# Eletromagnetismo - hayt - 6ª edição - solution, Exercícios de A Física de Dispositivos Energéticos. Centro Federal de Educação Tecnológica Celso Suckow da Fonseca (CEFET/RJ)

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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

2.31. In cylindrical coordinates with E(ρ, φ) = Eρ(ρ, φ)aρ+Eφ(ρ, φ)aφ , the differential equation describ- ing the direction lines is Eρ/Eφ = dρ/(ρdφ) in any constant-z plane. Derive the equation of the line passing through the point P(ρ = 4, φ = 10◦, z = 2) in the field E = 2ρ2 cos 3φaρ + 2ρ2 sin 3φaφ : Using the given information, we write

=

ρdφ = cot 3φ

Thus

ρ = cot 3φ dφ ⇒ ln ρ = 1

3 ln sin 3φ + lnC

or ρ = C(sin 3φ)1/3. Evaluate this at P to obtain C = 7.14. Finally,

ρ3 = 364 sin 3φ

26

CHAPTER 3

3.1. An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged (honorably) by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The penny is given a charge of +5 nC, and the nickel and dime are discharged. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground. The device is carefully disassembled with insulating gloves and tools.

a) What charges are found on each of the five metallic pieces? All coins were insulated during the entire procedure, so they will retain their original charges: Penny: +5 nC; nickel: 0; dime: 0. The penny’s charge will have induced an equal and opposite negative charge (-5 nC) on the inside wall of the can and lid. This left a charge layer of +5 nC on the outside surface which was neutralized by the ground connection. Therefore, the can retained a net charge of −5 nC after disassembly.

b) If the penny had been given a charge of +5 nC, the dime a charge of −2 nC, and the nickel a charge of −1 nC, what would the final charge arrangement have been? Again, since the coins are insulated, they retain their original charges. The charge induced on the inside wall of the can and lid is equal to negative the sum of the coin charges, or −2 nC. This is the charge that the can/lid contraption retains after grounding and disassembly.

3.2. A point charge of 12 nC is located at the origin. four uniform line charges are located in the x = 0 plane as follows: 80 nC/m at y = −1 and −5 m, −50 nC/m at y = −2 and −4 m.

a) Find D at P(0,−3, 2): Note that this point lies in the center of a symmetric arrangement of line charges, whose fields will all cancel at that point. Thus D arise from the point charge alone, and will be

D = 12 × 10 −9(−3ay + 2az)

4π(32 + 22)1.5 = −6.11 × 10 −11ay + 4.07 × 10−11az C/m2

= −61.1ay + 40.7az pC/m2

b) How much electric flux crosses the plane y = −3 and in what direction? The plane intercepts all flux that enters the −y half-space, or exactly half the total flux of 12 nC. The answer is thus 6 nC and in the −ay direction.

c) How much electric flux leaves the surface of a sphere, 4m in radius, centered at C(0,−3, 0)? This sphere encloses the point charge, so its flux of 12 nC is included. The line charge contributions are most easily found by translating the whole assembly (sphere and line charges) such that the sphere is centered at the origin, with line charges now at y = ±1 and ±2. The flux from the line charges will equal the total line charge that lies within the sphere. The length of each of the inner two line charges (at y = ±1) will be

h1 = 2r cos θ1 = 2(4) cos [

sin−1 (

1

4

)] = 1.94 m

That of each of the outer two line charges (at y = ±2) will be

h2 = 2r cos θ2 = 2(4) cos [

sin−1 (

2

4

)] = 1.73 m

27

3.2c. (continued) The total charge enclosed in the sphere (and the outward flux from it) is now

Ql +Qp = 2(1.94)(−50 × 10−9)+ 2(1.73)(80 × 10−9)+ 12 × 10−9 = 348 nC

3.3. The cylindrical surface ρ = 8 cm contains the surface charge density, ρs = 5e−20|z| nC/m2. a) What is the total amount of charge present? We integrate over the surface to find:

Q = 2 ∫ ∞

0

∫ 2π 0

5e−20z(.08)dφ dz nC = 20π(.08) (−1

20

) e−20z

∣∣∣∣∣ ∞

0

= 0.25 nC

b) How much flux leaves the surface ρ = 8 cm, 1 cm < z < 5cm, 30◦ < φ < 90◦? We just integrate the charge density on that surface to find the flux that leaves it.

 = Q′ = ∫ .05 .01

∫ 90◦ 30◦

5e−20z(.08) dφ dz nC = (

90 − 30 360

) 2π(5)(.08)

(−1 20

) e−20z

∣∣∣∣∣ .05

.01

= 9.45 × 10−3 nC = 9.45 pC

3.4. The cylindrical surfaces ρ = 1, 2, and 3 cm carry uniform surface charge densities of 20, −8, and 5 nC/m2, respectively.

a) How much electric flux passes through the closed surface ρ = 5 cm, 0 < z < 1 m? Since the densities are uniform, the flux will be

 = 2π(aρs1 + bρs2 + cρs3)(1 m) = 2π [(.01)(20)(.02)(8)+ (.03)(5)] × 10−9 = 1.2 nC

b) Find D at P(1 cm, 2 cm, 3 cm): This point lies at radius √

5 cm, and is thus inside the outermost charge layer. This layer, being of uniform density, will not contribute to D at P . We know that in cylindrical coordinates, the layers at 1 and 2 cm will produce the flux density:

D = aρ = aρs1 + bρs2 ρ

aρ

or

= (.01)(20)+ (.02)(−8).05

= 1.8 nC/m2

At P , φ = tan−1(2/1) = 63.4◦. Thus Dx = 1.8 cosφ = 0.8 and Dy = 1.8 sin φ = 1.6. Finally,

DP = (0.8ax + 1.6ay) nC/m2

28

3.5. Let D = 4xyax + 2(x2 + z2)ay + 4yzaz C/m2 and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 < x < 2, 0 < y < 3, 0 < z < 5 m: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. Why? First consider the planes at y = 0 and 3. The y component of D will penetrate those surfaces, but will be inward at y = 0 and outward at y = 3, while having the same magnitude in both cases. These fluxes will thus cancel. At the x = 0 plane, Dx = 0 and at the z = 0 plane, Dz = 0, so there will be no flux contributions from these surfaces. This leaves the 2 remaining surfaces at x = 2 and z = 5. The net outward flux becomes:

 = ∫ 5

0

∫ 3 0

D ∣∣ x=2 · ax dy dz+

∫ 3 0

∫ 2 0

D ∣∣ z=5 · az dx dy

= 5 ∫ 3

0 4(2)y dy + 2

∫ 3 0

4(5)y dy = 360 C

3.6. Two uniform line charges, each 20 nC/m, are located at y = 1, z = ±1 m. Find the total flux leaving a sphere of radius 2 m if it is centered at

a) A(3, 1, 0): The result will be the same if we move the sphere to the origin and the line charges to (0, 0,±1). The length of the line charge within the sphere is given by l = 4 sin[cos−1(1/2)] = 3.46. With two line charges, symmetrically arranged, the total charge enclosed is given by Q = 2(3.46)(20 nC/m) = 139 nC

b) B(3, 2, 0): In this case the result will be the same if we move the sphere to the origin and keep the charges where they were. The length of the line joining the origin to the midpoint of the line charge (in the yz plane) is l1 =

√ 2. The length of the line joining the origin to either endpoint

of the line charge is then just the sphere radius, or 2. The half-angle subtended at the origin by the line charge is then ψ = cos−1(√2/2) = 45◦. The length of each line charge in the sphere is then l2 = 2 × 2 sinψ = 2

√ 2. The total charge enclosed (with two line charges) is now

Q′ = 2(2√2)(20 nC/m) = 113 nC

3.7. Volume charge density is located in free space as ρv = 2e−1000r nC/m3 for 0 < r < 1 mm, and ρv = 0 elsewhere.

a) Find the total charge enclosed by the spherical surface r = 1 mm: To find the charge we integrate:

Q = ∫ 2π

0

π 0

.001 0

2e−1000r r2 sin θ dr dθ dφ

Integration over the angles gives a factor of 4π . The radial integration we evaluate using tables; we obtain

Q = 8π [−r2e−1000r

1000

∣∣∣.001 0

+ 2 1000

e−1000r

(1000)2 (−1000r − 1)

∣∣∣.001 0

] = 4.0 × 10−9 nC

b) By using Gauss’s law, calculate the value of Dr on the surface r = 1 mm: The gaussian surface is a spherical shell of radius 1 mm. The enclosed charge is the result of part a. We thus write 4πr2Dr = Q, or

Dr = Q 4πr2

= 4.0 × 10 −9

4π(.001)2 = 3.2 × 10−4 nC/m2

29

3.8. Uniform line charges of 5 nC/m ar located in free space at x = 1, z = 1, and at y = 1, z = 0. a) Obtain an expression for D in cartesian coordinates at P(0, 0, z). In general, we have

D(z) = ρs 2π

[ r1 − r′1

|r1 − r′1|2 + r2 − r

′ 2

|r2 − r′2|2 ]

where r1 = r2 = zaz, r′1 = ay , and r′2 = ax + az. Thus

D(z) = ρs 2π

[ [zaz ay] [1 + z2] +

[(z− 1)az ax] [1 + (z− 1)2]

]

= ρs 2π

[ −ax [1 + (z− 1)2] −

ay [1 + z2] +

( (z− 1)

[1 + (z− 1)2] + z

[1 + z2] )

az

]

b) Plot |D| vs. z at P , −3 < z < 10: Using part a, we find the magnitude of D to be

|D| = ρs 2π

[ 1

[1 + (z− 1)2]2 + 1

[1 + z2]2 + (

(z− 1) [1 + (z− 1)2] +

z

[1 + z2] )2]1/2

A plot of this over the specified range is shown in Prob3.8.pdf.

