# Solutions of engineering electromagnetics 6th edition william h. hayt, john a. buck.pdf, Past Exams for Electromagnetic Engineering

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CHAPTER 1

1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find: a) a unit vector in the direction of −M + 2N.

M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus

a = (26, 10, 4)|(26, 10, 4)| = (0.92, 0.36, 0.14)

b) the magnitude of 5ax + N − 3M: (5, 0, 0)+ (8, 7,−2)(−30, 12,−24) = (43,−5, 22), and |(43,−5, 22)| = 48.6.

c) |M||2N|(M + N): |(−10, 4,−8)||(16, 14,−4)|(−2, 11,−10) = (13.4)(21.6)(−2, 11,−10) = (−580.5, 3193,−2902)

1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,−2, 1): a) Specify the vector A extending from the origin to the point A.

A = (4, 3, 2) = 4ax + 3ay + 2az

b) Give a unit vector extending from the origin to the midpoint of line AB.

The vector from the origin to the midpoint is given by

M = (1/2)(A + B) = (1/2)(4 − 2, 3 + 0, 2 + 5) = (1, 1.5, 3.5) The unit vector will be

m = (1, 1.5, 3.5)|(1, 1.5, 3.5)| = (0.25, 0.38, 0.89)

c) Calculate the length of the perimeter of triangle ABC:

Begin with AB = (−6,−3, 3), BC = (9,−2,−4), CA = (3,−5,−1). Then

|AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.32

1.3. The vector from the origin to the point A is given as (6,−2,−4), and the unit vector directed from the origin toward point B is (2,−2, 1)/3. If points A and B are ten units apart, find the coordinates of point B.

With A = (6,−2,−4) and B = 13B(2,−2, 1), we use the fact that |B A| = 10, or |(6 − 23B)ax (2 − 23B)ay (4 + 13B)az| = 10 Expanding, obtain 36 − 8B + 49B2 + 4 − 83B + 49B2 + 16 + 83B + 19B2 = 100 or B2 − 8B − 44 = 0. Thus B = 8±

√ 64−176 2 = 11.75 (taking positive option) and so

B = 2 3 (11.75)ax − 2

3 (11.75)ay + 1

3 (11.75)az = 7.83ax − 7.83ay + 3.92az

1

1.4. given points A(8,−5, 4) and B(−2, 3, 2), find: a) the distance from A to B.

|B A| = |(−10, 8,−2)| = 12.96

b) a unit vector directed from A towards B. This is found through

aAB = B A|B A| = (−0.77, 0.62,−0.15)

c) a unit vector directed from the origin to the midpoint of the line AB.

a0M = (A + B)/2|(A + B)/2| = (3,−1, 3)

19 = (0.69,−0.23, 0.69)

d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3. Note that the midpoint, (3,−1, 3), as determined from part c happens to have z coordinate of 3. This is the point we are looking for.

1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az. Given two points, P(1, 2,−1) and Q(−2, 1, 3), find:

a) G at P : G(1, 2,−1) = (48, 36, 18) b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so

aG = (−48, 72, 162)|(−48, 72, 162)| = (−0.26, 0.39, 0.88)

c) a unit vector directed from Q toward P :

aQP = P Q|P Q| = (3,−1, 4)

26 = (0.59, 0.20,−0.78)

d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or 10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is

100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4

2

1.6. For the G field in Problem 1.5, make sketches of Gx , Gy , Gz and |G| along the line y = 1, z = 1, for 0 ≤ x ≤ 2. We find G(x, 1, 1) = (24x, 12x2 + 24, 18), from which Gx = 24x, Gy = 12x2 + 24, Gz = 18, and |G| = 6

√ 4x4 + 32x2 + 25. Plots are shown below.

1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region |x|, |y|, and |z| less than 2, find:

a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.

b) the region in which Ey = Ez: This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y, with |x| < 2, |y| < 2, |z| < 1.

c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x = y2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.

1.8. Two vector fields are F = −10ax+20x(y−1)ay and G = 2x2yax−4ay+zaz. For the point P(2, 3,−4), find:

a) |F|: F at (2, 3,−4) = (−10, 80, 0), so |F| = 80.6. b) |G|: G at (2, 3,−4) = (24,−4,−4), so |G| = 24.7. c) a unit vector in the direction of F G: F G = (−10, 80, 0)(24,−4,−4) = (−34, 84, 4). So

a = F G|F G| = (−34, 84, 4)

90.7 = (−0.37, 0.92, 0.04)

d) a unit vector in the direction of F + G: F + G = (−10, 80, 0)+ (24,−4,−4) = (14, 76,−4). So

a = F + G|F + G| = (14, 76,−4)

77.4 = (0.18, 0.98,−0.05)

3

1.9. A field is given as

G = 25 (x2 + y2) (xax + yay)

Find: a) a unit vector in the direction of G at P(3, 4,−2): Have Gp = 25/(9 + 16)× (3, 4, 0) = 3ax + 4ay ,

and |Gp| = 5. Thus aG = (0.6, 0.8, 0). b) the angle between G and ax at P : The angle is found through aG · ax = cos θ . So cos θ =

(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus θ = 53◦. c) the value of the following double integral on the plane y = 7:

∫ 4 0

∫ 2 0

G · aydzdx

∫ 4 0

∫ 2 0

25

x2 + y2 (xax + yay) · aydzdx = ∫ 4

0

∫ 2 0

25

x2 + 49 × 7 dzdx = ∫ 4

0

350

x2 + 49dx

= 350 × 1 7

[ tan−1

( 4

7

) − 0

] = 26

1.10. Use the definition of the dot product to find the interior angles at A and B of the triangle defined by the three points A(1, 3,−2), B(−2, 4, 5), and C(0,−2, 1):

a) Use RAB = (−3, 1, 7) and RAC = (−1,−5, 3) to form RAB · RAC = |RAB ||RAC | cos θA. Obtain 3 + 5 + 21 = √59√35 cos θA. Solve to find θA = 65.3◦.

b) Use RBA = (3,−1,−7) and RBC = (2,−6,−4) to form RBA · RBC = |RBA||RBC | cos θB . Obtain 6 + 6 + 28 = √59√56 cos θB . Solve to find θB = 45.9◦.

