1

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

(c) |rM| = 2 2 2( 1) 2 1 = 6 = 2.45

(d) aMP = 2 2 2

5 5

( 1) ( 5) ( 5)

x y z

a a a =

5 5

51

x y z a a a = –0.14ax – 0.7ay – 0.7az

(e) 2rP – 3rN = 2(–2ax – 3ay – 4az) – 3(3ax – 3ay) = –4ax – 6ay – 8az – 9ax + 9ay

= –13ax + 3ay – 8az

|2rP – 3rN| = 2 2 2( 13) 3 ( 8) = 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.

(a) S = 2 2 2

125(2 1) (4 2) (3 1)

(2 1) (4 2) (3 1)x y z

a a a

= 2 2 2

125( 2 4 )

1 2 4x y z

a a a =

125( 2 4 )

21x y z a a a

= 5.95ax + 11.9ay + 23.8az

(b) aS = 2 2 2

5.95 11.9 23.8

5.95 11.9 23.8

x y z

a a a =

5.95 11.9 23.8

27.27

x y z a a a

= 0.218ax + 0.436ay + 0.873az

(c) 1 = |S| =

2 2 2

2 2 2 2 2 2 2 2 2

125( 1) 125( 2) 125( 1)

( 1) ( 2) ( 1) ( 1) ( 2) ( 1) ( 1) ( 2) ( 1)

x y zx y z x y z x y z

1 = 2 2 2 2

2 2 2 2

125 [( 1) ( 2) ( 1)

[( 1) ( 2) ( 1) ]

x y zx y z

=

2 2 2

2 2 2

125 ( 1) ( 2) ( 1)

( 1) ( 2) ( 1)

x y zx y z

Transposing, 2 2 2

2 2 2

( 1) ( 2) ( 1)

( 1) ( 2) ( 1)

x y zx y z

= 125

Conjugating,

2 2 22 2 2

2 2 2 2 2 2

( 1) ( 2) ( 1)( 1) ( 2) ( 1)

( 1) ( 2) ( 1) ( 1) ( 2) ( 1)

x y zx y zx y z x y z

= 125

2 2 2( 1) ( 2) ( 1)x y z = 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

2

θBAC = 1cosAB AC

AB AC

R RR R

=

1

2 2 3 2 2 3

62cos

( 8) 4 ( 6) ( 9) 2 3

= 53.6°

(d) Analyzing the projection,

|RAB (L)| = RAB aAC = RABAC

AC

RR

= 2 2 2

62

( 9) 2 3 = 6.395

RAB (L) = |RAB (L)| aAC = |RAB (L)|AC

AC

RR

= 2 2 2

9 2 3(6.395)

( 9) 2 3

x y z

a a a

= –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) RAB × RAC = 8 4 6

9 2 3

x y z

a a a = [4(3) – (–6)(2)]ax + [(–6)(–9) – (–8)(3)]ay + [(–8)(2) – 4(–9)]az

= 24ax + 78ay + 20az

(b) A = 1

base height2

= 1

2AB BCR R

Finding the direction of vector RBC

RBC = (–3 – (–2))ax + (1 – 3)ay + (5 – (–4))az = –ax – 2ay + 9az

RAB × RBC = 8 4 6

1 2 9

x y z

a a a = 24ax + 78ay + 20az

|RAB × RBC| = 2 2 224 78 20 = 84

A = (1/2)(84) = 42

(c) AB ACR Ra = 24 78 20

84

x y z a a a = 0.286ax + 0.928ay + 0.238az

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 = 2 2x y

ϕ = tan–1 y/x

z = z

ρ = 2 2( 3.1) 2.6 = 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

3

|CD| = 2 2 2( 1.24) 6.59 ( 5) = 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

r = 2 2 2x y z

θ = 1coszr

ϕ = tan–1 y/x

r = 2 2 2( 3) 2 1 = 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

4

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| = 2 2 2(0.585 ( 3)) ( 1.607 2) (4.7 1) = 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 ϕay 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 = 2 2 211 4 ( 9) m = 14.76 m

(c) FAB = 0

1 2

24πϵAB

AB

Q Q

Ra =

1

20

2

4πϵAB

AB AB

Q Q

R R

R =

0

1 2

34πϵAB

AB

Q Q

RR

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,

= 6 6

9 3

( 20 10 )(50 10 )(11 4 9 )

4π(10 36π)(14.7648)x y z

a a a = (–2.7961 × 10–3)(11ax + 4ay – 9az)