3.9. A uniform volume charge density of 80µC/m3 is present throughout the region 8 mm < r < 10 mm. Let ρv = 0 for 0 < r < 8 mm.

a) Find the total charge inside the spherical surface r = 10 mm: This will be

Q = ∫ 2π

0

π 0

.010 .008

(80 × 10−6)r2 sin θ dr dθ dφ = 4π × (80 × 10−6) r 3

3

∣∣∣.010 .008

= 1.64 × 10−10 C = 164 pC

b) Find Dr at r = 10 mm: Using a spherical gaussian surface at r = 10, Gauss’ law is written as 4πr2Dr = Q = 164 × 10−12, or

Dr(10 mm) = 164 × 10 −12

4π(.01)2 = 1.30 × 10−7 C/m2 = 130 nC/m2

c) If there is no charge for r > 10 mm, find Dr at r = 20 mm: This will be the same computation as in part b, except the gaussian surface now lies at 20 mm. Thus

Dr(20 mm) = 164 × 10 −12

4π(.02)2 = 3.25 × 10−8 C/m2 = 32.5 nC/m2

3.10. Let ρs = 8µC/m2 in the region where x = 0 and −4 < z < 4 m, and let ρs = 0 elsewhere. Find D at P(x, 0, z), where x > 0: The sheet charge can be thought of as an assembly of infinitely-long parallel strips that lie parallel to the y axis in the yz plane, and where each is of thickness dz. The field from each strip is that of an infinite line charge, and so we can construct the field at P from a single strip as:

dDP = ρs dz

2π

r r′ |r r′|2

30

3.10 (continued) where r = xax + zaz and r′ = zaz We distinguish between the fixed coordinate of P , z, and the variable coordinate, z′, that determines the location of each charge strip. To find the net field at P , we sum the contributions of each strip by integrating over z′:

DP = ∫ 4 −4

8 × 10−6 dz(xax + (zz)az) 2π [x2 + (zz)2]

We can re-arrange this to determine the integral forms:

DP = 8 × 10 −6

2π

[ (xax + zaz)

∫ 4 −4

dz

(x2 + z2)− 2zz′ + (z)2 − az ∫ 4 −4

zdz

(x2 + z2)− 2zz′ + (z)2 ]

Using integral tables, we find

DP = 4 × 10 −6

π

[ (xax + zaz)1

x tan−1

( 2z′ − 2z

2x

)

− [

1

2 ln(x2 + z2 − 2zz′ + (z)2)+ 2z

2

1

x tan−1

( 2z′ − 2z

2x

)] az

]4 −4

which evaluates as

DP = 4 × 10 −6

π

{[ tan−1

( z+ 4 x

) − tan−1

( z− 4 x

)] ax + 1

2 ln

[ x2 + (z+ 4)2 x2 + (z− 4)2

] az

} C/m2

The student is invited to verify that for very small x or for a very large sheet (allowing z′ to approach infinity), the above expression reduces to the expected form, DP = ρs/2. Note also that the expression is valid for all x (positive or negative values).

3.11. In cylindrical coordinates, let ρv = 0 for ρ < 1 mm, ρv = 2 sin(2000πρ) nC/m3 for 1 mm < ρ < 1.5 mm, and ρv = 0 for ρ > 1.5 mm. Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed and will be constant over a cylindrical surface of a fixed radius. Gauss’ law applied to such a surface of unit length in z gives:

a) for ρ < 1 mm, = 0, since no charge is enclosed by a cylindrical surface whose radius lies within this range.

b) for 1 mm < ρ < 1.5 mm, we have

2πρDρ = 2π ρ .001

2 × 10−9 sin(2000πρ

= 4π × 10−9 [

1

(2000π)2 sin(2000πρ)ρ

2000π cos(2000πρ)

]ρ .001

or finally,

= 10 −15

2π2ρ

[ sin(2000πρ)+ 2π

[ 1 − 103ρ cos(2000πρ)

] ] C/m2 (1 mm < ρ < 1.5 mm)

31

3.11. (continued) c) for ρ > 1.5 mm, the gaussian cylinder now lies at radius ρ outside the charge distribution, so

the integral that evaluates the enclosed charge now includes the entire charge distribution. To accomplish this, we change the upper limit of the integral of part b from ρ to 1.5 mm, finally obtaining:

= 2.5 × 10 −15

πρ C/m2 (ρ > 1.5 mm)

3.12. A nonuniform volume charge density, ρv = 120r C/m3, lies within the spherical surface r = 1 m, and ρv = 0 everywhere else.

a) Find Dr everywhere. For r < 1 m, we apply Gauss’ law to a spherical surface of radius r within this range to find

4πr2Dr = 4π r

0 120r (r )2 dr ′ = 120πr4

Thus Dr = (30r2) for r < 1 m. For r > 1 m, the gaussian surface lies outside the charge distribution. The set up is the same, except the upper limit of the above integral is 1 instead of r . This results in Dr = (30/r2) for r > 1 m.

b) What surface charge density, ρs2, should be on the surface r = 2 such that Dr,r=2− = 2Dr,r=2+? At r = 2−, we have Dr,r=2− = 30/22 = 15/2, from part a. The flux density in the region r > 2 arising from a surface charge at r = 2 is found from Gauss’ law through

4πr2Drs = 4π(2)2ρs2 ⇒ Drs = 4ρs2 r2

The total flux density in the region r > 2 arising from the two distributions is

DrT = 30 r2

+ 4ρs2 r2

Our requirement that Dr,r=2− = 2Dr,r=2+ becomes

30

22 = 2

( 30

22 + ρs2

) ⇒ ρs2 = −15

4 C/m2

c) Make a sketch of Dr vs. r for 0 < r < 5 m with both distributions present. With both charges, Dr(r < 1) = 30r2, Dr(1 < r < 2) = 30/r2, and Dr(r > 2) = 15/r2. These are plotted on the next page.

32

.

3.13. Spherical surfaces at r = 2, 4, and 6 m carry uniform surface charge densities of 20 nC/m2,−4 nC/m2, and ρs0, respectively.

a) Find D at r = 1, 3 and 5 m: Noting that the charges are spherically-symmetric, we ascertain that D will be radially-directed and will vary only with radius. Thus, we apply Gauss’ law to spherical shells in the following regions: r < 2: Here, no charge is enclosed, and so Dr = 0.

2 < r < 4 : 4πr2Dr = 4π(2)2(20 × 10−9) Dr = 80 × 10 −9

r2 C/m2

So Dr(r = 3) = 8.9 × 10−9 C/m2.

4 < r < 6 : 4πr2Dr = 4π(2)2(20 × 10−9)+ 4π(4)2(−4 × 10−9) Dr = 16 × 10 −9

r2

So Dr(r = 5) = 6.4 × 10−10 C/m2. b) Determine ρs0 such that D = 0 at r = 7 m. Since fields will decrease as 1/r2, the question could

be re-phrased to ask for ρs0 such that D = 0 at all points where r > 6 m. In this region, the total field will be

Dr(r > 6) = 16 × 10 −9

r2 + ρs0(6)

2

r2

Requiring this to be zero, we find ρs0 = −(4/9)× 10−9 C/m2.

3.14. If ρv = 5 nC/m3 for 0 < ρ < 1 mm and no other charges are present: a) find for ρ < 1 mm: Applying Gauss’ law to a cylindrical surface of unit length in z, and of

radius ρ < 1 mm, we find

2πρDρ = πρ2(5 × 10−9) = 2.5 ρ × 10−9 C/m2

33

3.14b. find for ρ > 1 mm: The Gaussian cylinder now lies outside the charge, so

2πρDρ = π(.001)2(5 × 10−9) = 2.5 × 10 −15

ρ C/m2

c) What line charge ρL at ρ = 0 would give the same result for part b? The line charge field will be

Dr = ρL 2πρ

= 2.5 × 10 −15

ρ (part b)

Thus ρL = 5π × 10−15 C/m. In all answers, ρ is expressed in meters.