1.11. Given the points M(0.1,−0.2,−0.1), N(−0.2, 0.1, 0.3), and P(0.4, 0, 0.1), find: a) the vector RMN : RMN = (−0.2, 0.1, 0.3)(0.1,−0.2,−0.1) = (−0.3, 0.3, 0.4). b) the dot product RMN · RMP : RMP = (0.4, 0, 0.1) (0.1,−0.2,−0.1) = (0.3, 0.2, 0.2). RMN ·

RMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05. c) the scalar projection of RMN on RMP :

RMN · aRMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2)√ 0.09 + 0.04 + 0.04 =

0.05√ 0.17

= 0.12

d) the angle between RMN and RMP :

θM = cos−1 (

RMN · RMP |RMN ||RMP |

) = cos−1

( 0.05√

0.34 √

0.17

) = 78◦

4

1.12. Given points A(10, 12,−6), B(16, 8,−2), C(8, 1,−4), and D(−2,−5, 8), determine: a) the vector projection of RAB + RBC on RAD: RAB + RBC = RAC = (8, 1, 4) (10, 12,−6) =

(−2,−11, 10) Then RAD = (−2,−5, 8) (10, 12,−6) = (−12,−17, 14). So the projection will be:

629

] (−12,−17, 14)

629 = (−6.7,−9.5, 7.8)

b) the vector projection of RAB +RBC on RDC : RDC = (8,−1, 4)(−2,−5, 8) = (10, 6,−4). The projection is:

(RAC · aRDC)aRDC = [ (−2,−11, 10) · (10, 6,−4)

152

] (10, 6,−4)

152 = (−8.3,−5.0, 3.3)

c) the angle between RDA and RDC : Use RDA = −RAD = (12, 17,−14) and RDC = (10, 6,−4). The angle is found through the dot product of the associated unit vectors, or:

θD = cos−1(aRDA · aRDC) = cos−1 ( (12, 17,−14) · (10, 6,−4)

629 √

152

) = 26◦

1.13. a) Find the vector component of F = (10,−6, 5) that is parallel to G = (0.1, 0.2, 0.3):

F||G = F · G|G|2 G = (10,−6, 5) · (0.1, 0.2, 0.3)

0.01 + 0.04 + 0.09 (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)

b) Find the vector component of F that is perpendicular to G:

FpG = F F||G = (10,−6, 5)(0.93, 1.86, 2.79) = (9.07,−7.86, 2.21)

c) Find the vector component of G that is perpendicular to F:

GpF = GG||F = GG · F|F|2 F = (0.1, 0.2, 0.3)− 1.3

100 + 36 + 25 (10,−6, 5) = (0.02, 0.25, 0.26)

1.14. The four vertices of a regular tetrahedron are located at O(0, 0, 0), A(0, 1, 0), B(0.5 √

3, 0.5, 0), and C(

√ 3/6, 0.5,

√ 2/3).

a) Find a unit vector perpendicular (outward) to the face ABC: First find

RBA × RBC = [(0, 1, 0)(0.5 √

3, 0.5, 0)] × [(

3/6, 0.5,

2/3)(0.5 √

3, 0.5, 0)]

= (−0.5 √

3, 0.5, 0)× (− √

3/3, 0,

2/3) = (0.41, 0.71, 0.29) The required unit vector will then be:

RBA × RBC |RBA × RBC | = (0.47, 0.82, 0.33)

b) Find the area of the face ABC:

Area = 1 2 |RBA × RBC | = 0.43

5

1.15. Three vectors extending from the origin are given as r1 = (7, 3,−2), r2 = (−2, 7,−3), and r3 = (0, 2, 3). Find:

a) a unit vector perpendicular to both r1 and r2:

ap12 = r1 × r2|r1 × r2| = (5, 25, 55)

60.6 = (0.08, 0.41, 0.91)

b) a unit vector perpendicular to the vectors r1 − r2 and r2 − r3: r1 − r2 = (9,−4, 1) and r2 − r3 = (−2, 5,−6). So r1 − r2 × r2 − r3 = (19, 52, 32). Then

ap = (19, 52, 32)|(19, 52, 32)| = (19, 52, 32)

63.95 = (0.30, 0.81, 0.50)

c) the area of the triangle defined by r1 and r2:

Area = 1 2 |r1 × r2| = 30.3

d) the area of the triangle defined by the heads of r1, r2, and r3:

Area = 1 2 |(r2 − r1)× (r2 − r3)| = 1

2 |(−9, 4,−1)× (−2, 5,−6)| = 32.0

1.16. Describe the surfaces defined by the equations:

a) r · ax = 2, where r = (x, y, z): This will be the plane x = 2. b) |r × ax | = 2: r × ax = (0, z,y), and |r × ax | =

z2 + y2 = 2. This is the equation of a cylinder,

centered on the x axis, and of radius 2.

1.17. Point A(−4, 2, 5) and the two vectors, RAM = (20, 18,−10) and RAN = (−10, 8, 15), define a triangle. a) Find a unit vector perpendicular to the triangle: Use

ap = RAM × RAN|RAM × RAN | = (350,−200, 340)

527.35 = (0.664,−0.379, 0.645)

The vector in the opposite direction to this one is also a valid answer.

b) Find a unit vector in the plane of the triangle and perpendicular to RAN :

aAN = (−10, 8, 15)√ 389

= (−0.507, 0.406, 0.761)

Then

apAN = ap × aAN = (0.664,−0.379, 0.645)× (−0.507, 0.406, 0.761) = (−0.550,−0.832, 0.077) The vector in the opposite direction to this one is also a valid answer.

c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit vector in the required direction is (1/2)(aAM + aAN), where

aAM = (20, 18,−10)|(20, 18,−10)| = (0.697, 0.627,−0.348)

6

1.17c. (continued) Now

1

2 (aAM + aAN) = 1

2 [(0.697, 0.627,−0.348)+ (−0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207)

Finally,

abis = (0.095, 0.516, 0.207)|(0.095, 0.516, 0.207)| = (0.168, 0.915, 0.367)

1.18. Given points A(ρ = 5, φ = 70◦, z = −3) and B(ρ = 2, φ = −30◦, z = 1), find: a) unit vector in cartesian coordinates at A toward B: A(5 cos 70◦, 5 sin 70◦,−3) = A(1.71, 4.70,−3), In

the same manner, B(1.73,−1, 1). So RAB = (1.73,−1, 1) (1.71, 4.70,−3) = (0.02,−5.70, 4) and therefore

aAB = (0.02,−5.70, 4)|(0.02,−5.70, 4)| = (0.003,−0.82, 0.57)

b) a vector in cylindrical coordinates at A directed toward B: aAB · aρ = 0.003 cos 70◦ − 0.82 sin 70◦ = −0.77. aAB · aφ = −0.003 sin 70◦ − 0.82 cos 70◦ = −0.28. Thus

aAB = −0.77aρ − 0.28aφ + 0.57az

.

c) a unit vector in cylindrical coordinates at B directed toward A: Use aBA = (−0, 003, 0.82,−0.57). Then aBA ·aρ = −0.003 cos(−30◦)+0.82 sin(−30◦) = −0.43, and aBA · aφ = 0.003 sin(−30◦)+ 0.82 cos(−30◦) = 0.71. Finally,

aBA = −0.43aρ + 0.71aφ − 0.57az

1.19 a) Express the field D = (x2 + y2)−1(xax + yay) in cylindrical components and cylindrical variables: Have x = ρ cosφ, y = ρ sin φ, and x2 + y2 = ρ2. Therefore

D = 1 ρ (cosφax + sin φay)

Then

= D · aρ = 1 ρ

[ cosφ(ax · aρ)+ sin φ(ay · aρ)

] = 1 ρ

[ cos2 φ + sin2 φ

] = 1

ρ

and

= D · aφ = 1 ρ

[ cosφ(ax · aφ)+ sin φ(ay · aφ)

] = 1 ρ

[cosφ(− sin φ)+ sin φ cosφ] = 0

Therefore

D = 1 ρ

aρ

7

1.19b. Evaluate D at the point where ρ = 2, φ = 0.2π , and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5aρ . To express this in cartesian, we use