= 30.76ax + 11.184ay – 25.16az mN

(d) FAB = 0

1 2

34πϵAB

AB

Q Q

RR =

6 6

12 3

( 20 10 )(50 10 )(11 4 9 )

4π(8.854 10 )(14.7648)x y z

a a a

= (–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| = 2 2 2( 0.25) 0.3 ( 0.15) m = 0.4183 m

EA = 0

24πϵA

AO

AO

Q

Ra =

2

A AO

AO AO

Q

R R

R =

034πϵ

AAO

AO

Q

RR

04πϵ

5

= 6

30

0.3 10( 0.25 0.3 0.15 )

4πϵ (0.4183)x y z

a a a V/m

= 9.21 11.05 5.53x y z a a a 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

|RBO| = 2 2 20.1 ( 0.08) ( 0.12) m = 0.1755 m

EB = 0

34πϵB

BO

BO

Q

RR =

30

60.5 10(0.1 0.08 0.12 )

4πϵ (0.1755)x y z

a a a V/m

= 83.13 66.51 99.76x y z a a a 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 = 1.14 5.7 3.99x y z a a a 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 = 10.79 5.18 16.41x y z a a a kV/m

E = 11.9ax – 0.52ay – 12.4az kV/m

D2.3. Evaluate the sums: (a)5

20

1 ( 1);

1

m

m m

(b)

4

2 1.51

(0.1) 1.

(4 )

m

m m

(a) 0 1 2 3 4 5

2 2 2 2 2 2

1 ( 1) 1 ( 1) 1 ( 1) 1 ( 1) 1 ( 1) 1 ( 1)

0 1 1 1 2 1 3 1 4 1 5 1

= 2.52

(b) 1 2 3 4

2 1.5 2 1.5 2 1.5 2 1.5

(0.1) 1 (0.1) 1 (0.1) 1 (0.1) 1

(4 1 ) (4 2 ) (4 3 ) (4 4 )

= 0.176

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,

Q = vvolρ dv =

0.2 0.2 0.2

3 3 30.1 0.1 0.1

1dx dy dz

x y z = 0.2 0.2 0.2

3 3 3

0.1 0.1 0.1x dx y dy z dz

=

0.2 0.2 0.22 2 2

0.1 0.1 0.12 2 2

x y z

= 112.5 C

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

2 2

0 0 2

( sin 0.6 )ρ z ρ dρ d dz = 0.1 π 4

3 2

0 0 2

sin 0.6ρ dρ d z dz

=

0.1 4π4 3

00 2

cos0.6

4 0.6 3

ρ z

= 4 3 30.1 cos0.6(180 ) cos0.6(0 ) 4 2

04 0.6 0.6 3 3

= 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,

Q = 2π π

2 2 2

0 0 2

( )( sin )re r r θ dθ d dr

= 2π π

2

0 0 2

sin rd θ dθ e dr

= 2π π 2

0 0 0cos 2rθ e

= (2π – 0)(–cos 180° – (–cos 0°))(–e–∞ 2 – (–e 0 2))

6

= 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).

(a) E = 2πϵ0

Lρ

ρ

ρa where ρ is just a radical distance R, then

= 02π

Lρ

R R

R =

202π

Lρ

RR

Finding Ex and Ey at Point A since infinite uniform line charges lie along x and y axes,

Rx = 4az

Ex = 9

20

5 10(4 )

2π 4z

a = 22.469az V/m

Ry = 4az

Ey = 9

20

5 10(4 )

2πϵ 4z

a = 22.469az V/m

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

Ex =

9

22 2

0

5 10(3 4 )

2πϵ 3 4y z

a a = 10.785ay + 14.38az V/m

Ey =

0

9

22

5 10(4 )

2πϵ 4z

a = 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/m2 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,

Ez = –4 = s

N

ρa =

9

0

3 10

2ϵz

a = 169.41az V/m

Ez = 1 = 9

0

6 10

2ϵz

a = 338.82az V/m

Ez = 4 = 9

0

8 10

2ϵz

a = –451.76az V/m

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

E = (a) 8x

y

ax +2

2

4x

yay; (b) 2e5x[y(5x + 1)ax + xay].

