2 θ + sin3 θ) is solved by ‘integration by substitution’ and trigonometric
identity yielding,
 

 1 0.003 
π2
π2
 cos3 θ  cos θ 0  0
= 45 000ϵ0   3
 3r 0.002 


= 45 000ϵ0 [1 (3(0.003)3 )  1 (3(0.002)3 )](2)(π 2) = 36.7 J
CHAPTER 5
D5.1. Given the vector current density J = 10ρ2zaρ – 4ρ cos2 ϕ aϕ mA/m2: (a) find the current
density at P(ρ = 3, ϕ = 30°, z = 2); (b) determine the total current flowing outward through the
circular band ρ = 3, 0 < ϕ < 2π, 2 < z < 2.8.
(a) J = 10(32)(2) aρ – 4(3)(cos 30°)2 aϕ mA/m2 = 180aρ – 9 aϕ mA/m2
(b) I =

S
J dS where dS = ρ dϕ dz aρ
2π
2.8
0
2
= (1103 )  (10 ρ2 zaρ – 4 ρcos2a ) ( ρ d dz aρ  0a ) = (1103 )10ρ3  d 
S
  z
2π
= 0.01ρ3  0
2
2.8
22

ρ 3
z dz
= 0.01(27)(2π)(1.92) = 3.26 A
D5.2. Current density is given in cylindrical coordinates as J = –106z1.5az A/m2 in the region 0 ≤ ρ
≤ 20 μm, J = 0. (a) Find the total current crossing the surface z = 0.1 m in the az direction. (b) If
the charge velocity is 2 × 106 m/s at z = 0.1 m, find ρv there. (c) If the volume charge density at
z = 0.15 m is –2000 C/m3, find the charge velocity.
(a) I =  J dS where dS = ρ dϕ dρ az
S
=