D = 0.5(aρ · ax)ax + 0.5(aρ · ay)ay = 0.5 cos 36◦ax + 0.5 sin 36◦ay = 0.41ax + 0.29ay

1.20. Express in cartesian components: a) the vector at A(ρ = 4, φ = 40◦, z = −2) that extends to B(ρ = 5, φ = −110◦, z = 2): We

have A(4 cos 40◦, 4 sin 40◦,−2) = A(3.06, 2.57,−2), and B(5 cos(−110◦), 5 sin(−110◦), 2) = B(−1.71,−4.70, 2) in cartesian. Thus RAB = (−4.77,−7.30, 4).

b) a unit vector at B directed toward A: Have RBA = (4.77, 7.30,−4), and so

aBA = (4.77, 7.30,−4)|(4.77, 7.30,−4)| = (0.50, 0.76,−0.42)

c) a unit vector at B directed toward the origin: Have rB = (−1.71,−4.70, 2), and so −rB = (1.71, 4.70,−2). Thus

a = (1.71, 4.70,−2)|(1.71, 4.70,−2)| = (0.32, 0.87,−0.37)

1.21. Express in cylindrical components:

a) the vector from C(3, 2,−7) to D(−1,−4, 2): C(3, 2,−7) C(ρ = 3.61, φ = 33.7◦, z = −7) and D(−1,−4, 2) D(ρ = 4.12, φ = −104.0◦, z = 2). Now RCD = (−4,−6, 9) and = RCD · aρ = −4 cos(33.7) − 6 sin(33.7) = −6.66. Then = RCD · aφ = 4 sin(33.7)− 6 cos(33.7) = −2.77. So RCD = −6.66aρ − 2.77aφ + 9az

b) a unit vector at D directed toward C: RCD = (4, 6,−9) and = RDC · aρ = 4 cos(−104.0) + 6 sin(−104.0) = −6.79. Then = RDC · aφ = 4[− sin(−104.0)] + 6 cos(−104.0) = 2.43. So RDC = −6.79aρ + 2.43aφ − 9az Thus aDC = −0.59aρ + 0.21aφ − 0.78az

c) a unit vector at D directed toward the origin: Start with rD = (−1,−4, 2), and so the vector toward the origin will be −rD = (1, 4,−2). Thus in cartesian the unit vector is a = (0.22, 0.87,−0.44). Convert to cylindrical: = (0.22, 0.87,−0.44) · aρ = 0.22 cos(−104.0)+ 0.87 sin(−104.0) = −0.90, and = (0.22, 0.87,−0.44) · aφ = 0.22[− sin(−104.0)] + 0.87 cos(−104.0) = 0, so that finally, a = −0.90aρ − 0.44az.

1.22. A field is given in cylindrical coordinates as

F = [

40

ρ2 + 1 + 3(cosφ + sin φ) ]

aρ + 3(cosφ − sin φ)aφ − 2az

where the magnitude of F is found to be:

|F| = √

F · F = [

1600

2 + 1)2 + 240

ρ2 + 1 (cosφ + sin φ)+ 22 ]1/2

8

Sketch |F|: a) vs. φ with ρ = 3: in this case the above simplifies to

|F= 3)| = |Fa| = [38 + 24(cosφ + sin φ)]1/2

b) vs. ρ with φ = 0, in which:

|F= 0)| = |Fb| = [

1600

2 + 1)2 + 240

ρ2 + 1 + 22 ]1/2

c) vs. ρ with φ = 45◦, in which

|F= 45◦)| = |Fc| = [

1600

2 + 1)2 + 240

√ 2

ρ2 + 1 + 22 ]1/2

9

1.23. The surfaces ρ = 3, ρ = 5, φ = 100◦, φ = 130◦, z = 3, and z = 4.5 define a closed surface. a) Find the enclosed volume:

Vol = ∫ 4.5

3

∫ 130◦ 100◦

∫ 5 3

ρ dρ dφ dz = 6.28

NOTE: The limits on the φ integration must be converted to radians (as was done here, but not shown).

b) Find the total area of the enclosing surface:

Area = 2 ∫ 130◦

100◦

∫ 5 3

ρ dρ dφ + ∫ 4.5

3

∫ 130◦ 100◦

3 dφ dz

+ ∫ 4.5

3

∫ 130◦ 100◦

5 dφ dz + 2 ∫ 4.5

3

∫ 5 3

dρ dz = 20.7

c) Find the total length of the twelve edges of the surfaces:

Length = 4 × 1.5 + 4 × 2 + 2 × [

30◦

360◦ × 2π × 3 + 30

360◦ × 2π × 5

] = 22.4

d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(ρ = 3, φ = 100◦, z = 3) and B(ρ = 5, φ = 130◦, z = 4.5). Performing point transformations to cartesian coordinates, these become A(x = −0.52, y = 2.95, z = 3) and B(x = −3.21, y = 3.83, z = 4.5). Taking A and B as vectors directed from the origin, the requested length is

Length = |B A| = |(−2.69, 0.88, 1.5)| = 3.21

1.24. At point P(−3, 4, 5), express the vector that extends from P to Q(2, 0,−1) in: a) rectangular coordinates.

RPQ = Q P = 5ax − 4ay − 6az Then |RPQ| =

√ 25 + 16 + 36 = 8.8

b) cylindrical coordinates. At P , ρ = 5, φ = tan−1(4/− 3) = −53.1◦, and z = 5. Now, RPQ · aρ = (5ax − 4ay − 6az) · aρ = 5 cosφ − 4 sin φ = 6.20

RPQ · aφ = (5ax − 4ay − 6az) · aφ = −5 sin φ − 4 cosφ = 1.60 Thus

RPQ = 6.20aρ + 1.60aφ − 6az and |RPQ| =

√ 6.202 + 1.602 + 62 = 8.8

c) spherical coordinates. At P , r = √9 + 16 + 25 = √50 = 7.07, θ = cos−1(5/7.07) = 45◦, and φ = tan−1(4/− 3) = −53.1◦.

RPQ · ar = (5ax − 4ay − 6az) · ar = 5 sin θ cosφ − 4 sin θ sin φ − 6 cos θ = 0.14 RPQ · aθ = (5ax − 4ay − 6az) · aθ = 5 cos θ cosφ − 4 cos θ sin φ (−6) sin θ = 8.62

RPQ · aφ = (5ax − 4ay − 6az) · aφ = −5 sin φ − 4 cosφ = 1.60

10

1.24. (continued)

Thus RPQ = 0.14ar + 8.62aθ + 1.60aφ

and |RPQ| = √

0.142 + 8.622 + 1.602 = 8.8 d) Show that each of these vectors has the same magnitude. Each does, as shown above.