(a) dy

dx =

y

x

E

E =

2 24

8

x y

x y =

2

x

y

02ϵ

ϵ ϵ

ϵ

7

2ydy = –xdx 2ydy = xdx

y2 = 2 2

2 2

x C

For P(x = 1, y = 4, z = –2),

42 = 2 21

2 2

C ; C 2 = 33

x2 + 2y2 = 33

(b) dy

dx =

(5 1)

x

y x

ydy = 5 1

xdx

x

Letting u = 5x + 1 and du = 5 dx, and then x = (u – 1)/5, 2

2

y =

1 1

5 5

u du

u

= 1

25

udu

u

=

1 1

25 25du du

u =

1( ln )

25u u

2

2

y =

21[(5 1) ln 5 1]

25 2

Cx x

For P(x = 1, y = 4, z = –2), 24

2 =

21[(5(1) 1) ln 5(1) 1]

25 2

C ; C 2 = 15.66

2

2

y =

15.660.04(5 1) 0.04ln 5 1

2x x

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 = 24π

r

Q

ra

dΨ = 2

2( sin )

4πr r

Qr θ dθ d

ra a = sin

4π

Qθ dθ d

Ψ = π 2 π 2

0 0

sin4π

Qθ dθ d =

π 2 π 2

0 0

sin4π

Qθ dθ d =

π/2 π/2

0 0( cos ) ( )

4π

Qθ

= 660 10

( cos90 ( cos0 ))((π 2) 0)4π

= 7.5 µC

(b) Deriving Gauss’s law of the cylinder,

Q = S d D S = SD ρ d dz =

π

0 0

L

SD ρ d dz =2π

0 0

L

SD ρ z = 2πSD ρL

DS = Dρ = 2π

Q

ρL or Dρ =

2π

Q

ρLaρ

Solving Ψ,

dΨ = 2π

Q

ρLaρ (ρ dϕ dz)aρ =

2π

Q

ρLρ dϕ dz =

2π

0 02π

LQρ d dz

ρL = (2π )

2π

QρL

ρL

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,

dΨ = π

0 02π

LQρ d dz

ρL = (π )

2π

QρL

ρL

Ψ = 2Q = 30 µC

8

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.

(a) D = ϵ0E = 24π

R

Q

Ra =

34π

Q

RR

RQP = (2 – (–2))ax + (–3 – 3)ay + (6 – (–6))az = 4ax – 6ay + 12az

RQP = 14

DQP = 34π

QP

QP

Q

RR =

3

3

55 10

4π14

(4ax – 6ay + 12az) = 6.38ax – 9.57ay + 19.14az µC/m2

(b) D = ϵ0E = 2π

LR

ρ

Ra =

22π

Lρ

RR

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

Rx = 45

Dx = 22π

Lx

x

ρ

RR =

3

2

20 10

2π 45

(–3ay + 6az) = –212ay + 424az µC/m2

(c) D = ϵ0E = 2

SR

ρa

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 = ( 2)S zρ a = (120 2) μ za = 60az µC/m2

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 = 0D = –9 20(0 10 ).3 rr a = –9 2

0( (0.3 10 2 ) )r a = 135.5ar V/m

(b) Q = d D S = 2π π

9 2 2

0 00.3 10 sin r rr r θ dθ d a a = 9 2 20.3 10 (4π )r r

= 9 40.3 10 (4π )r = 9 40.3 10 (4π(3 )) = 305 nC

(c) Same procedure in (b),

Ψ = Q = 9 41.2 10 πr = 9 41.2 10 π(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,

2 25 (5 3) = 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,

ϵ ϵ ϵ

9

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).

(a) D = 1

2( )4π

r

D

Q

ra =

6

2 2

0.25 10

4π(0.5 10 )r

a = 796ar µC

(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

D = 1 2

2( )4π

r

D

Q Q

r

a =

6 6

2 2

0.25 10 2.513 10

4π(1.5 10 )r

a = 977ar µC

(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

D = 1 2 3

2( )4π

r

D

Q Q Q

r

a =

6 6 6

2 2

0.25 10 2.513 10 2.443 10

4π(2.5 10 )r

a = 40.8ar µC

(d) 0 = D = 1 2 3 4

2( )4π

r

D

Q Q Q Q

r

a =

1 2 3

2( )4π

Sr

D

Q Q Q ρ S

r

a =

6 2( )

2( )

0.32 10 4π

4π

S Qr

D

ρ r

r

a

0 = 6 2( )(0.32 10 4 π )S Q rρ r a

–0.32 × 10–6 = 4ρSπr(Q)2

ρS = 6

2( )