S
–106 z1.5az
ρ dϕ dz az = –106 z1.5 
20106
0
2π
ρ dρ  d
0
= –10 (0.1 )(2 10 )(2π) = –39.7 µA
6
1.5
10
(b) J = ρvv ; ρv = J z vz where they represent as z component.
15
ρv = [ –106 (0.11.5 )] (2 106 ) = –15.8 mC/m3
(c) Same formula in (b),
vz = J z ρv = [–106 (0.151.5 )] (2000) = 29.0 m/s
D5.3. Find the magnitude of the current density in a sample of silver for which σ = 6.17 × 107 S/m
and μe = 0.0056 m2/V ∙ s if (a) the drift velocity is 1.5 μm/s; (b) the electric field intensity is
1 mV/m; (c) the sample is a cube 2.5 mm on a side having a voltage of 0.4 mV between opposite
faces; (d) the sample is a cube 2.5 mm on a side carrying a total current of 0.5 A.
(a) J = σE
Since vd = –µeE, then J =  σvd μe .
Since we only find the value of the magnitude of the current density J, which we know that
vectors are denoted by bold letters, we consider J and other vectors as the same dot unit vector.
The negative is canceled because they refer to the direction.
J = σvd μe = [(6.17 107 )(1.5 106 )] 0.0056 = 16.5 kA/m2
(b) J = σE = (6.17 × 107)(1 × 10–3) = 61.7 kA/m2
(c) J = σV L = [(6.17 107 )(0.4 103 )] (2.5 103 ) = 9.9 MA/m2
(d) J = I S = 0.5 (2.5 103 )2 = 80 kA/m2
D5.4. A copper conductor has a diameter of 0.6 in. and it is 1200 ft long. Assume that it carries a
total dc current of 50 A. (a) Find the total resistance of the conductor. (b) What current density
exists in it? (c) What is the dc voltage between the conductor ends? (d) How much power is
dissipated in the wire?
(a) We must convert the English units to metric systems. A 1200 feet is about 365.76 m. A 0.6 in
is about 0.01524 m. The radius is half of the diameter; so r = 7.62 × 10–3 m.
The conductivity of the copper is 5.8 × 107 S/m.
The area of the circle is πr2.
R = L σS = 365.76 [(5.8 107 )π( 7.62  10–3 )2 ] = 0.035 Ω
(b) J = I S = 50 [π(7.62 103 )2 ] = 2.74 × 105 A/m2
(c) V = IR = 50(0.035) = 1.75 V
With regards from the answer of the textbook, V = 50(0.03457) = 1.73 V
(d) P = VI = 1.75(50) = 87.5 W
With regards from the answer of the textbook, P = 1.7285(50) = 86.4 W
D5.5. Given the potential field in free space, V = 100 sinh 5x sin 5y V, and a point P(0.1, 0.2, 0.3),
find at P: (a) V; (b) E; (c) |E|; (d) |ρs| if it is known that P lies on a conductor surface.
(a) The function sinh(x) is kind of a hyperbolic trigonometry where sinh(x) = (ex  e x ) 2 . Be
careful that the points are in radians, not degrees.
V = 100 sinh 5(0.1) sin 5(0.2) = 43.8 V
(b) Using back the formula in Chapter 4,
E = V =  (V x)ax  (V y)ay  (V z )az 
=  ((100sinh 5x sin 5 y) x)ax  ((100sinh 5x sin 5 y) y)ay  0az 
The derivative of sinh x is cosh x or (ex  e x ) 2 . But sinh 5x dx is 5 cosh 5x.
= 500cosh 5x sin 5 y ax  500sinh 5x cos5 y ay
= 500cosh 5(0.1)sin 5(0.2) ax  500sinh 5(0.1)cos5(0.2) ay
= –474ax – 140.8ay V/m
(c) E =
(474)2  (140.8)2 = 495 V/m
(d) ρS = ϵ0EN = ϵ0(495) = 4.38 nC/m2
D5.6. A perfectly conducting plane is located in free space at x = 4, and a uniform infinite line
charge of 40 nC/m lies along the line x = 6, y = 3. Let V = 0 at the conducting plane. At P(7, –1, 5)
find: (a) V; (b) E.
It is really good if we refer to answer first in (b) because it is difficult to solve a voltage with lack
of given variables like the charge Q and the electric field E. In (b), a method of image is required.
We are going to mirror the infinite line charge ρL including the distance (or radial vector) R without
16
the conducting plane. For more details, see Section 5.5. The book is explained that the radial vector
from positive (or negative if ever) line charge to point P is symmetric where the plane is located.
For example, if the line charge (not the plane) is at z = 3 and the plane is at z = 0, then we put an
image line charge at z = –3 which the plane is removed.
(b) Locating the radial vector R from the given location of line charge and P,
R+ = ax – 4ay and
R– = –5ax – 4ay
How did we obtain them? Well, draw a two-dimensional rectangular coordinates. Locate P at
x = 7 and y = –1. Notice that no z axis is included because the conducting plane activates on x
axis and the line charge activates on x and y axes. The plane is at x = 4 while the line charge is
x = 6 and y = 3. Therefore, the image line charge is symmetrically at x = 2 and y = 3. We draw
the plane, line charge and image line charge with horizontal line.
We can now find the radial vector R. For R+, the line charge is one unit away from P(x = 7),
so 7 – 6 = 1; the line charge is four units away from P(y = 3), so –1 – 3 = –4. The image line
charge reacts on negative distance because we deal with directions no matter where plane is
placed. For R–, the image line charge is five units away from P(x = 7), so –(7 – 2) = –5; the
image line charge is four units away from P(y = 3) is –(–1 – 3) = 4. Finding their magnitudes,
R+ = 17 = 4.123 and R– = 41 = 6.403.
Knowing each electric fields,
ρL
40 109
aR =
E+ =
(ax  4a y ) = 42.294ax – 169.177ay
2
2πϵ0 R
2πϵ0 17