1.25. Given point P(r = 0.8, θ = 30◦, φ = 45◦), and

E = 1 r2

( cosφ ar + sin φ

sin θ aφ

)

a) Find E at P : E = 1.10aρ + 2.21aφ . b) Find |E| at P : |E| = √1.102 + 2.212 = 2.47. c) Find a unit vector in the direction of E at P :

aE = E|E| = 0.45ar + 0.89aφ

1.26. a) Determine an expression for ay in spherical coordinates at P(r = 4, θ = 0.2π, φ = 0.8π): Use ay · ar = sin θ sin φ = 0.35, ay · aθ = cos θ sin φ = 0.48, and ay · aφ = cosφ = −0.81 to obtain

ay = 0.35ar + 0.48aθ − 0.81aφ

b) Express ar in cartesian components at P : Find x = r sin θ cosφ = −1.90, y = r sin θ sin φ = 1.38, and z = r cos θ = −3.24. Then use ar · ax = sin θ cosφ = −0.48, ar · ay = sin θ sin φ = 0.35, and ar · az = cos θ = 0.81 to obtain

ar = −0.48ax + 0.35ay + 0.81az

1.27. The surfaces r = 2 and 4, θ = 30◦ and 50◦, and φ = 20◦ and 60◦ identify a closed surface. a) Find the enclosed volume: This will be

Vol = ∫ 60◦

20◦

∫ 50◦ 30◦

∫ 4 2

r2 sin θdrdθdφ = 2.91

where degrees have been converted to radians. b) Find the total area of the enclosing surface:

Area = ∫ 60◦

20◦

∫ 50◦ 30◦

(42 + 22) sin θdθdφ + ∫ 4

2

∫ 60◦ 20◦

r(sin 30◦ + sin 50◦)drdφ

+ 2 ∫ 50◦

30◦

∫ 4 2

rdrdθ = 12.61

c) Find the total length of the twelve edges of the surface:

Length = 4 ∫ 4

2 dr + 2

∫ 50◦ 30◦

(4 + 2)dθ + ∫ 60◦

20◦ (4 sin 50◦ + 4 sin 30◦ + 2 sin 50◦ + 2 sin 30◦)dφ

= 17.49

11

1.27. (continued)

d) Find the length of the longest straight line that lies entirely within the surface: This will be from A(r = 2, θ = 50◦, φ = 20◦) to B(r = 4, θ = 30◦, φ = 60◦) or

A(x = 2 sin 50◦ cos 20◦, y = 2 sin 50◦ sin 20◦, z = 2 cos 50◦)

to B(x = 4 sin 30◦ cos 60◦, y = 4 sin 30◦ sin 60◦, z = 4 cos 30◦)

or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B A = (−0.44, 1.21, 2.18) and

Length = |B A| = 2.53

1.28. a) Determine the cartesian components of the vector from A(r = 5, θ = 110◦, φ = 200◦) to B(r = 7, θ = 30◦, φ = 70◦): First transform the points to cartesian: xA = 5 sin 110◦ cos 200◦ = −4.42, yA = 5 sin 110◦ sin 200◦ = −1.61, and zA = 5 cos 110◦ = −1.71; xB = 7 sin 30◦ cos 70◦ = 1.20, yB = 7 sin 30◦ sin 70◦ = 3.29, and zB = 7 cos 30◦ = 6.06. Now

RAB = B A = 5.62ax + 4.90ay + 7.77az

b) Find the spherical components of the vector at P(2,−3, 4) extending to Q(−3, 2, 5): First, RPQ = Q P = (−5, 5, 1). Then at P , r = √4 + 9 + 16 = 5.39, θ = cos−1(4/√29) = 42.0◦, and φ = tan−1(−3/2) = −56.3◦. Now

RPQ · ar = −5 sin(42◦) cos(−56.3◦)+ 5 sin(42◦) sin(−56.3◦)+ 1 cos(42◦) = −3.90

RPQ · aθ = −5 cos(42◦) cos(−56.3◦)+ 5 cos(42◦) sin(−56.3◦)− 1 sin(42◦) = −5.82 RPQ · aφ = −(−5) sin(−56.3◦)+ 5 cos(−56.3◦) = −1.39

So finally, RPQ = −3.90ar − 5.82aθ − 1.39aφ

c) If D = 5ar − 3aθ + 4aφ , find D · aρ at M(1, 2, 3): First convert aρ to cartesian coordinates at the specified point. Use aρ = (aρ · ax)ax + (aρ · ay)ay . At A(1, 2, 3), ρ =

√ 5, φ = tan−1(2) = 63.4◦,

r = √14, and θ = cos−1(3/√14) = 36.7◦. So aρ = cos(63.4◦)ax + sin(63.4◦)ay = 0.45ax + 0.89ay . Then

(5ar − 3aθ + 4aφ) · (0.45ax + 0.89ay) = 5(0.45) sin θ cosφ + 5(0.89) sin θ sin φ − 3(0.45) cos θ cosφ − 3(0.89) cos θ sin φ + 4(0.45)(− sin φ) + 4(0.89) cosφ = 0.59

1.29. Express the unit vector ax in spherical components at the point: a) r = 2, θ = 1 rad, φ = 0.8 rad: Use

ax = (ax · ar )ar + (ax · aθ )aθ + (ax · aφ)aφ = sin(1) cos(0.8)ar + cos(1) cos(0.8)aθ + (− sin(0.8))aφ = 0.59ar + 0.38aθ − 0.72aφ

12

1.29 (continued) Express the unit vector ax in spherical components at the point:

b) x = 3, y = 2, z = −1: First, transform the point to spherical coordinates. Have r = √14, θ = cos−1(−1/√14) = 105.5◦, and φ = tan−1(2/3) = 33.7◦. Then

ax = sin(105.5◦) cos(33.7◦)ar + cos(105.5◦) cos(33.7◦)aθ + (− sin(33.7◦))aφ = 0.80ar − 0.22aθ − 0.55aφ

c) ρ = 2.5, φ = 0.7 rad, z = 1.5: Again, convert the point to spherical coordinates. r = √ ρ2 + z2 =√

8.5, θ = cos−1(z/r) = cos−1(1.5/√8.5) = 59.0◦, and φ = 0.7 rad = 40.1◦. Now

ax = sin(59◦) cos(40.1◦)ar + cos(59◦) cos(40.1◦)aθ + (− sin(40.1◦))aφ = 0.66ar + 0.39aθ − 0.64aφ

1.30. Given A(r = 20, θ = 30◦, φ = 45◦) and B(r = 30, θ = 115◦, φ = 160◦), find: a) |RAB |: First convert A and B to cartesian: Have xA = 20 sin(30◦) cos(45◦) = 7.07, yA =

20 sin(30◦) sin(45◦) = 7.07, and zA = 20 cos(30◦) = 17.3. xB = 30 sin(115◦) cos(160◦) = −25.6, yB = 30 sin(115◦) sin(160◦) = 9.3, and zB = 30 cos(115◦) = −12.7. Now RAB = RB RA = (−32.6, 2.2,−30.0), and so |RAB | = 44.4.

b) |RAC |, given C(r = 20, θ = 90◦, φ = 45◦). Again, converting C to cartesian, obtain xC = 20 sin(90◦) cos(45◦) = 14.14, yC = 20 sin(90◦) sin(45◦) = 14.14, and zC = 20 cos(90◦) = 0. So RAC = RC RA = (7.07, 7.07,−17.3), and |RAC | = 20.0.

c) the distance from A to C on a great circle path: Note that A and C share the same r and φ coordinates; thus moving from A to C involves only a change in θ of 60◦. The requested arc length is then

distance = 20 × [

60

( 2π

360

)] = 20.9

13

CHAPTER 2

2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for  = 0: Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the fifth charge will be on the z axis at location z = 4√2, which puts it at 8cm distance from the other four. By symmetry, the force on the fifth charge will be z-directed, and will be four times the z component of force produced by each of the four other charges.