0.32 10

4π Qr

=

6

2 2

0.32 10

4π(3 10 )

= –28.3 µC/m2

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 D S where dS = dx dy az

= 3 2

4 2 4 2 3

1 0(8 4 16 )x y x zxyz x z x yz dxdy a a a a =

3 22 3

1 0

16x yz dxdy

= 3 2

3 2

1 016z y dy x dx = z3 3

2

1( 2)y

23

0(16 3)x = [16z3(4) (8 3) ]z = 2 = 1365 pC

(b) E = 0D = [((8xyz4ax + 4x2z4ay + 16x2yz3az)–12

0 10 ) ]x = 2, y = –1, z = 3

= –146.4ax + 146.4ay – 195.2az V/m

(c) Q = Charge enclosed in volume Δv Δx y zD D D

vx y z

Finding the partial derivatives of each term,

xD x = 4(8 )xyz x = (8yz4)x = 2, y = –1, z = 3 = –648

yD y = 2 4(4 )x z y = 0

zD z = 2 3(16 )x yz 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 = xD x + yD y + zD z

Finding the partial derivatives of each term,

xD x = 2(2 )xyz y x = 2yz

ϵ ϵ

10

yD y = 2( 2 )x z xy y = –2x

zD z = 2( )x y z = 0

div D = (2yz – 2x)x = 2, y = 3, z = –1 = –10

(b) div D = 1 ( ) 1ρ zρD D D

ρ ρ ρ z

Finding the partial derivatives of each term, 2 2 2

2 22, 110 , 1

1 ( ) 1 (2 sin ) = = (4 sin ) =

ρρ z

ρD ρ zz

ρ ρ ρ ρ

3.532

22

2, 110 , 1

1 1 ( sin 2 ) = = (2 cos 2 ) =ρ z

D ρzz

ρ ρ

–1.532

2 22 2

2, 110 , 1

(2 sin ) = = (2 sin ) =

zρ z

D ρ zρ

z ρ

7.064

div D = 3.532 + (–1.532) + 7.064 = 9.06

(c) div D = 2

2

1 ( ) 1 (sin ) 1

sin sin

r θr D θD D

r r r θ θ r θ

2 3

1.5, 30 , 502 2

1 ( ) 1 (2 sin cos ) = = (6sin cos ) =

rr θ

r D r θθ

r r r r

1.928

1 (sin ) 1 ( sin cos cos ) =

sin sin

θ θθD r θ θ D

r θ θ r θ θ

Applying product rule of the derivative of sin θ cos θ,

= 2 21.5, 30 , 50

cos(cos sin ) =

sinr θθ θ

θ

0.643

1.5, 30 , 50

1 1 ( sin ) = = ( cos sin ) =

sin sinr θ

D rθ

r θ r θ

–1.286

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)

D = 4xy

zax +

22x

zay –

2

2

2x y

zaz; (b) D = z sin ϕ aρ + z cos ϕ aϕ + ρ sin ϕ az; (c) D = sin θ sin ϕ ar +

cos θ sin ϕ aθ + cos ϕ aϕ.

Note: The volume charge density ρv is equal to div D.