3.15. Volume charge density is located as follows: ρv = 0 for ρ < 1 mm and for ρ > 2 mm, ρv = 4ρ µC/m3 for 1 < ρ < 2 mm.

a) Calculate the total charge in the region 0 < ρ < ρ1, 0 < z < L, where 1 < ρ1 < 2 mm: We find

Q = ∫ L

0

∫ 2π 0

ρ1 .001

4ρ ρ dρ dφ dz = 8πL 3

[ρ31 − 10−9] µC

where ρ1 is in meters.

b) Use Gauss’ law to determine at ρ = ρ1: Gauss’ law states that 2πρ1LDρ = Q, where Q is the result of part a. Thus

Dρ(ρ1) = 431 − 10−9)

3ρ1 µC/m2

where ρ1 is in meters.

c) Evaluate at ρ = 0.8 mm, 1.6 mm, and 2.4 mm: At ρ = 0.8 mm, no charge is enclosed by a cylindrical gaussian surface of that radius, so Dρ(0.8mm) = 0. At ρ = 1.6 mm, we evaluate the part b result at ρ1 = 1.6 to obtain:

Dρ(1.6mm) = 4[(.0016) 3 − (.0010)3]

3(.0016) = 3.6 × 10−6 µC/m2

At ρ = 2.4, we evaluate the charge integral of part a from .001 to .002, and Gauss’ law is written as

2πρLDρ = 8πL 3

[(.002)2 − (.001)2] µC from which Dρ(2.4mm) = 3.9 × 10−6 µC/m2.

3.16. Given the electric flux density, D = 2xy ax + x2 ay + 6z3 az C/m2: a) use Gauss’ law to evaluate the total charge enclosed in the volume 0 < x, y, z < a: We call the

surfaces at x = a and x = 0 the front and back surfaces respectively, those at y = a and y = 0 the right and left surfaces, and those at z = a and z = 0 the top and bottom surfaces. To evaluate the total charge, we integrate D · n over all six surfaces and sum the results:

 = Q = ∮

D · n da = ∫ a

0

a 0

2ay dy dz︸ ︷︷ ︸ front

+ ∫ a

0

a 0

−2(0)y dy dz︸ ︷︷ ︸ back

+ ∫ a

0

a 0

x2 dx dz︸ ︷︷ ︸ left

+ ∫ a

0

a 0

x2 dx dz︸ ︷︷ ︸ right

+ ∫ a

0

a 0

−6(0)3 dx dy︸ ︷︷ ︸ bottom

+ ∫ a

0

a 0

6a3 dx dy︸ ︷︷ ︸ top

34

3.16a. (continued) Noting that the back and bottom integrals are zero, and that the left and right integrals cancel, we evaluate the remaining two (front and top) to obtain Q = 6a5 + a4.

b) use Eq. (8) to find an approximate value for the above charge. Evaluate the derivatives at P(a/2, a/2, a/2): In this application, Eq. (8) states that Q

.= (∇ · D∣∣ P )v. We find ∇ · D =

2x+18z2, which when evaluated at P becomes ∇ ·D∣∣ P = a+4.5a2. Thus Q .= (a+4.5a2)a3 =

4.5a5 + a4

c) Show that the results of parts a and b agree in the limit as a → 0. In this limit, both expressions reduce to Q = a4, and so they agree.

3.17. A cube is defined by 1 < x, y, z < 1.2. If D = 2x2yax + 3x2y2ay C/m2: a) apply Gauss’ law to find the total flux leaving the closed surface of the cube. We call the surfaces

at x = 1.2 and x = 1 the front and back surfaces respectively, those at y = 1.2 and y = 1 the right and left surfaces, and those at z = 1.2 and z = 1 the top and bottom surfaces. To evaluate the total charge, we integrate D · n over all six surfaces and sum the results. We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces. The fluxes through the remaining four are

 = Q = ∮

D · n da = ∫ 1.2

1

∫ 1.2 1

2(1.2)2y dy dz︸ ︷︷ ︸ front

+ ∫ 1.2

1

∫ 1.2 1

−2(1)2y dy dz︸ ︷︷ ︸ back

+ ∫ 1.2

1

∫ 1.2 1

−3x2(1)2 dx dz︸ ︷︷ ︸ left

+ ∫ 1.2

1

∫ 1.2 1

3x2(1.2)2 dx dz︸ ︷︷ ︸ right

= 0.1028 C

b) evaluate ∇ · D at the center of the cube: This is

∇ · D = [ 4xy + 6x2y

] (1.1,1.1)

= 4(1.1)2 + 6(1.1)3 = 12.83

c) Estimate the total charge enclosed within the cube by using Eq. (8): This is

Q .= ∇ · D∣∣center ×v = 12.83 × (0.2)3 = 0.1026 Close!

3.18. Let a vector field by given by G = 5x4y4z4 ay . Evaluate both sides of Eq. (8) for this G field and the volume defined by x = 3 and 3.1, y = 1 and 1.1, and z = 2 and 2.1. Evaluate the partial derivatives at the center of the volume. First find

∇ · G = ∂Gy ∂y

= 20x4y3z4

The center of the cube is located at (3.05,1.05,2.05), and the volume is v = (0.1)3 = 0.001. Eq. (8) then becomes

 .= 20(3.05)4(1.05)3(2.05)4(0.001) = 35.4

35

3.19. A spherical surface of radius 3 mm is centered at P(4, 1, 5) in free space. Let D = xax C/m2. Use the results of Sec. 3.4 to estimate the net electric flux leaving the spherical surface: We use 

.= ∇ ·Dv, where in this case ∇ · D = (∂/∂x)x = 1 C/m3. Thus

 .= 4

3 π(.003)3(1) = 1.13 × 10−7 C = 113 nC

3.20. A cube of volume a3 has its faces parallel to the cartesian coordinate surfaces. It is centered at P(3,−2, 4). Given the field D = 2x3ax C/m2:

a) calculate div D at P : In the present case, this will be

∇ · D = ∂Dx ∂x

= dDx dx

= 54 C/m3

b) evaluate the fraction in the rightmost side of Eq. (13) for a = 1 m, 0.1 m, and 1 mm: With the field having only an x component, flux will pentrate only the two surfaces at x = 3 ± a/2, each of which has surface area a2. The cube volume is v = a3. The equation reads:

D · dS v

= 1 a3

[ 2 (

3 + a 2

)3 a2 − 2

( 3 − a

2

)3 a2 ] = 2

a

[ (3 + a

2 )3 − (3 − a

2 )3 ]

evaluating the above formula at a = 1 m, .1 m, and 1 mm, yields respectively

54.50, 54.01, and 54.00 C/m3,

thus demonstrating the approach to the exact value as v gets smaller.

3.21. Calculate the divergence of D at the point specified if a) D = (1/z2) [10xyz ax + 5x2z ay + (2z3 − 5x2y) az] at P(−2, 3, 5): We find

∇ · D = [

10y

z + 0 + 2 + 10x

2y

z3

] (−2,3,5)

= 8.96

b) D = 5z2aρ + 10ρz az at P(3,−45◦, 5): In cylindrical coordinates, we have

∇ · D = 1 ρ

∂ρ (ρDρ)+ 1

ρ

∂Dφ

∂φ + ∂Dz

∂z = [

5z2

ρ + 10ρ

] (3,−45◦,5)

= 71.67

c) D = 2r sin θ sin φ ar + r cos θ sin φ aθ + r cosφ aφ at P(3, 45◦,−45◦): In spherical coordinates, we have

∇ · D = 1 r2

∂r (r2Dr)+ 1

r sin θ

∂θ (sin θDθ)+ 1

r sin θ

∂Dφ

∂φ

= [

6 sin θ sin φ + cos 2θ sin φ sin θ

− sin φ sin θ

] (3,45◦,−45◦)

= −2

36

3.22. Let D = 8ρ sin φ aρ + 4ρ cosφ aφ C/m2. a) Find div D: Using the divergence formula for cylindrical coordinates (see problem 3.21), we find

∇ · D = 12 sin φ. b) Find the volume charge density at P(2.6, 38◦,−6.1): Since ρv = ∇ · D, we evaluate the result of

part a at this point to find ρvP = 12 sin 38◦ = 7.39 C/m3. c) How much charge is located inside the region defined by 0 < ρ < 1.8, 20◦ < φ < 70◦,

2.4 < z < 3.1? We use

Q = ∫ vol

ρvdv = ∫ 3.1

2.4

∫ 70◦ 20◦

∫ 1.8 0

12 sin φρ dρ dφ dz = −(3.1 − 2.4)12 cosφ ∣∣∣70◦ 20◦

ρ2

2

∣∣∣1.8 0

= 8.13 C

3.23. a) A point charge Q lies at the origin. Show that div D is zero everywhere except at the origin. For a point charge at the origin we know that D = Q/(4πr2) ar . Using the formula for divergence in spherical coordinates (see problem 3.21 solution), we find in this case that

∇ · D = 1 r2

d

dr

( r2

Q

4πr2

) = 0

The above is true provided r > 0. When r = 0, we have a singularity in D, so its divergence is not defined.

b) Replace the point charge with a uniform volume charge density ρv0 for 0 < r < a. Relate ρv0 to Q and a so that the total charge is the same. Find div D everywhere: To achieve the same net charge, we require that (4/3)πa3ρv0 = Q, so ρv0 = 3Q/(4πa3) C/m3. Gauss’ law tells us that inside the charged sphere

4πr2Dr = 4 3 πr3ρv0 = Qr

3

a3

Thus

Dr = Qr 4πa3

C/m2 and ∇ · D = 1 r2

d

dr

( Qr3

4πa3

) = 3Q

4πa3

as expected. Outside the charged sphere, D = Q/(4πr2) ar as before, and the divergence is zero.