F = 4√ 2 × q

2

4π0d2 = 4√

2 × (10

−8)2

4π(8.85 × 10−12)(0.08)2 = 4.0 × 10 −4 N

2.2. A charge Q1 = 0.1 µC is located at the origin, while Q2 = 0.2 µC is at A(0.8,−0.6, 0). Find the locus of points in the z = 0 plane at which the x component of the force on a third positive charge is zero.

To solve this problem, the z coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates (x, y, 0). We take its magnitude to be Q3. The vector directed from the first charge to the third is R13 = xax + yay ; the vector directed from the second charge to the third is R23 = (x − 0.8)ax + (y + 0.6)ay . The force on the third charge is now

F3 = Q3 4π0

[ Q1R13 |R13|3 +

Q2R23 |R23|3

]

= Q3 × 10 −6

4π0

[ 0.1(xax + yay) (x2 + y2)1.5 +

0.2[(x − 0.8)ax + (y + 0.6)ay] [(x − 0.8)2 + (y + 0.6)2]1.5

]

We desire the x component to be zero. Thus,

0 = [

0.1xax (x2 + y2)1.5 +

0.2(x − 0.8)ax [(x − 0.8)2 + (y + 0.6)2]1.5

]

or x[(x − 0.8)2 + (y + 0.6)2]1.5 = 2(0.8 − x)(x2 + y2)1.5

2.3. Point charges of 50nC each are located at A(1, 0, 0), B(−1, 0, 0), C(0, 1, 0), and D(0,−1, 0) in free space. Find the total force on the charge at A.

The force will be:

F = (50 × 10 −9)2

4π0

[ RCA

|RCA|3 + RDA

|RDA|3 + RBA

|RBA|3 ]

where RCA = ax ay , RDA = ax + ay , and RBA = 2ax . The magnitudes are |RCA| = |RDA| = √

2, and |RBA| = 2. Substituting these leads to

F = (50 × 10 −9)2

4π0

[ 1

2 √

2 + 1

2 √

2 + 2

8

] ax = 21.5ax µN

where distances are in meters.

14

2.4. Let Q1 = 8 µC be located at P1(2, 5, 8) while Q2 = −5 µC is at P2(6, 15, 8). Let  = 0. a) Find F2, the force on Q2: This force will be

F2 = Q1Q2 4π0

R12 |R12|3 =

(8 × 10−6)(−5 × 10−6) 4π0

(4ax + 10ay) (116)1.5

= (−1.15ax − 2.88ay)mN

b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be:

F3 = Q3 4π0

[ Q1R13 |R13|3 +

Q2R23 |R23|3

] where R13 = (x − 2)ax + (y − 5)ay and R23 = (x − 6)ax + (y − 15)ay . Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q2. The slope of this vector is (15 − 5)/(6 − 2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5x, 8). With this restriction, the force becomes:

F3 = Q3 4π0

[ 8[(x − 2)ax + 2.5(x − 2)ay]

[(x − 2)2 + (2.5)2(x − 2)2]1.5 − 5[(x − 6)ax + 2.5(x − 6)ay]

[(x − 6)2 + (2.5)2(x − 6)2]1.5 ]

where we require the term in large brackets to be zero. This leads to

8(x − 2)[((2.5)2 + 1)(x − 6)2]1.5 − 5(x − 6)[((2.5)2 + 1)(x − 2)2]1.5 = 0 which reduces to

8(x − 6)2 − 5(x − 2)2 = 0 or

x = 6 √

8 − 2√5√ 8 −√5 = 21.1

The coordinates of P3 are thus P3(21.1, 52.8, 8)

2.5. Let a point charge Q125 nC be located at P1(4,−2, 7) and a charge Q2 = 60 nC be at P2(−3, 4,−2). a) If  = 0, find E at P3(1, 2, 3): This field will be

E = 10 −9

4π0

[ 25R13 |R13|3 +

60R23 |R23|3

]

where R13 = −3ax+4ay−4az and R23 = 4ax−2ay+5az. Also, |R13| = √

41 and |R23| = √

45. So

E = 10 −9

4π0

[ 25 × (−3ax + 4ay − 4az)

(41)1.5 + 60 × (4ax − 2ay + 5az)

(45)1.5

] = 4.58ax − 0.15ay + 5.51az

b) At what point on the y axis is Ex = 0? P3 is now at (0, y, 0), so R13 = −4ax + (y + 2)ay − 7az and R23 = 3ax + (y − 4)ay + 2az. Also, |R13| =

√ 65 + (y + 2)2 and |R23| =

√ 13 + (y − 4)2.

Now the x component of E at the new P3 will be:

Ex = 10 −9

4π0

[ 25 × (−4)

[65 + (y + 2)2]1.5 + 60 × 3

[13 + (y − 4)2]1.5 ]

To obtain Ex = 0, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic:

0.48y2 + 13.92y + 73.10 = 0 which yields the two values: y = −6.89,−22.11

15

2.6. Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0,−1) in free space. a) Find E at P(0.5, 0, 0): This will be

EP = 120 × 10 −9

4π0

[ RAP

|RAP |3 + RBP

|RBP |3 ]

where RAP = 0.5ax az and RBP = 0.5ax + az. Also, |RAP | = |RBP | = √

1.25. Thus:

EP = 120 × 10 −9ax

4π(1.25)1.50 = 772 V/m

b) What single charge at the origin would provide the identical field strength? We require

Q0

4π0(0.5)2 = 772

from which we find Q0 = 21.5 nC.

2.7. A 2 µC point charge is located at A(4, 3, 5) in free space. Find , , and Ez at P(8, 12, 2). Have

EP = 2 × 10 −6

4π0

RAP |RAP |3 =

2 × 10−6 4π0

[ 4ax + 9ay − 3az

(106)1.5

] = 65.9ax + 148.3ay − 49.4az

Then, at point P , ρ = √82 + 122 = 14.4, φ = tan−1(12/8) = 56.3◦, and z = z. Now, = Ep · aρ = 65.9(ax · aρ)+ 148.3(ay · aρ) = 65.9 cos(56.3◦)+ 148.3 sin(56.3◦) = 159.7

and

= Ep · aφ = 65.9(ax · aφ)+ 148.3(ay · aφ) = −65.9 sin(56.3◦)+ 148.3 cos(56.3◦) = 27.4 Finally, Ez = −49.4

2.8. Given point charges of −1 µC at P1(0, 0, 0.5) and P2(0, 0,−0.5), and a charge of 2 µC at the origin, find E at P(0, 2, 1) in spherical components, assuming  = 0. The field will take the general form:

EP = 10 −6

4π0

[ − R1|R1|3 +

2R2 |R2|3 −

R3 |R3|3

]

where R1, R2, R3 are the vectors toP from each of the charges in their original listed order. Specifically, R1 = (0, 2, 0.5), R2 = (0, 2, 1), and R3 = (0, 2, 1.5). The magnitudes are |R1| = 2.06, |R2| = 2.24, and |R3| = 2.50. Thus

EP = 10 −6

4π0

[−(0, 2, 0.5) (2.06)3

+ 2(0, 2, 1) (2.24)3

(0, 2, 1.5) (2.50)3

] = 89.9ay + 179.8az

Now, at P , r = √5, θ = cos−1(1/√5) = 63.4◦, and φ = 90◦. So Er = EP · ar = 89.9(ay · ar )+ 179.8(az · ar ) = 89.9 sin θ sin φ + 179.8 cos θ = 160.9

= EP · aθ = 89.9(ay · aθ )+ 179.8(az · aθ ) = 89.9 cos θ sin φ + 179.8(− sin θ) = −120.5 = EP · aφ = 89.9(ay · aφ)+ 179.8(az · aφ) = 89.9 cosφ = 0

16

2.9. A 100 nC point charge is located at A(−1, 1, 3) in free space. a) Find the locus of all points P(x, y, z) at which Ex = 500 V/m: The total field at P will be:

EP = 100 × 10 −9

4π0

RAP |RAP |3

where RAP = (x + 1)ax + (y − 1)ay + (z − 3)az, and where |RAP | = [(x + 1)2 + (y − 1)2 + (z− 3)2]1/2. The x component of the field will be

Ex = 100 × 10 −9

4π0

[ (x + 1)

[(x + 1)2 + (y − 1)2 + (z− 3)2]1.5 ] = 500 V/m

And so our condition becomes:

(x + 1) = 0.56 [(x + 1)2 + (y − 1)2 + (z− 3)2]1.5

b) Find y1 if P(−2, y1, 3) lies on that locus: At point P , the condition of part a becomes

3.19 = [ 1 + (y1 − 1)2

]3 from which (y1 − 1)2 = 0.47, or y1 = 1.69 or 0.31

2.10. Charges of 20 and -20 nC are located at (3, 0, 0) and (−3, 0, 0), respectively. Let  = 0. Determine |E| at P(0, y, 0): The field will be

EP = 20 × 10 −9

4π0

[ R1

|R1|3 − R2

|R2|3 ]

where R1, the vector from the positive charge to point P is (−3, y, 0), and R2, the vector from the negative charge to point P , is (3, y, 0). The magnitudes of these vectors are |R1| = |R2| =√

9 + y2. Substituting these into the expression for EP produces

EP = 20 × 10 −9

4π0

[ −6ax (9 + y2)1.5

]

from which

|EP | = 1079 (9 + y2)1.5 V/m

2.11. A charge Q0 located at the origin in free space produces a field for which Ez = 1 kV/m at point P(−2, 1,−1).

a) Find Q0: The field at P will be

EP = Q0 4π0

[−2ax + ay az 61.5

]

Since the z component is of value 1 kV/m, we find Q0 = −4π061.5 × 103 = −1.63 µC.

17

2.11. (continued)

b) Find E at M(1, 6, 5) in cartesian coordinates: This field will be:

EM = −1.63 × 10 −6

4π0

[ ax + 6ay + 5az [1 + 36 + 25]1.5

]

or EM = −30.11ax − 180.63ay − 150.53az. c) Find E at M(1, 6, 5) in cylindrical coordinates: At M , ρ = √1 + 36 = 6.08, φ = tan−1(6/1) =

80.54◦, and z = 5. Now

= EM · aρ = −30.11 cosφ − 180.63 sin φ = −183.12

= EM · aφ = −30.11(− sin φ)− 180.63 cosφ = 0 (as expected) so that EM = −183.12aρ − 150.53az.

d) Find E at M(1, 6, 5) in spherical coordinates: At M , r = √1 + 36 + 25 = 7.87, φ = 80.54◦ (as before), and θ = cos−1(5/7.87) = 50.58◦. Now, since the charge is at the origin, we expect to obtain only a radial component of EM . This will be:

Er = EM · ar = −30.11 sin θ cosφ − 180.63 sin θ sin φ − 150.53 cos θ = −237.1

2.12. The volume charge density ρv = ρ0e−|x|−|y|−|z| exists over all free space. Calculate the total charge present: This will be 8 times the integral of ρv over the first octant, or

Q = 8 ∫ ∞

0

∫ ∞ 0

∫ ∞ 0

ρ0e xyz dx dy dz = 8ρ0

2.13. A uniform volume charge density of 0.2 µC/m3 (note typo in book) is present throughout the spherical shell extending from r = 3 cm to r = 5 cm. If ρv = 0 elsewhere:

a) find the total charge present throughout the shell: This will be

Q = ∫ 2π

0

π 0

.05 .03

0.2 r2 sin θ dr dθ dφ = [

4π(0.2) r3

3

].05 .03

= 8.21 × 10−5 µC = 82.1 pC

b) find r1 if half the total charge is located in the region 3 cm < r < r1: If the integral over r in part a is taken to r1, we would obtain[

4π(0.2) r3

3

]r1 .03

= 4.105 × 10−5

Thus

r1 = [

3 × 4.105 × 10−5 0.2 × 4π + (.03)

3

]1/3 = 4.24 cm

18

2.14. Let

ρv = 5e−0.1ρ (π − |φ|) 1 z2 + 10 µC/m

3

in the region 0 ≤ ρ ≤ 10, −π < φ < π , all z, and ρv = 0 elsewhere. a) Determine the total charge present: This will be the integral of ρv over the region where it exists; specifically,

Q = ∫ ∞ −∞

π π

∫ 10 0

5e−0.1ρ (π − |φ|) 1 z2 + 10ρ dρ dφ dz

which becomes

Q = 5 [ e−0.1ρ

(0.1)2 (−0.1 − 1)

]10 0

∫ ∞ −∞

2 ∫ π

0 φ) 1

z2 + 10dφ dz

or

Q = 5 × 26.4 ∫ ∞ −∞

π2 1

z2 + 10 dz

Finally,

Q = 5 × 26.4 × π2 [

1√ 10

tan−1 (

z√ 10

)]∞ −∞

= 5(26.43

√ 10

= 1.29 × 103 µC = 1.29 mC

b) Calculate the charge within the region 0 ≤ ρ ≤ 4, −π/2 < φ < π/2, −10 < z < 10: With the limits thus changed, the integral for the charge becomes:

Q′ = ∫ 10 −10

2 ∫ π/2

0

∫ 4 0

5e−0.1ρ (π φ) 1 z2 + 10ρ dρ dφ dz

Following the same evaulation procedure as in part a, we obtain Q′ = 0.182 mC.