(a) xD x = (4 )xy z x = 4y z

yD y = 2(2 )x z y = 0

zD z = 2 2( 2 )x y z z = 2 34x y z

ρv = 4y z + 2 34x y z = 3 2 2(4 )( )y z x z

(b) (1 ) ( ) = (1 ) ( sin ) = ( sin )ρρ ρD ρ ρ ρz ρ z ρ

(1 ) = (1 ) ( cos ) = ( sin )ρ D ρ z z ρ

= ( sin ) = 0zD z ρ z

ρv = ( sin )z ρ + ( sin )z ρ = 0

(c) 2 2 2 2(1 ) ( ) = (1 ) ( sin sin ) = (2sin sin )rr r D r r r θ r θ r

(1 sin ) (sin ) = (1 sin ) (sin cos sin ) θr θ θD θ r θ θ θ θ

= 2 2(sin sin )(cos sin )r θ θ θ

(1 sin )r θ D = (1 sin ) (cos )r θ = sin ( sin )r θ

ρv = (2sin sin ) r + 2 2(sin sin )(cos sin )r + ( sin ( sin ))r

= 2 2 22sin sin cos sin sin sin sin

sinr

=

2 2sin sin cos sin sin

sinr

Since 2 2sin cos = 1,

= 2 2sin (sin cos ) sin

sinr

=

0

sinr = 0

11

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: S

d D S ,

S

d D S = (6 sin ( 2) 1.5 co ( )s ( 2)S

ρ ρρ d dz dρ dρ ρ z a a a a

= 6 sin ( 2) 1.5 cos( 2) SS

ρ ρ d dρdz dzρ

= π 5 2 5

0 0 0 0

6 sin ( 2) 1.5 cos( 2) ρ dρ dρ d dz ρ z

= 5 5

0 0 0

22

0 (6 sin ( 2) cos( 2))1.5

π

ρ d ρdz dρ dz

= 2π 5 52

0 0 0024 2cos 2 (1.5cos 2) 2z ρ z

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

dv D but solving first the , D

D = div D = (1 ) ( )ρρ ρD ρ + (1 )ρ D + zD z

= 2(1 ) (6 sin 2)ρ ρ ρ + (1 ) (1.5 cos 2)ρ ρ + 0

= 12sin 2 + ( 1.5 2)sin 2 = 11.25sin 2

vol

dv D = 2 π 5

0 0 0(11.25sin 2)ρ dρ d dz =

2 π 5

0 0 011.25 (sin 2)ρ dρ d dz

= 2 52

0 0011.25 2 2cos 2

πρ z = 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) W =

A

B

Q d E L where dL = dx ax + dy ay + dz az

= 0

1

4 (5 0 0 ) ( )x y z x y zdx dy dz a a a a a a =0

14 5 dx =

0

120 x = –20(–1) = 20 J

(b) Same procedure in (a),

W = 0

14 5 x dx =

02

120( 2)x = –20(–0.5) = 10 J

(c) Same procedure in (a) and (b),

12

W = (0,2)

(1,0)4 (5 5 )x dx y dy =

(0,2)2 2

(1,0)

202 2

x y

= –20(1.5) = –30 J

D4.3. We will see later that a time-varying E field need not be conservative. (If it is not

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),

W1 =

2

1

3 ( 0 0 ) x y z xdxy a a a a =3

13y dx = 2

13

3y

y x

= –9 J

For segment (2, 3, 5) to (2, 0, 5),

W2 =

2

1

3 ( 0 0 ) x y z ydyy a a a a = 0

For segment (2, 0, 5) to (2, 0, 3),

W3 =

2

1

3 ( 0 0 ) x y z zdzy a a a a = 0

Adding the total work, W = W1 + W2 + W3 = –9 J

(b) Same procedure in (a). For segment (1, 2, 5) to (1, 3, 3),

W1 = 1

13y dx = 2

13

3y

y x

= 0

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).

(a) VMN = M

Nd E L =

(2,6, 1)2

( 3, 3,2)(6 6 4 )x dx y dy dz

=

(2,6, 1)2 2

( 3, 3,2)(2 3 4 )x y z

= (70 81 12) = –139 V

(b) VMQ = M

Qd E L =

(2,6, 1)2 2

(4, 2, 35)(2 3 4 )x y z

= ( 112 96 136) = –120 V

VMQ = VM – VQ

VM = VMQ + VQ = –120 V + 0 = –120 V

(c) VNP = N

Pd E L =

( 3, 3,2)2 2

(1,2, 4)(2 3 4 )x y z

= ( 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,

VPM = 0

1 1

4πϵ P M

Q

R R

RP = 2 2 2( 2) 3 ( 1) = 14

RM = 2 2 26 5 4 = 77

VPM = 0

915 10 1 1

4πϵ 14 77

= 20.67 V

VPM = VP – VM

VP = VPM + VM = 20.67 V + 0 = 20.67 V

(b) Same procedure in (a),

RP = 14

RM =

VPM = 9

0

15 10 1 1

4πϵ 14

= 36 V

final

13

VP = VPM + VM = 36 V + 0 = 36 V

(c) Same procedure in (a) and (b),

RP = 14

RM = 2 22 4 = 20

VPM = 0

915 10 1 1

4πϵ 14 20

= 5.89 V

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.