3.24. Inside the cylindrical shell, 3 < ρ < 4 m, the electric flux density is given as

D = 5− 3)3 aρ C/m2

a) What is the volume charge density at ρ = 4 m? In this case we have

ρv = ∇ · D = 1 ρ

d

dρ (ρDρ) = 1

ρ

d

[5ρ(ρ − 3)3] = 5− 3)

2

ρ (4ρ − 3) C/m3

Evaluating this at ρ = 4 m, we find ρv(4) = 16.25 C/m3

b) What is the electric flux density at ρ = 4 m? We evaluate the given D at this point to find D(4) = 5 aρ C/m2

37

3.24c. How much electric flux leaves the closed surface 3 < ρ < 4, 0 < φ < 2π , −2.5 < z < 2.5? We note that D has only a radial component, and so flux would leave only through the cylinder sides. Also, D does not vary with φ or z, so the flux is found by a simple product of the side area and the flux density. We further note that D = 0 at ρ = 3, so only the outer side (at ρ = 4) will contribute. We use the result of part b, and write the flux as

 = [2.5 − (−2.5)]2π(4)(5) = 200π C

d) How much charge is contained within the volume used in part c? By Gauss’ law, this will be the same as the net outward flux through that volume, or again, 200π C.

3.25. Within the spherical shell, 3 < r < 4 m, the electric flux density is given as

D = 5(r − 3)3 ar C/m2

a) What is the volume charge density at r = 4? In this case we have

ρv = ∇ · D = 1 r2

d

dr (r2Dr) = 5

r (r − 3)2(5r − 6) C/m3

which we evaluate at r = 4 to find ρv(r = 4) = 17.50 C/m3. b) What is the electric flux density at r = 4? Substitute r = 4 into the given expression to

find D(4) = 5 ar C/m2

c) How much electric flux leaves the sphere r = 4? Using the result of part b, this will be  = 4π(4)2(5) = 320π C

d) How much charge is contained within the sphere, r = 4? From Gauss’ law, this will be the same as the outward flux, or again, Q = 320π C.

3.26. Given the field

D = 5 sin θ cosφ r

ar C/m2,

find: a) the volume charge density: Use

ρv = ∇ · D = 1 r2

d

dr (r2Dr) = 5 sin θ cosφ

r2 C/m3

b) the total charge contained within the region r < 2 m: To find this, we integrate over the volume:

Q = ∫ 2π

0

π 0

∫ 2 0

5 sin θ cosφ

r2 r2 sin θ dr dθ dφ

Before plunging into this one notice that the φ integration is of cosφ from zero to 2π . This yields a zero result, and so the total enclosed charge is Q = 0.

c) the value of D at the surface r = 2: Substituting r = 2 into the given field produces

D(r = 2) = 5 2

sin θ cosφ ar C/m2

38

3.26d. the total electric flux leaving the surface r = 2 Since the total enclosed charge is zero (from part b), the net outward flux is also zero, from Gauss’ law.

3.27. Let D = 5.00r2ar mC/m2 for r ≤ 0.08 m and D = 0.205 ar/r2 µC/m2 for r ≥ 0.08 m (note error in problem statement).

a) Find ρv for r = 0.06 m: This radius lies within the first region, and so

ρv = ∇ · D = 1 r2

d

dr (r2Dr) = 1

r2

d

dr (5.00r4) = 20r mC/m3

which when evaluated at r = 0.06 yields ρv(r = .06) = 1.20 mC/m3. b) Find ρv for r = 0.1 m: This is in the region where the second field expression is valid. The 1/r2

dependence of this field yields a zero divergence (shown in Problem 3.23), and so the volume charge density is zero at 0.1 m.

c) What surface charge density could be located at r = 0.08 m to cause D = 0 for r > 0.08 m? The total surface charge should be equal and opposite to the total volume charge. The latter is

Q = ∫ 2π

0

π 0

.08 0

20r(mC/m3) r2 sin θ dr dθ dφ = 2.57 × 10−3 mC = 2.57µC

So now

ρs = − [

2.57

4π(.08)2

] = −32µC/m2

3.28. The electric flux density is given as D = 20ρ3 aρ C/m2 for ρ < 100µm, and k aρ/ρ for ρ > 100µm. a) Find k so that D is continuous at ρ = 100µm: We require

20 × 10−12 = k 10−4

k = 2 × 10−15 C/m

b) Find and sketch ρv as a function of ρ: In cylindrical coordinates, with only a radial component of D, we use

ρv = ∇ · D = 1 ρ

∂ρ (ρDρ) = 1

ρ

∂ρ (20ρ4) = 80ρ2 C/m3 (ρ < 100µm)

For ρ > 100µm, we obtain

ρv = 1 ρ

∂ρ (ρ

k

ρ ) = 0

The sketch of ρv vs. ρ would be a parabola, starting at the origin, reaching a maximum value of 8 × 10−7 C/m3 at ρ = 100 µm. The plot is zero at larger radii.

3.29. In the region of free space that includes the volume 2 < x, y, z < 3,

D = 2 z2

(yz ax + xz ay − 2xy az) C/m2

a) Evaluate the volume integral side of the divergence theorem for the volume defined above: In cartesian, we find ∇ · D = 8xy/z3. The volume integral side is now∫

vol

∇ · D dv = ∫ 3

2

∫ 3 2

∫ 3 2

8xy

z3 dxdydz = (9 − 4)(9 − 4)

( 1

4 − 1

9

) = 3.47 C

39

3.29b. Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at x = 3 and x = 2 the front and back surfaces respectively, those at y = 3 and y = 2 the right and left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces. To evaluate the surface integral side, we integrate D · n over all six surfaces and sum the results. Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other. The same is true for the left and right surfaces, since Dy does not vary with y. This leaves only the top and bottom surfaces, where the fluxes are:∮

D · dS = ∫ 3

2

∫ 3 2

−4xy 32

dxdy︸ ︷︷ ︸ top

− ∫ 3

2

∫ 3 2

−4xy 22

dxdy︸ ︷︷ ︸ bottom

= (9 − 4)(9 − 4) (

1

4 − 1

9

) = 3.47 C

3.30. If D = 15ρ2 sin 2φ aρ + 10ρ2 cos 2φ aφ C/m2, evaluate both sides of the divergence theorem for the region 1 < ρ < 2 m, 1 < φ < 2 rad, 1 < z < 2 m: Taking the surface integral side first, the six sides over which the flux must be evaluated are only four, since there is no z component of D. We are left with the sides at φ = 1 and φ = 2 rad (left and right sides, respectively), and those at ρ = 1 and ρ = 2 (back and front sides). We evaluate∮

D · dS = ∫ 2

1

∫ 2 1

15(2)2 sin(2φ) (2)dφdz︸ ︷︷ ︸ front

− ∫ 2

1

∫ 2 1

15(1)2 sin(2φ) (1)dφdz︸ ︷︷ ︸ back

− ∫ 2

1

∫ 2 1

10ρ2 cos(2) dρdz︸ ︷︷ ︸ left

+ ∫ 2

1

∫ 2 1

10ρ2 cos(4) dρdz︸ ︷︷ ︸ right

= 6.93 C

For the volume integral side, we first evaluate the divergence of D, which is

∇ · D = 1 ρ

∂ρ (15ρ3 sin 2φ)+ 1

ρ

∂φ (10ρ2 cos 2φ) = 25ρ sin 2φ

Next ∫ vol

∇ · D dv = ∫ 2

1

∫ 2 1

∫ 2 1

25ρ sin(2φ) ρdρ dφ dz = 25 3 ρ3 ∣∣∣2 1

[− cos(2φ) 2

]2 1 = 6.93 C

3.31. Given the flux density

D = 16 r

cos(2θ) aθ C/m2,

use two different methods to find the total charge within the region 1 < r < 2 m, 1 < θ < 2 rad, 1 < φ < 2 rad: We use the divergence theorem and first evaluate the surface integral side. We are evaluating the net outward flux through a curvilinear “cube”, whose boundaries are defined by the specified ranges. The flux contributions will be only through the surfaces of constant θ , however, since D has only a θ component. On a constant-theta surface, the differential area is da = r sin θdrdφ, where θ is fixed at the surface location. Our flux integral becomes∮

D · dS = − ∫ 2

1

∫ 2 1

16

r cos(2) r sin(1) drdφ︸ ︷︷ ︸

θ=1

+ ∫ 2

1

∫ 2 1

16

r cos(4) r sin(2) drdφ︸ ︷︷ ︸

θ=2 = −16 [cos(2) sin(1)− cos(4) sin(2)] = −3.91 C

40

3.31. (continued) We next evaluate the volume integral side of the divergence theorem, where in this case,

∇ · D = 1 r sin θ

d

dθ (sin θ Dθ) = 1

r sin θ

d

[ 16

r cos 2θ sin θ

] = 16

r2

[ cos 2θ cos θ

sin θ − 2 sin 2θ

]