2.15. A spherical volume having a 2 µm radius contains a uniform volume charge density of 1015 C/m3.

a) What total charge is enclosed in the spherical volume? This will be Q = (4/3)π(2 × 10−6)3 × 1015 = 3.35 × 10−2 C.

b) Now assume that a large region contains one of these little spheres at every corner of a cubical grid 3mm on a side, and that there is no charge between spheres. What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, the average density becomes

ρv,avg = 3.35 × 10 −2

(0.003)3 = 1.24 × 106 C/m3

2.16. The region in which 4 < r < 5, 0 < θ < 25◦, and 0.9π < φ < 1.1π contains the volume charge density of ρv = 10(r − 4)(r − 5) sin θ sin(φ/2). Outside the region, ρv = 0. Find the charge within the region: The integral that gives the charge will be

Q = 10 ∫ 1.1π .9π

∫ 25◦ 0

∫ 5 4

(r − 4)(r − 5) sin θ sin(φ/2) r2 sin θ dr dθ dφ

19

2.16. (continued) Carrying out the integral, we obtain

Q = 10 [ r5

5 − 9 r

4

4 + 20 r

3

3

]5 4

[ 1

2 θ − 1

4 sin(2θ)

]25◦ 0

[ −2 cos

( θ

2

)]1.1π .9π

= 10(−3.39)(.0266)(.626) = 0.57 C

2.17. A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If  = 0: a) Find E at P(1, 2, 3): This will be

EP = ρl 2π0

RP |RP |2

where RP = (1, 2, 3)(1,−2, 5) = (0, 4,−2), and |RP |2 = 20. So

EP = 16 × 10 −9

2π0

[ 4ay − 2az

20

] = 57.5ay − 28.8az V/m

b) Find E at that point in the z = 0 plane where the direction of E is given by (1/3)ay (2/3)az: With z = 0, the general field will be

Ez=0 = ρl 2π0

[ (y + 2)ay − 5az (y + 2)2 + 25

]

We require |Ez| = −|2Ey |, so 2(y + 2) = 5. Thus y = 1/2, and the field becomes:

Ez=0 = ρl 2π0

[ 2.5ay − 5az (2.5)2 + 25

] = 23ay − 46az

2.18. Uniform line charges of 0.4 µC/m and −0.4 µC/m are located in the x = 0 plane at y = −0.6 and y = 0.6 m respectively. Let  = 0.

a) Find E at P(x, 0, z): In general, we have

EP = ρl 2π0

[ R+P |R+P | −

RP |RP |

]

where R+P and RP are, respectively, the vectors directed from the positive and negative line charges to the point P , and these are normal to the z axis. We thus have R+P = (x, 0, z) (0,.6, z) = (x, .6, 0), and RP = (x, 0, z)(0, .6, z) = (x,.6, 0). So

EP = ρl 2π0

[ xax + 0.6ay x2 + (0.6)2 −

xax − 0.6ay x2 + (0.6)2

] = 0.4 × 10

−6

2π0

[ 1.2ay

x2 + 0.36 ] = 8.63ay

x2 + 0.36 kV/m

20

2.18. (continued) b) Find E at Q(2, 3, 4): This field will in general be:

EQ = ρl 2π0

[ R+Q |R+Q| −

RQ |RQ|

]

where R+Q = (2, 3, 4)(0,.6, 4) = (2, 3.6, 0), and RQ = (2, 3, 4)(0, .6, 4) = (2, 2.4, 0). Thus

EQ = ρl 2π0

[ 2ax + 3.6ay 22 + (3.6)2 −

2ax + 2.4ay 22 + (2.4)2

] = −625.8ax − 241.6ay V/m

2.19. A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P(1, 2, 3) if the charge extends from

a) −∞ < z < ∞: With the infinite line, we know that the field will have only a radial component in cylindrical coordinates (or x and y components in cartesian). The field from an infinite line on the z axis is generally E = [ρl/(2π0ρ)]aρ . Therefore, at point P :

EP = ρl 2π0

RzP |RzP |2 =

(2 × 10−6) 2π0

ax + 2ay 5

= 7.2ax + 14.4ay kV/m

where RzP is the vector that extends from the line charge to point P , and is perpendicular to the z axis; i.e., RzP = (1, 2, 3)(0, 0, 3) = (1, 2, 0).

b) −4 ≤ z ≤ 4: Here we use the general relation

EP = ∫

ρldz

4π0

r r′ |r r′|3

where r = ax + 2ay + 3az and r′ = zaz. So the integral becomes

EP = (2 × 10 −6)

4π0

∫ 4 −4

ax + 2ay + (3 − z)az [5 + (3 − z)2]1.5 dz

Using integral tables, we obtain:

EP = 3597 [ (ax + 2ay)(z− 3)+ 5az

(z2 − 6z+ 14) ]4 −4

V/m = 4.9ax + 9.8ay + 4.9az kV/m

The student is invited to verify that when evaluating the above expression over the limits −∞ < z < ∞, the z component vanishes and the x and y components become those found in part a.

2.20. Uniform line charges of 120 nC/m lie along the entire extent of the three coordinate axes. Assuming free space conditions, find E at P(−3, 2,−1): Since all line charges are infinitely-long, we can write:

EP = ρl 2π0

[ RxP

|RxP |2 + RyP

|RyP |2 + RzP

|RzP |2 ]

where RxP , RyP , and RzP are the normal vectors from each of the three axes that terminate on point P . Specifically, RxP = (−3, 2,−1) (−3, 0, 0) = (0, 2,−1), RyP = (−3, 2,−1) (0, 2, 0) = (−3, 0,−1), and RzP = (−3, 2,−1)(0, 0,−1) = (−3, 2, 0). Substituting these into the expression for EP gives

EP = ρl 2π0

[ 2ay az

5 + −3ax az

10 + −3ax + 2ay

13

] = −1.15ax + 1.20ay − 0.65az kV/m

21

2.21. Two identical uniform line charges with ρl = 75 nC/m are located in free space at x = 0, y = ±0.4 m. What force per unit length does each line charge exert on the other? The charges are parallel to the z axis and are separated by 0.8 m. Thus the field from the charge at y = −0.4 evaluated at the location of the charge at y = +0.4 will be E = [ρl/(2π0(0.8))]ay . The force on a differential length of the line at the positive y location is dF = dqE = ρldzE. Thus the force per unit length acting on the line at postive y arising from the charge at negative y is

F = ∫ 1

0

ρ2l dz

2π0(0.8) ay = 1.26 × 10−4 ay N/m = 126 ay µN/m

The force on the line at negative y is of course the same, but with −ay .