(a) V (r) = 0

( )

4πϵLρ ' dL'

'r

r r where dLʹ = ρ dϕ

Since the distance is converted in rectangular coordinates, we can mentally solve it where ϕ = 0°.

|r – rʹ| = 2 2(0 2.5) (2 0) = 3.20

V (r) = 9

2

00

π12 10

4πϵ (3.20)

ρ d =

92π

00 (3.20)

12 10

4πϵρ

d

= 9

2π

200

.5(3.20)

12 10

4πϵ ρ

ρ

= 529 V

(b) V (r) = 0

1

14πϵQ

r r

|r – rʹ| = 2 2 2(0 1) (0 2) (2 ( 1)) = 14

V (r) = 9

0

18 10

4πϵ 14

= 43.2 V

(c) Same formula in (a) but now dLʹ = dx. Solving the distance |r – rʹ| where x is symmetric,

|r – rʹ| = 2 2 2(0 ) (0 2.5) (2 0)x = 2 10.25x

V (r) = 2

91

10

12 10

4πϵ 10.25xdx

=

2

91

10

1

10.

2 10 1

4πϵ 25dx

x

Integrating 21 10.25x is solved by trigonometric substitution and yielding,

=

19 2

10

12 10 10.25ln

4πϵ 10.25

x x

= 9

0

12 100.615

4π

= 66.3 V

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

ϵ

14

(b) E = 1

= ρ z

V V VV

ρ ρ ρ

a a a

= 2 2 2 2

100 100 200cos sin cos

1 1 ( 1)ρ z

zρ

z z z

a a a = –(10aρ – 17.32aϕ – 24az)

= –10aρ + 17.32aϕ + 24az V/m

(c) E = 2 2 2( 10) 17.32 24 = 31.2 V/m

(d) dV dN = E = 31.2 V/m

(e) E = ( ) NdV dN a

aN = dV dN

E

= –10 17.

31.2

32 24ρ z

a a a = 0.32aρ – 0.55aϕ – 0.77az

(f) D = ϵ0E = ϵ 2 2 2

02

100 100 200cos sin cos

1 1 ( 1)ρ z

zρ

z z z

a a a

ρv = D = div D = (1 ) ( )ρρ ρD ρ + (1 )ρ D + zD z

= 0 + 0 + 0

2 2[ ( 1) ]200ϵ ρcos

z z

z

Applying product rule of the derivative of 2 2( 1)z z and yielding,

= 0

3,

2

2

60

2 2 3

, 2

1 4200ϵ ρcos

( 1) ( 1)ρ z

z

z z

= –234 pC/m3

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°.

(a) V = 0

2

1

4πϵ'

''

r rp

r rr r =

3

0

1

4 ϵπ ''

p r r

r r

r – rʹ = (2 – 0)ax + (3 – 0)ay + (4 – 0)az = 2ax + 3ay + 4az

|r – rʹ| = 2 2 22 3 4 = 29

V =

9

3

0

3 – 2 41

( )(1 10 ) (2 3 )

4πϵ 29x y z x y z

a a a a a a =

9

3

0

4 10

4πϵ 29

= 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| = 2 2 20.958 0.803 2.165 = 2.5

V = 9

03

( )(1 10 ) (0.958

3 – 22.1650.803 )

4πϵ

2.5

x y zx y z

a a aa a a =

9

30

3.43 10

4πϵ 2.5

= 1.97 V

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

V = 9

03

(1.368 )(1 10 ) 6

4πϵ 43.759x z

z

a aa =

9

40

22.55 10

4

πϵ

= 3.167 V = 3.17 V

(b) E = 3

0

(2cos sin )4πϵ

r

Qd

r a a

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,

V = 0

2

cos

4πϵQd

r

; Qd =

204π

cos

r V

Substituting,

ϵ

15

E = 2

0 3

0

(4πϵ ) cos (2cos sin )4πϵ

rr

r

a a = (2 tan )r

Vr

a a

All variables are given in (a). Solving,

= 3.167 (2 tan 20 )

4r a a = 1.58ar + 0.29aθ V/m

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 = ( ) (1 )( ) (1 sin )( )rV r r V r V a a a