We now evaluate:

vol

∇ · D dv = ∫ 2

1

∫ 2 1

∫ 2 1

16

r2

[ cos 2θ cos θ

sin θ − 2 sin 2θ

] r2 sin θ drdθdφ

The integral simplifies to

∫ 2 1

∫ 2 1

∫ 2 1

16[cos 2θ cos θ − 2 sin 2θ sin θ ] drdθdφ = 8 ∫ 2

1 [3 cos 3θ − cos θ ] = −3.91 C

3.32. If D = 2r ar C/m2, find the total electric flux leaving the surface of the cube, 0 < x, y, z < 0.4: This is where the divergence theorem really saves you time! First find

∇ · D = 1 r2

d

dr (r2 × 2r) = 6

Then the net outward flux will be ∫ vol

∇ · D dv = 6(0.4)3 = 0.38 C

41

CHAPTER 4

4.1. The value of E at P(ρ = 2, φ = 40◦, z = 3) is given as E = 100aρ −200aφ+300az V/m. Determine the incremental work required to move a 20µC charge a distance of 6 µm:

a) in the direction of aρ : The incremental work is given by dW = −q E · dL, where in this case, dL = aρ = 6 × 10−6 aρ . Thus

dW = −(20 × 10−6 C)(100 V/m)(6 × 10−6 m) = −12 × 10−9 J = −12 nJ

b) in the direction of aφ : In this case dL = 2 aφ = 6 × 10−6 aφ , and so

dW = −(20 × 10−6)(−200)(6 × 10−6) = 2.4 × 10−8 J = 24 nJ

c) in the direction of az: Here, dL = dz az = 6 × 10−6 az, and so

dW = −(20 × 10−6)(300)(6 × 10−6) = −3.6 × 10−8 J = −36 nJ

d) in the direction of E: Here, dL = 6 × 10−6 aE , where

aE = 100aρ − 200aφ + 300az [1002 + 2002 + 3002]1/2 = 0.267 aρ − 0.535 aφ + 0.802 az

Thus

dW = −(20 × 10−6)[100aρ − 200aφ + 300az] · [0.267 aρ − 0.535 aφ + 0.802 az](6 × 10−6) = −44.9 nJ

e) In the direction of G = 2 ax − 3 ay + 4 az: In this case, dL = 6 × 10−6 aG, where

aG = 2ax − 3ay + 4az [22 + 32 + 42]1/2 = 0.371 ax − 0.557 ay + 0.743 az

So now

dW = −(20 × 10−6)[100aρ − 200aφ + 300az] · [0.371 ax − 0.557 ay + 0.743 az](6 × 10−6) = −(20 × 10−6) [37.1(aρ · ax)− 55.7(aρ · ay)− 74.2(aφ · ax)+ 111.4(aφ · ay) + 222.9] (6 × 10−6)

where, at P , (aρ · ax) = (aφ · ay) = cos(40◦) = 0.766, (aρ · ay) = sin(40◦) = 0.643, and (aφ · ax) = − sin(40◦) = −0.643. Substituting these results in

dW = −(20 × 10−6)[28.4 − 35.8 + 47.7 + 85.3 + 222.9](6 × 10−6) = −41.8 nJ

42

4.2. Let E = 400ax − 300ay + 500az in the neighborhood of point P(6, 2,−3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by:

a) ax + ay + az: We write

dW = −qE · dL = −4(400ax − 300ay + 500az) · (ax + ay + az)√ 3

(10−3)

= − (4 × 10 −3)

3 (400 − 300 + 500) = −1.39 J

b) −2ax + 3ay az: The computation is similar to that of part a, but we change the direction:

dW = −qE · dL = −4(400ax − 300ay + 500az) · (−2ax + 3ay az)√ 14

(10−3)

= − (4 × 10 −3)

14 (−800 − 900 − 500) = 2.35 J

4.3. If E = 120 aρ V/m, find the incremental amount of work done in moving a 50µm charge a distance of 2 mm from:

a) P(1, 2, 3) toward Q(2, 1, 4): The vector along this direction will be Q P = (1,−1, 1) from which aPQ = [ax ay + az]/

√ 3. We now write

dW = −qE · dL = −(50 × 10−6) [

120aρ · (ax ay + az√ 3

] (2 × 10−3)

= −(50 × 10−6)(120) [(aρ · ax)(aρ · ay)] 1√ 3 (2 × 10−3)

At P , φ = tan−1(2/1) = 63.4◦. Thus (aρ · ax) = cos(63.4) = 0.447 and (aρ · ay) = sin(63.4) = 0.894. Substituting these, we obtain dW = 3.1µJ.

b) Q(2, 1, 4) toward P(1, 2, 3): A little thought is in order here: Note that the field has only a radial component and does not depend on φ or z. Note also that P and Q are at the same radius (

√ 5)

from the z axis, but have different φ and z coordinates. We could just as well position the two points at the same z location and the problem would not change. If this were so, then moving along a straight line between P and Q would thus involve moving along a chord of a circle whose radius is

√ 5. Halfway along this line is a point of symmetry in the field (make a sketch to see

this). This means that when starting from either point, the initial force will be the same. Thus the answer is dW = 3.1µJ as in part a. This is also found by going through the same procedure as in part a, but with the direction (roles of P and Q) reversed.

4.4. Find the amount of energy required to move a 6-C charge from the origin to P(3, 1,−1) in the field E = 2xax − 3y2ay + 4az V/m along the straight-line path x = −3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path.

W = −q

E · dL = −6 ∫ (2xax − 3y2ay + 4az) · (dxax + dyay + dzaz)

= −6 ∫ 3

0 2xdx + 6

∫ 1 0

3y2dy − 6 ∫ −1

0 4dz = −24 J

43

4.5. Compute the value of ∫ P A

G · dL for G = 2yax with A(1,−1, 2) and P(2, 1, 2) using the path: a) straight-line segments A(1,−1, 2) to B(1, 1, 2) to P(2, 1, 2): In general we would have

P A

G · dL = ∫ P A

2y dx

The change in x occurs when moving between B and P , during which y = 1. Thus ∫ P A

G · dL = ∫ P B

2y dx = ∫ 2

1 2(1)dx = 2

b) straight-line segmentsA(1,−1, 2) toC(2,−1, 2) toP(2, 1, 2): In this case the change in x occurs when moving from A to C, during which y = −1. Thus

P A

G · dL = ∫ C A

2y dx = ∫ 2

1 2(−1)dx = −2

4.6. Let G = 4xax+2zay+2yaz. Given an initial pointP(2, 1, 1) and a final pointQ(4, 3, 1), find ∫

G ·dL using the path: a) straight line: y = x − 1, z = 1; b) parabola: 6y = x2 + 2, z = 1: With G as given, the line integral will be

G · dL =

∫ 4 2

4x dx + ∫ 3

1 2z dy +

∫ 1 1

2y dz

Clearly, we are going nowhere in z, so the last integral is zero. With z = 1, the first two evaluate as ∫

G · dL = 2x2 ∣∣∣4 2 + 2y

∣∣∣3 1 = 28

The paths specified in parts a and b did not play a role, meaning that the integral between the specified points is path-independent.

4.7. Repeat Problem 4.6 for G = 3xy3ax + 2zay . Now things are different in that the path does matter: a) straight line: y = x − 1, z = 1: We obtain:

G · dL =

∫ 4 2

3xy2 dx + ∫ 3

1 2z dy =

∫ 4 2

3x(x − 1)2 dx + ∫ 3

1 2(1) dy = 90

b) parabola: 6y = x2 + 2, z = 1: We obtain: ∫

G · dL = ∫ 4

2 3xy2 dx +

∫ 3 1

2z dy = ∫ 4

2

1

12 x(x2 + 2)2 dx +

∫ 3 1

2(1) dy = 82

44

4.8. A point chargeQ1 is located at the origin in free space. Find the work done in carrying a chargeQ2 from: (a) B(rB, θB, φB) to C(rA, θB, φB) with θ and φ held constant; (b) C(rA, θB, φB) to D(rA, θA, φB) with r and φ held constant; (c) D(rA, θA, φB) to A(rA, θA, φA) with r and θ held constant: The general expression for the work done in this instance is

W = −Q2 ∫

E · dL = −Q2 ∫

Q1

4π0r2 ar · (drar + rdθaθ + r sin θdφaφ) = −Q1Q2

4π0

dr

r2

We see that only changes in r will produce non-zero results. Thus for part a we have

W = −Q1Q2 4π0

rA rB

dr

r2 = Q1Q2

4π0

[ 1

rA − 1

rB

] J

The answers to parts b and c (involving paths over which r is held constant) are both 0.