2.22. A uniform surface charge density of 5 nC/m2 is present in the region x = 0, −2 < y < 2, and all z. If  = 0, find E at:

a) PA(3, 0, 0): We use the superposition integral:

E = ∫ ∫

ρsda

4π0

r r′ |r r′|3

where r = 3ax and r′ = yay + zaz. The integral becomes:

EPA = ρs 4π0

∫ ∞ −∞

∫ 2 −2

3ax yay zaz [9 + y2 + z2]1.5 dy dz

Since the integration limits are symmetric about the origin, and since the y and z components of the integrand exhibit odd parity (change sign when crossing the origin, but otherwise symmetric), these will integrate to zero, leaving only the x component. This is evident just from the symmetry of the problem. Performing the z integration first on the x component, we obtain (using tables):

Ex,PA = 3ρs 4π0

∫ 2 −2

dy

(9 + y2)

[ z

9 + y2 + z2

]∞ −∞

= 3ρs 2π0

∫ 2 −2

dy

(9 + y2)

= 3ρs 2π0

( 1

3

) tan−1

(y 3

) ∣∣∣2−2 = 106 V/m The student is encouraged to verify that if the y limits were −∞ to ∞, the result would be that of the infinite charged plane, or Ex = ρs/(20).

b) PB(0, 3, 0): In this case, r = 3ay , and symmetry indicates that only a y component will exist. The integral becomes

Ey,PB = ρs 4π0

∫ ∞ −∞

∫ 2 −2

(3 − y) dy dz [(z2 + 9)− 6y + y2]1.5 =

ρs

2π0

∫ 2 −2

(3 − y) dy (3 − y)2

= − ρs 2π0

ln(3 − y) ∣∣∣2−2 = 145 V/m

22

2.23. Given the surface charge density, ρs = 2µC/m2, in the region ρ < 0.2 m, z = 0, and is zero elsewhere, find E at:

a) PA(ρ = 0, z = 0.5): First, we recognize from symmetry that only a z component of E will be present. Considering a general point z on the z axis, we have r = zaz. Then, with r′ = ρaρ , we obtain r r′ = zaz ρaρ . The superposition integral for the z component of E will be:

Ez,PA = ρs

4π0

∫ 2π 0

∫ 0.2 0

z ρ dρ dφ

2 + z2)1.5 = − 2πρs 4π0

z

[ 1√

z2 + ρ2

]0.2 0

= ρs 20

z

[ 1√ z2

− 1√ z2 + 0.4

]

With z = 0.5 m, the above evaluates as Ez,PA = 8.1 kV/m. b) With z at −0.5 m, we evaluate the expression for Ez to obtain Ez,PB = −8.1 kV/m.

2.24. Surface charge density is positioned in free space as follows: 20 nC/m2 at x = −3, −30 nC/m2 at y = 4, and 40 nC/m2 at z = 2. Find the magnitude of E at the three points, (4, 3,−2), (−2, 5,−1), and (0, 0, 0). Since all three sheets are infinite, the field magnitude associated with each one will be ρs/(20), which is position-independent. For this reason, the net field magnitude will be the same everywhere, whereas the field direction will depend on which side of a given sheet one is positioned. We take the first point, for example, and find

EA = 20 × 10 −9

20 ax + 30 × 10

−9

20 ay − 40 × 10

−9

20 az = 1130ax + 1695ay − 2260az V/m

The magnitude of EA is thus 3.04 kV/m. This will be the magnitude at the other two points as well.

2.25. Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2, 0, 6); uniform line charge density, 3nC/m at x = −2, y = 3; uniform surface charge density, 0.2 nC/m2 at x = 2. The sum of the fields at the origin from each charge in order is:

E = [ (12 × 10−9)

4π0

(−2ax − 6az) (4 + 36)1.5

] + [ (3 × 10−9)

2π0

(2ax − 3ay) (4 + 9)

] − [ (0.2 × 10−9)ax

20

] = −3.9ax − 12.4ay − 2.5az V/m

2.26. A uniform line charge density of 5 nC/m is at y = 0, z = 2 m in free space, while −5 nC/m is located at y = 0, z = −2 m. A uniform surface charge density of 0.3 nC/m2 is at y = 0.2 m, and −0.3 nC/m2 is at y = −0.2 m. Find |E| at the origin: Since each pair consists of equal and opposite charges, the effect at the origin is to double the field produce by one of each type. Taking the sum of the fields at the origin from the surface and line charges, respectively, we find:

E(0, 0, 0) = −2 × 0.3 × 10 −9

20 ay − 2 × 5 × 10

−9

2π0(2) az = −33.9ay − 89.9az

so that |E| = 96.1 V/m.

23

2.27. Given the electric field E = (4x − 2y)ax (2x + 4y)ay , find: a) the equation of the streamline that passes through the point P(2, 3,−4): We write

dy

dx = Ey

Ex = −(2x + 4y)

(4x − 2y) Thus

2(x dy + y dx) = y dy x dx or

2 d(xy) = 1 2 d(y2)− 1

2 d(x2)

So

C1 + 2xy = 1 2 y2 − 1

2 x2

or y2 − x2 = 4xy + C2

Evaluating at P(2, 3,−4), obtain:

9 − 4 = 24 + C2, or C2 = −19

Finally, at P , the requested equation is

y2 − x2 = 4xy − 19

b) a unit vector specifying the direction of E at Q(3,−2, 5): Have EQ = [4(3)+ 2(2)]ax − [2(3)− 4(2)]ay = 16ax + 2ay . Then |E| =

√ 162 + 4 = 16.12 So

aQ = 16ax + 2ay 16.12

= 0.99ax + 0.12ay

2.28. Let E = 5x3 ax − 15x2y ay , and find: a) the equation of the streamline that passes through P(4, 2, 1): Write

dy

dx = Ey

Ex = −15x

2y

5x3 = −3y

x

So dy

y = −3 dx

x ⇒ ln y = −3 ln x + lnC

Thus

y = e−3 ln xelnC = C x3

At P , have 2 = C/(4)3 ⇒ C = 128. Finally, at P ,

y = 128 x3

24

2.28. (continued) b) a unit vector aE specifying the direction of E at Q(3,−2, 5): At Q, EQ = 135ax + 270ay , and

|EQ| = 301.9. Thus aE = 0.45ax + 0.89ay . c) a unit vector aN = (l, m, 0) that is perpendicular to aE atQ: Since this vector is to have no z compo-

nent, we can find it through aN = ±(aE×az). Performing this, we find aN = ±(0.89ax − 0.45ay).

2.29. If E = 20e−5y (cos 5xax − sin 5xay), find: a) |E| at P(π/6, 0.1, 2): Substituting this point, we obtain EP = −10.6ax − 6.1ay , and so |EP | =

12.2.

b) a unit vector in the direction of EP : The unit vector associated with E is just ( cos 5xax − sin 5xay

) ,

which evaluated at P becomes aE = −0.87ax − 0.50ay . c) the equation of the direction line passing through P : Use

dy

dx = − sin 5x

cos 5x = − tan 5x dy = − tan 5x dx

Thus y = 15 ln cos 5x + C. Evaluating at P , we find C = 0.13, and so

y = 1 5

ln cos 5x + 0.13

2.30. Given the electric field intensity E = 400yax + 400xay V/m, find: a) the equation of the streamline passing through the point A(2, 1,−2): Write:

dy

dx = Ey

Ex = x

y x dx = y dy

Thus x2 = y2 + C. Evaluating at A yields C = 3, so the equation becomes

x2

3 − y

2

3 = 1

b) the equation of the surface on which |E| = 800 V/m: Have |E| = 400 √ x2 + y2 = 800. Thus

x2 + y2 = 4, or we have a circular-cylindrical surface, centered on the z axis, and of radius 2. c) A sketch of the part a equation would yield a parabola, centered at the origin, whose axis is the

positive x axis, and for which the slopes of the asymptotes are ±1. d) A sketch of the trace produced by the intersection of the surface of part b with the z = 0 plane

would yield a circle centered at the origin, of radius 2.

25