= [( (200 ) ) 0 0 ]rr r a a a = 2(200 ) rr a

E = 22(200 )r = 2200 r

2vol

00.5 E dv = 22 2

vol

00.5 200 ( sin )r r θ dθ d dr

= 0.003 π 2 π 22

0.00

02 0 0

20 000 (1 ) sinr dr θ dθ d

= 0.003 π 2 π 2

0.002 00 020 000 1 cosr θ = 0 20 000 (500 3)(π 2) = 46.4 μJ

(b) Same procedure in (a),

E = 2 2 [(300cos ) ] (1 ) [(300cos ) ] 0 r θθ r r r θ r θ a a a

= 3(600cos ) rr a + 3(300sin ) r a

E = 23 3 2[(600cos ) ] [(300sin ) ]r r = 2 6 2 6

(360 000cos ) (90 000sin )r r

2vol

00.5 E dv = 2 6 2 6 2

vol00.5 (360 000cos ) (90 000sin ) ( sin )θ r θ r r θ dθ d dr

= 4 2 3

vo0

l45 000 (1 )(4sin cos sin ) r θ θ θ dθ d dr

= 0.003 π 2 π 24 2 3

0.00 0 0

02

45 000 (1 ) (4sin cos sin ) r dr θ θ θ dθ d Integrating (4sin θ cos2 θ + sin3 θ) is solved by ‘integration by substitution’ and trigonometric

identity yielding,

= 0.003

π 2 π 233 00

0.0 20

0

145 000ϵ cos cos3

θ θr

= 03 3

45 000 [ 1 (3(0.003) ) 1 (3(0.002) )](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

d J S where dS = ρ dϕ dz aρ

= 2 2

3 (10 – 4(1 10 ) ) ( 0 )ρ ρ

Sρ z ρcos ρ d dz a a a a =

2π 2.83 3

0 2(1 10 )10ρ z dzd

= 2.82π3 20 2 3

0.01 2ρ

ρ z

= 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 = S

d J S where dS = ρ dϕ dρ az

= 6 1.5–10 S

z z a ρ dϕ dz az = 620 10 2π

0

50

6 1.–10 ρ dρ dz

= 1 106 .5(0.1 )(–10 2 10 )(2π) = –39.7 µA

(b) J = ρvv ; ρv = z zJ v where they represent as z component.

ϵ ϵ

ϵ

ϵ ϵ

ϵ ϵ

ϵ

ϵ

ϵ

16

ρv = 1.5 66[ )] (2–10 (0 ). 101 = –15.8 mC/m3

(c) Same formula in (b), vz = z vJ ρ = 6 1.5(0.15 )] ([–1 )0 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 = d eσ μ v .

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 = d eσv μ = 7 6[(6.17 10 )(1.5 10 )] 0.0056 = 16.5 kA/m2

(b) J = σE = (6.17 × 107)(1 × 10–3) = 61.7 kA/m2

(c) J = σV L = 7 3 3[(6.17 10 )(0.4 10 )] (2.5 10 ) = 9.9 MA/m2

(d) J = I S = 3 20.5 (2.5 10 ) = 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 = 237 –365.76 [(5.8 10 )π 7.62 10( ) ] = 0.035 Ω

(b) J = I S = 3 250 [π(7.62 10 ) ] = 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) = ( ) 2x xe e . 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 = ( ) ( ) ( )x y zV x V y V z a a a

= ( (100sinh5 sin5 ) ) ( (100sinh5 sin5 ) ) 0x y zx y x x y y a a a

The derivative of sinh x is cosh x or ( ) 2x xe e . But sinh 5x dx is 5 cosh 5x.

= 500cosh5 sin5 500sinh5 cos5x yx y x y a a

= 500cosh5(0.1)sin5(0.2) 500sinh5(0.1)cos5(0.2)x y a a

= –474ax – 140.8ay V/m

(c) E = 2 2( 474) ( 140.8) = 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

17

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,

E+ = 2πϵ0

LR

ρR

a =

9

2

0

40 10( 4 )

2πϵ 17x y

a a = 42.294ax – 169.177ay

E– = 2πϵ0

LR

ρR

a =

9

2

0

40 10( 5 4 )

2πϵ 41x y

a a = –87.683ax + 70.147ay

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 = 2( 45.39) (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 = 0 rD ϵ ϵ = 90(8 10 ) 3.8 = 238 V/m

(b) D = ϵ0E + P ; P = D – ϵ0E

P = (8 × 10–9) – 238ϵ0 = 5.89 nC/m2

(c) P = 0

1

1lim

n v

iv iv

p

We have one dipole moment p and no limit of the average volume Δv. Generalizing the formula,

P = vp ; 1 v = P p = 9 295.89 10 10 = 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.

ϵ

18

DN1 = D1z = 270 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 = 2 2( 30) 50 = 58.31 nC/m2

(d) D1 = 2 2 2( 30) 50 70 = 91.1 nC/m2

(e) DN1 = D1 cos θ1

θ1 = cos–1 1 1ND D = cos–1 9 9(70 10 91.1 10 ) = 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.

1 2t tD D = 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