E– =

ρL
40 109
aR =
(5ax  4a y ) = –87.683ax + 70.147ay
2
2πϵ0R
2πϵ0 41


We cannot get the exact answer from the textbook. Adding together,
E = E+ + E– = –45.39ax – 99.03ay V/m
(a) We know that the plane stands at x axis. Finding the voltage V where only the electric field E
and length for P are extracted at x direction,
V = ExLx =
(45.39)2 (7) = 317.73 V
D5.7. Using the values given in this section for the electron and hole mobilities in silicon at
300 K, and assuming hole and electron charge densities are 0.0029 C/m3 and –0.0029 C/m3,
respectively, find: (a) the component of the conductivity due to holes; (b) the component of the
conductivity due to electrons; (c) the conductivity.
The electron and hole mobilities of pure silicon is 0.12 and 0.025.
(a) σh = ρhµh = (0.0029)0.025 = 72.5 µS/m
(b) σe = –ρeµe = –(–0.0029)0.12 = 348 µS/m
(c) σ = σh + σe = 421 µS/m
D5.8. A slab of dielectric material has a relative dielectric constant of 3.8 and contains a uniform
electric flux density of 8 nC/m2. If the material is lossless, find: (a) E; (b) P; (c) the average number
of dipoles per cubic meter if the average dipole moment is 10–29 C ∙ m.
(a) D = ϵE where ϵ = ϵ0ϵr
E = D ϵ0ϵr = (8 109 ) 3.8ϵ0 = 238 V/m
(b) D = ϵ0E + P ; P = D – ϵ0E
P = (8 × 10–9) – 238ϵ0 = 5.89 nC/m2
1 nv
pi
(c) P = lim

v 0 v
i 1
We have one dipole moment p and no limit of the average volume Δv. Generalizing the formula,
P = p v ; 1 v = P p = 5.89 109 1029 = 5.89 × 1020 m–3
D5.9. Let Region 1 (z < 0) be composed of a uniform dielectric material for which ϵr = 3.2, while
Region 2 (z > 0) is characterized by ϵr = 2. Let D1 = –30ax + 50ay + 70az nC/m2 and find: (a) DN1;
(b) Dt1; (c) Dt1; (d) D1; (e) θ1; (f) P1.
(a) Since the Region 1 affects the z axis, only Dz is referred.
17
DN1 = D1z =
702 nC/m2 = 70 nC/m2
(b) While the normal component DN refers to the mentioned region where it locates, then the rest
which does not affect the region is Dt1.
Dt1 = D1 – DN1 = [(–30ax + 50ay + 70az) – 70az] nC/m2 = –30ax + 50ay nC/m2
(c) Dt1 =
(30)2  502 = 58.31 nC/m2
(d) D1 =
(30)2  502  702 = 91.1 nC/m2
(e) DN1 = D1 cos θ1
θ1 = cos–1 DN 1 D1 = cos–1 (70 109 91.1109 ) = 39.8°
(f) D1 = ϵ0E1 + P1 where D = ϵ0ϵrE
P1 = D1 – ϵ0E1 = D1(1 – (1/ϵr1)) = (–30ax + 50ay + 70az)(1 – (1/3.2)) nC/m2
= –20.6ax + 34.4ay + 48.1az nC/m2
D5.10. Continue Problem D5.9 by finding: (a) DN2; (b) Dt2; (c) D2; (d) P2; (e) θ2.
(a) DN1 = DN2 = 70az nC/m2
(b) Dt1 is not equal to Dt2 because of their relative permittivity ϵr.
Dt1 Dt 2 = ϵr1/ϵr2
Dt2 = Dt1ϵr2 ϵr1 = [(–30ax + 50ay)2] 3.2 nC/m2 = –18.75ax + 31.25ay nC/m2
(c) D1 is not equal to D2 because D2 is the cause of the refraction.
D2 = Dt2 + DN2 = –18.75ax + 31.25ay + 70az nC/m2
(d) P2 = D2(1 – (1 ϵr2)) = (–18.75ax + 31.25ay + 70az)(1 – (1/2))nC/m2
= –9.38ax + 15.63ay + 35az nC/m2
(e) tan θ1 tan θ2 = ϵr1 ϵr2
θ2 = tan–1 [(ϵr2 tan θ1) ϵr1] = tan–1 [(2 tan 39.8°) 3.2] = 27.5°
Prepared by @kimonbalu © Dec 2014
18