4.9. A uniform surface charge density of 20 nC/m2 is present on the spherical surface r = 0.6 cm in free space.

a) Find the absolute potential at P(r = 1 cm, θ = 25◦, φ = 50◦): Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter. The potential function for r > 0.6 cm will be that of a point charge of Q = 4πa2ρs , or

V (r) = 4π(0.6 × 10 −2)2(20 × 10−9)

4π0r = 0.081

r V with r in meters

At r = 1 cm, this becomes V (r = 1 cm) = 8.14 V b) Find VAB given points A(r = 2 cm, θ = 30◦, φ = 60◦) and B(r = 3 cm, θ = 45◦, φ = 90◦):

Again, the angles do not matter because of the spherical symmetry. We use the part a result to obtain

VAB = VA VB = 0.081 [

1

0.02 − 1

0.03

] = 1.36 V

4.10. Given a surface charge density of 8 nC/m2 on the plane x = 2, a line charge density of 30 nC/m on the line x = 1, y = 2, and a 1-µC point charge at P(−1,−1, 2), find VAB for points A(3, 4, 0) and B(4, 0, 1): We need to find a potential function for the combined charges. That for the point charge we know to be

Vp(r) = Q 4π0r

Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have

Vl(ρ) = − ∫

ρl

2π0ρ dρ + C1 = − ρl

2π0 ln(ρ)+ C1

For the sheet charge, we have

Vs(x) = − ∫

ρs

20 dx + C2 = − ρs

20 x + C2

45

4.10. (continued) The total potential function will be the sum of the three. Combining the integration con- stants, we obtain:

V = Q 4π0r

ρl 2π0

ln(ρ)ρs 20

x + C

The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r , ρ, and x are scalar distances from the charges, and will be treated as such here. For point A we have rA =

(3 − (−1))2 + (4 − (−1))2 + (−2)2 = √45, ρA =√

(3 − 1)2 + (4 − 2)2 = √8, and its distance from the sheet charge is xA = 3 − 2 = 1. The potential at A is then

VA = 10 −6

4π0 √

45 − 30 × 10

−9

2π0 ln √

8 − 8 × 10 −9

20 (1)+ C

At point B, rB = √ (4 − (−1))2 + (0 − (−1))2 + (1 − 2)2 = √27,

ρB = √ (4 − 1)2 + (0 − 2)2 = √13, and the distance from the sheet charge is xB = 4 − 2 = 2.

The potential at A is then

VB = 10 −6

4π0 √

27 − 30 × 10

−9

2π0 ln √

13 − 8 × 10 −9

20 (2)+ C

Then

VA VB = 10 −6

4π0

[ 1√ 45

− 1√ 27

] − 30 × 10

−9

2π0 ln

(√ 8

13

) − 8 × 10

−9

20 (1 − 2) = 193 V

4.11. Let a uniform surface charge density of 5 nC/m2 be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4, and a point charge of 2µC be present at P(2, 0, 0). If V = 0 at M(0, 0, 5), find V at N(1, 2, 3): We need to find a potential function for the combined charges which is zero at M . That for the point charge we know to be

Vp(r) = Q 4π0r

Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have

Vl(ρ) = − ∫

ρl

2π0ρ dρ + C1 = − ρl

2π0 ln(ρ)+ C1

For the sheet charge, we have

Vs(z) = − ∫

ρs

20 dz+ C2 = − ρs

20 z+ C2

The total potential function will be the sum of the three. Combining the integration constants, we obtain:

V = Q 4π0r

ρl 2π0

ln(ρ)ρs 20

z+ C

46

4.11. (continued) The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r , ρ, and z are scalar distances from the charges, and will be treated as such here. To evaluate the constant, C, we first look at point M , where VT = 0. At M , r = √22 + 52 = √29, ρ = 1, and z = 5. We thus have

0 = 2 × 10 −6

4π0 √

29 − 8 × 10

−9

2π0 ln(1)− 5 × 10

−9

20 5 + C C = −1.93 × 103 V

At point N , r = √1 + 4 + 9 = √14, ρ = √2, and z = 3. The potential at N is thus

VN = 2 × 10 −6

4π0 √

14 − 8 × 10

−9

2π0 ln(

√ 2)− 5 × 10

−9

20 (3)− 1.93 × 103 = 1.98 × 103 V = 1.98 kV

4.12. Three point charges, 0.4µC each, are located at (0, 0,−1), (0, 0, 0), and (0, 0, 1), in free space. a) Find an expression for the absolute potential as a function of z along the line x = 0, y = 1:

From a point located at position z along the given line, the distances to the three charges are R1 =

(z− 1)2 + 1, R2 =

z2 + 1, and R3 =

(z+ 1)2 + 1. The total potential will be

V (z) = q 4π0

[ 1

R1 + 1

R2 + 1

R3

]

Using q = 4 × 10−7 C, this becomes

V (z) = (3.6 × 103) [

1√ (z− 1)2 + 1

+ 1√ z2 + 1 +

1√ (z+ 1)2 + 1

] V

b) Sketch V (z). The sketch will show that V maximizes to a value of 8.68 × 103 at z = 0, and then monotonically decreases with increasing |z| symmetrically on either side of z = 0.

4.13. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.5 mm on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them? This will be the magnitude of the charge times the potential difference between the finishing and starting positions, or

W = (4 × 10 −12)2

2π0

[ 1

2.5 − 1

5

] × 104 = 5.76 × 10−10 J = 576 pJ

4.14. two 6-nC point charges are located at (1, 0, 0) and (−1, 0, 0) in free space. a) Find V at P(0, 0, z): Since the charges are positioned symmetrically about the z axis, the potential

at z will be double that from one charge. This becomes:

V (z) = (2) q 4π0

z2 + 1 =

q

2π0 √ z2 + 1

b) Find Vmax : It is clear from the part a result that V will maximize at z = 0, or vmax = q/(2π0) = 108 V.

47

4.14. (continued) c) Calculate |dV/dz| on the z axis: Differentiating the part a result, we find

∣∣∣dV dz

∣∣∣ = qz π0(z2 + 1)3/2 V/m

d) Find |dV/dz|max : To find this we need to differentiate the part c result and find its zero:

d

dz

∣∣∣dV dz

∣∣∣ = q(1 − 2z2) π0(z2 + 1)5/2 = 0 ⇒ z = ±

1√ 2

Substituting z = 1/√2 into the part c result, we find ∣∣∣dV dz

∣∣∣ max

= q√ 2π0(3/2)3/2

= 83.1 V/m

4.15. Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = −1, y = 2 in free space. If the potential at the origin is 100 V, find V at P(4, 1, 3): The net potential function for the two charges would in general be:

V = − ρl 2π0

ln(R1)ρl 2π0

ln(R2)+ C

At the origin, R1 = R2 = √

5, and V = 100 V. Thus, with ρl = 8 × 10−9,

100 = −2 (8 × 10 −9)

2π0 ln(

√ 5)+ C C = 331.6 V

At P(4, 1, 3), R1 = |(4, 1, 3)(1, 1, 2)| = √

10 and R2 = |(4, 1, 3)(−1, 2, 3)| = √

26. Therefore

VP = − (8 × 10 −9)

2π0

[ ln(

√ 10)+ ln(

√ 26) ] + 331.6 = −68.4 V

48

4.16. Uniform surface charge densities of 6, 4, and 2 nC/m2 are present at r = 2, 4, and 6 cm, respectively, in free space.

a) Assume V = 0 at infinity, and find V (r). We keep in mind the definition of absolute potential as the work done in moving a unit positive charge from infinity to location r . At radii outside all three spheres, the potential will be the same as that of a point charge at the origin, whose charge is the sum of the three sphere charges:

V (r) (r > 6 cm) = q1 + q2 + q3 4π0r

= [4π(.02) 2(6)+ 4π(.04)2(4)+ 4π(.06)2(2)] × 10−9

4π0r

= (96 + 256 + 288× 10 −13

4π(8.85 × 10−12)r = 1.81

r V where r is in meters

As the unit charge is moved inside the outer sphere to positions 4 < r < 6 cm, the outer sphere contribution to the energy is fixed at its value at r = 6. Therefore,

V (r) (4 < r < 6 cm) = q1 + q2 4π0r

+ q3 4π0(.06)

= 0.994 r

+ 13.6 V

In moving inside the sphere at r = 4 cm, the contribution from that sphere becomes fixed at its potential function at r = 4:

V (r) (2 < r < 4 cm) = q1 4π0r

+ q2 4π0(.04)

+ q3 4π0(.06)

= 0.271 r

+ 31.7 V

Finally, using the same reasoning, the potential inside the inner sphere becomes

V (r) (r < 2 cm) = 0.271 .02

+ 31.7 = 45.3 V

b) Calculate V at r = 1, 3, 5, and 7 cm: Using the results of part a, we substitute these distances (in meters) into the appropriate formulas to obtain: V (1) = 45.3 V, V (3) = 40.7 V, V (5) = 33.5 V, and V (7) = 25.9 V.

c) Sketch V versus r for 0 < r < 10 cm.

49

4.17. Uniform surface charge densities of 6 and 2 nC/m2 are present at ρ = 2 and 6 cm respectively, in free space. Assume V = 0 at ρ = 4 cm, and calculate V at:

a) ρ = 5 cm: Since V = 0 at 4 cm, the potential at 5 cm will be the potential difference between points 5 and 4:

V5 = − ∫ 5

4 E · dL = −

∫ 5 4

aρsa

0ρ dρ = − (.02)(6 × 10

−9) 0

ln

( 5

4

) = −3.026 V

b) ρ = 7 cm: Here we integrate piecewise from ρ = 4 to ρ = 7:

V7 = − ∫ 6

4

aρsa

0ρ dρ

∫ 7 6

(aρsa + bρsb) 0ρ

With the given values, this becomes

V7 = − [ (.02)(6 × 10−9)

0

] ln

( 6

4

) − [ (.02)(6 × 10−9)+ (.06)(2 × 10−9)

0

] ln

( 7

6

) = −9.678 V

4.18. A nonuniform linear charge density, ρL = 8/(z2 + 1) nC/m lies along the z axis. Find the potential at P(ρ = 1, 0, 0) in free space if V = 0 at infinity: This last condition enables us to write the potential at P as a superposition of point charge potentials. The result is the integral:

VP = ∫ ∞ −∞

ρLdz

4π0R

where R = √z2 + 1 is the distance from a point z on the z axis to P . Substituting the given charge distribution and R into the integral gives us

VP = ∫ ∞ −∞

8 × 10−9dz 4π0(z2 + 1)3/2 =

2 × 10−9 π0

zz2 + 1

∣∣∣∞−∞ = 144 V

4.19. The annular surface, 1 cm < ρ < 3 cm, z = 0, carries the nonuniform surface charge density ρs = 5ρ nC/m2. Find V at P(0, 0, 2 cm) if V = 0 at infinity: We use the superposition integral form:

VP = ∫ ∫

ρs da

4π0|r r′| where r = zaz and r′ = ρaρ . We integrate over the surface of the annular region, with da = ρ dρ dφ. Substituting the given values, we find

VP = ∫ 2π

0

.03 .01

(5 × 10−92 dρ dφ 4π0

ρ2 + z2

Substituting z = .02, and using tables, the integral evaluates as

VP = [ (5 × 10−9)

20

] [ ρ

2

ρ2 + (.02)2 − (.02)

2

2 ln+

ρ2 + (.02)2)

].03 .01

= .081 V

50

4.20. Fig. 4.11 shows three separate charge distributions in the z = 0 plane in free space. a) find the total charge for each distribution: Line charge along the y axis:

Q1 = ∫ 5

3 π × 10−9dy = 2π × 10−9 C = 6.28 nC

Line charge in an arc at radius ρ = 3:

Q2 = ∫ 70◦

10◦ (10−9) 3 = 4.5 × 10−9 (70 − 10) 2π

360 = 4.71 × 10−9 C = 4.71 nC

Sheet charge:

Q3 = ∫ 70◦

10◦

∫ 3.5 1.6

(10−9) ρ dρ dφ = 5.07 × 10−9 C = 5.07 nC

b) Find the potential at P(0, 0, 6) caused by each of the three charge distributions acting alone: Line charge along y axis:

VP1 = ∫ 5

3

ρLdL

4π0R = ∫ 5

3

π × 10−9dy 4π0

y2 + 62

= 10 3

4 × 8.854 ln(y + √ y2 + 62)

∣∣∣5 3 = 7.83 V

Line charge in an arc a radius ρ = 3:

VP2 = ∫ 70◦

10◦

(1.5 × 10−9) 3 4π0

√ 32 + 62 =

Q2

4π0 √

45 = 6.31 V

Sheet charge:

VP3 = ∫ 70◦

10◦

∫ 3.5 1.6

(10−9) ρ dρ dφ 4π0

ρ2 + 62

= 60 × 10 −9

4π(8.854 × 10−12 (

2π

360

)∫ 3.5 1.6

ρ dρρ2 + 36

= 9.42 √ ρ2 + 36

∣∣∣3.5 1.6

= 6.93 V

c) Find VP : This will be the sum of the three results of part b, or

VP = VP1 + VP 2 + VP 3 = 7.83 + 6.31 + 6.93 = 21.1 V

4.21. Let V = 2xy2z3 + 3 ln(x2 + 2y2 + 3z2)V in free space. Evaluate each of the following quantities at P(3, 2,−1):

a) V : Substitute P directly to obtain: V = −15.0 V b) |V |. This will be just 15.0 V. c) E: We have

E ∣∣∣ P = −∇V

∣∣∣ P = −

[( 2y2z3 + 6x

x2 + 2y2 + 3z2 )

ax + (

4xyz3 + 12y x2 + 2y2 + 3z2

) ay

+ (

6xy2z2 + 18z x2 + 2y2 + 3z2

) az

] P

= 7.1ax + 22.8ay − 71.1az V/m

51

4.21d. |E|P : taking the magnitude of the part c result, we find |E|P = 75.0 V/m. e) aN : By definition, this will be

aN ∣∣∣ P = − E|E| = −0.095 ax − 0.304 ay + 0.948 az

f) D: This is D ∣∣∣ P = 0E

∣∣∣ P = 62.8 ax + 202 ay − 629 az pC/m2.

4.22. It is known that the potential is given as V = 80r0.6 V. Assuming free space conditions, find: a) E: We use

E = −∇V = −dV dr

ar = −(0.6)80r−0.4 ar = −48r−0.4 ar V/m

b) the volume charge density at r = 0.5 m: Begin by finding

D = 0E = −48r−0.40 ar C/m2

We next find

ρv = ∇ · D = 1 r2

d

dr

( r2Dr

) = 1

r2

d

dr

( −480r1.6

) = −76.80

r1.4 C/m3

Then at r = 0.5 m,

ρv(0.5) = −76.8(8.854 × 10 −12)

(0.5)1.4 = −1.79 × 10−9 C/m3 = −1.79 nC/m3

c) the total charge lying within the surface r = 0.6: The easiest way is to use Gauss’ law, and integrate the flux density over the spherical surface r = 0.6. Since the field is constant at constant radius, we obtain the product:

Q = 4π(0.6)2(−480(0.6)−0.4) = −2.36 × 10−9 C = −2.36 nC

4.23. It is known that the potential is given as V = 80ρ.6 V. Assuming free space conditions, find: a) E: We find this through

E = −∇V = −dV dρ

aρ = −48ρ.4 V/m

b) the volume charge density at ρ = .5 m: Using D = 0E, we find the charge density through

ρv

∣∣∣ .5 = [∇ · D].5 =

( 1

ρ

) d

( ρDρ

) ∣∣∣ .5 = −28.80ρ−1.4

∣∣∣ .5 = −673 pC/m3

52

4.23c. the total charge lying within the closed surface ρ = .6, 0 < z < 1: The easiest way to do this calculation is to evaluate at ρ = .6 (noting that it is constant), and then multiply by the cylinder area: Using part a, we have

∣∣∣ .6 = −480(.6).4 = −521 pC/m2. ThusQ = −2π(.6)(1)521×10−12 C = −1.96 nC.

4.24. Given the potential field V = 80r2 cos θ and a point P(2.5, θ = 30◦, φ = 60◦) in free space, find at P : a) V : Substitute the coordinates into the function and find VP = 80(2.5)2 cos(30) = 433 V. b) E:

E = −∇V = −∂V ∂r

ar − 1 r

∂V

∂θ aθ = −160r cos θar + 80r sin θaθ V/m

Evaluating this at P yields Ep = −346ar + 100aθ V/m. c) D: In free space, DP = 0EP = (−346ar + 100aθ )0 = −3.07 ar + 0.885 aθ nC/m2. d) ρv:

ρv = ∇ · D = 0∇ · E = 0 [

1

r2

∂r

( r2Er

) + 1

r2 sin θ

∂θ (Eθ sin θ)

]

Substituting the components of E, we find

ρv = [ −160 cos θ

r2 3r2 + 1

r sin θ 80r(2 sin θ cos θ)

] 0 = −3200 cos θ = −2.45 nC/m3

with θ = 30◦. e) dV/dN : This will be just |E| evaluated at P , which is

dV

dN

∣∣∣ P = | − 346ar + 100aθ | =

(346)2 + (100)2 = 360 V/m

f) aN : This will be

aN = − EP|EP | = − [

−346ar + 100aθ(346)2 + (100)2

] = 0.961 ar − 0.278 aθ

4.25. Within the cylinder ρ = 2, 0 < z < 1, the potential is given by V = 100 + 50ρ + 150ρ sin φ V. a) Find V , E, D, and ρv at P(1, 60◦, 0.5) in free space: First, substituting the given point, we find

VP = 279.9 V. Then,

E = −∇V = −∂V ∂ρ

aρ − 1 ρ

∂V

∂φ aφ = − [50 + 150 sin φ] aρ − [150 cosφ] aφ

Evaluate the above at P to find EP = −179.9aρ − 75.0aφ V/m Now D = 0E, so DP = −1.59aρ .664aφ nC/m2. Then

ρv = ∇ ·D = (

1

ρ

) d

( ρDρ

)+ 1 ρ

∂Dφ

∂φ = [ − 1 ρ (50 + 150 sin φ)+ 1

ρ 150 sin φ

] 0 = −50

ρ 0 C

At P , this is ρvP = −443 pC/m3.

53

4.25b. How much charge lies within the cylinder? We will integrate ρv over the volume to obtain:

Q = ∫ 1

0

∫ 2π 0

∫ 2 0

−500 ρ

ρ dρ dφ dz = −2π(50)0(2) = −5.56 nC

4.26. A dipole having Qd/(4π0) = 100 V · m2 is located at the origin in free space and aligned so that its moment is in the az direction. a) Sketch |V (r = 1, θ, φ = 0)| versus θ on polar graph paper (homemade if you wish). b) Sketch |E(r = 1, θ, φ = 0)| versus θ on polar graph paper:

V = Qd cos θ 4π0r2

= 100 cos θ r2

⇒ |V (r = 1, θ, φ = 0)| = |100 cos θ |

E = Qd 4π0r3

(2 cos θ ar + sin θ aθ ) = 100 r3

(2 cos θ ar + sin θ aθ )

|E(r = 1, θ, φ = 0)| = 100 (

4 cos2 θ + sin2 θ )1/2 = 100 (1 + 3 cos2 θ)1/2

These results are plotted below:

54

4.27. Two point charges, 1 nC at (0, 0, 0.1) and −1 nC at (0, 0,−0.1), are in free space. a) Calculate V at P(0.3, 0, 0.4): Use

VP = q 4π0|R+| −

q

4π0|R−|

where R+ = (.3, 0, .3) and R− = (.3, 0, .5), so that |R+| = 0.424 and |R−| = 0.583. Thus

VP = 10 −9

4π0

[ 1

.424 − 1

.583

] = 5.78 V

b) Calculate |E| at P : Use

EP = q(.3ax + .3az) 4π0(.424)3

q(.3ax + .5az) 4π0(.583)3

= 10 −9

4π0

[ 2.42ax + 1.41az

] V/m

Taking the magnitude of the above, we find |EP | = 25.2 V/m. c) Now treat the two charges as a dipole at the origin and find V at P : In spherical coordinates, P

is located at r = √.32 + .42 = .5 and θ = sin−1(.3/.5) = 36.9◦. Assuming a dipole in far-field, we have

VP = qd cos θ 4π0r2

= 10 −9(.2) cos(36.9◦)

4π0(.5)2 = 5.76 V

4.28. A dipole located at the origin in free space has a moment p2 × 10−9 az C · m. At what points on the line y = z, x = 0 is:

a) || = 1 mV/m? We note that the line y = z lies at θ = 45◦. Begin with

E = 2 × 10 −9

4π0r3 (2 cos θ ar + sin θ aθ ) = 10

−9

2 √

2π0r3 (2ar + aθ ) at θ = 45◦

from which

= 10 −9

2π0r3 = 10−3 V/m (required) r3 = 1.27 × 10−4 or r = 23.3 m

The y and z values are thus y = z = ±23.3/√2 = ±16.5 m b) |Er | = 1 mV/m? From the above field expression, the radial component magnitude is twice that

of the theta component. Using the same development, we then find

Er = 2 10 −9

2π0r3 = 10−3 V/m (required) r3 = 2(1.27 × 10−4) or r = 29.4 m

The y and z values are thus y = z = ±29.4/√2 = ±20.8 m

55

4.29. A dipole having a moment p = 3ax − 5ay + 10az nC · m is located at Q(1, 2,−4) in free space. Find V at P(2, 3, 4): We use the general expression for the potential in the far field:

V = p · (r r )

4π0|r r′|3

where r r′ = P Q = (1, 1, 8). So

VP = (3ax − 5ay + 10az) · (ax + ay + 8az)× 10 −9

4π0[12 + 12 + 82]1.5 = 1.31 V

4.30. A dipole, having a moment p = 2az nC · m is located at the origin in free space. Give the magnitude of E and its direction aE in cartesian components at r = 100 m, φ = 90◦, and θ =: a) 0◦; b) 30◦; c) 90◦. Begin with

E = p 4π0r3

[2 cos θ ar + sin θ aθ ]

from which

|E| = p 4π0r3

[ 4 cos2 θ + sin2 θ

]1/2 = p 4π0r3

[ 1 + 3 cos2 θ

]1/2 Now

Ex = E · ax = p 4π0r3

[2 cos θ ar · ax + sin θ aθ · ax] = p 4π0r3

[3 cos θ sin θ cosφ]

then

Ey = E · ay = p 4π0r3

[ 2 cos θ ar · ay + sin θ aθ · ay

] = p 4π0r3

[3 cos θ sin θ sin φ]

and

Ez = E · az = p 4π0r3

[ 2 cos θ ar · az + sin θ aθ · az

] = p 4π0r3

[ 2 cos2 θ − sin2 θ

] Since φ is given as 90◦, Ex = 0, and the field magnitude becomes

|E= 90◦)| = √ E2y + E2z =

p

4π0r3

[ 9 cos2 θ sin2 θ + (2 cos2 θ − sin2 θ)2

]1/2 Then the unit vector becomes (again at φ = 90◦):

aE = 3 cos θ sin θ ay + (2 cos 2 θ − sin2 θ) az[

9 cos2 θ sin2 θ + (2 cos2 θ − sin2 θ)2]1/2 Now with r = 100 m and p = 2 × 10−9,

p

4π0r3 = 2 × 10

−9

4π(8.854 × 10−12)106 = 1.80 × 10 −5

Using the above formulas, we find at θ = 0◦, |E| = (1.80 × 10−5)(2) = 36.0µV/m and aE = az. At θ = 30◦, we find |E| = (1.80 × 10−5)[1.69 + 1.56]1/2 = 32.5µV/m and aE = (1.30ay + 1.25az)/1.80 = 0.72 ax + 0.69 az. At θ = 90◦, |E| = (1.80×10−5)(1) = 18.0 µV/m and aE = −az.

56

4.31. A potential field in free space is expressed as V = 20/(xyz)V. a) Find the total energy stored within the cube 1 < x, y, z < 2. We integrate the energy density over

the cube volume, where wE = (1/2)0E · E, and where

E = −∇V = 20 [

1

x2yz ax + 1

xy2z ay + 1

xyz2 az

] V/m

The energy is now

WE = 2000 ∫ 2

1

∫ 2 1

∫ 2 1

[ 1

x4y2z2 + 1

x2y4z2 + 1

x2y2z4

] dx dy dz

The integral evaluates as follows:

WE = 2000 ∫ 2

1

∫ 2 1

[ − (

1

3

) 1

x3y2z2 − 1

xy4z2 − 1

xy2z4

]2 1 dy dz

= 2000 ∫ 2

1

∫ 2 1

[( 7

24

) 1

y2z2 + (

1

2

) 1

y4z2 + (

1

2

) 1

y2z4

] dy dz

= 2000 ∫ 2

1

[ − (

7

24

) 1

yz2 − (

1

6

) 1

y3z2 − (

1

2

) 1

yz4

]2 1 dz

= 2000 ∫ 2

1

[( 7

48

) 1

z2 + (

7

48

) 1

z2 + (

1

4

) 1

z4

] dz

= 2000(3) [

7

96

] = 387 pJ

b) What value would be obtained by assuming a uniform energy density equal to the value at the center of the cube? At C(1.5, 1.5, 1.5) the energy density is

wE = 2000(3) [

1

(1.5)4(1.5)2(1.5)2

] = 2.07 × 10−10 J/m3

This, multiplied by a cube volume of 1, produces an energy value of 207 pJ.

4.32. In the region of free space where 2 < r < 3, 0.4π < θ < 0.6π , 0 < φ < π/2, let E = k/r2 ar . a) Find a positive value for k so that the total energy stored is exactly 1 J: The energy is found through

WE = ∫ v

1

2 0E

2 dv = ∫ π/2

0

∫ 0.6π 0.4π

∫ 3 2

1

2 0 k2

r2 r2 sin θ dr dθ dφ

= π 2 (− cos θ)

∣∣∣.6π .4π

( 1

2

) 0k

2 ( −1 r

) ∣∣∣3 2 = 0.616π

24 0k

2 = 1 J

Solve for k to find k = 1.18 × 106 V · m.

57

4.32b. Show that the surface θ = 0.6π is an equipotential surface: This will be the surface of a cone, centered at the origin, along which E, in the ar direction, will exist. Therefore, the given surface cannot be an equipotential (the problem was ill-conceived). Only a surface of constant r could be an equipotential in this field.

c) Find VAB , given points A(2, θ = π/2, φ = π/3) and B(3, π/2, π/4): Use

VAB = − ∫ A B

E · dL = − ∫ 3

2

k

r2 ar · ar dr = k

( 1

2 − 1

3

) = k

6

Using the result of part a, we find VAB = (1.18 × 106)/6 = 197 kV.

4.33. A copper sphere of radius 4 cm carries a uniformly-distributed total charge of 5µC in free space. a) Use Gauss’ law to find D external to the sphere: with a spherical Gaussian surface at radius r , D

will be the total charge divided by the area of this sphere, and will be ar -directed. Thus

D = Q 4πr2

ar = 5 × 10 −6

4πr2 ar C/m2

b) Calculate the total energy stored in the electrostatic field: Use

WE = ∫ vol

1

2 D · E dv =

∫ 2π 0

π 0

∫ ∞ .04

1

2

(5 × 10−6)2 16π20r4

r2 sin θ dr dθ dφ

= (4π) (

1

2

) (5 × 10−6)2

16π20

∫ ∞ .04

dr

r2 = 25 × 10

−12

8π0

1

.04 = 2.81 J

c) Use WE = Q2/(2C) to calculate the capacitance of the isolated sphere: We have

C = Q 2

2WE = (5 × 10

−6)2

2(2.81) = 4.45 × 10−12 F = 4.45 pF

4.34. Given the potential field in free space, V = 80φ V (note that aphi should not be present), find: a) the energy stored in the region 2 < ρ < 4 cm, 0 < φ < 0.2π , 0 < z < 1 m: First we find

E = −∇V = − 1 ρ