Your Name (required) Your Email (required) Subject. Your Country. Nov 12, 2014 - Engineering electromagnetics 7th edition - william h. Hayt - solution manual. CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and. Internet Archive BookReader Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual.

• Engineering Electromagnetics Solutions Manual

CHAPTER 1 1.1. Given the vectors M = −10 a x + 4 a y − 8 a z and N = 8 a x + 7 a y − 2 a z, find: a) a unit vector in the direction of − M + 2 N. − M + 2 N = 10 a x − 4 a y + 8 a z + 16 a x + 14 a y − 4 a z = (26, 10, 4 ) Thus a = (26, 10, 4 ) (26, 10, 4 ) = (0.92, 0.36, 0.14 ) b) the magnitude of 5 a x + N − 3 M: (5, 0, 0 )+ (8, 7,−2 )− (−30, 12,−24 ) = (43,−5, 22 ), and (43,−5, 22 ) = 48.6. C) M 2 N ( 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 ) = 4 a x + 3 a y + 2 a z 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 = 13 B(2,−2, 1 ), we use the fact that B − A = 10, or (6 − 23 B) a x − (2 − 23 B) a y − (4 + 13 B) a z = 10 Expanding, obtain 36 − 8 B + 49 B2 + 4 − 83 B + 49 B2 + 16 + 83 B + 19 B2 = 100 or B2 − 8 B − 44 = 0. Thus B = 8± √ 64−176 2 = 11.75 (taking positive option) and so B = 2 3 (11.75 ) a x − 2 3 (11.75 ) a y + 1 3 (11.75 ) a z = 7.83 a x − 7.83 a y + 3.92 a z 1.

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 a AB = 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. A0 M = ( 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. A vector field is specified as G = 24 xy a x + 12 (x2 + 2 ) a y + 18 z2 a z.

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 a G = (−48, 72, 162 ) (−48, 72, 162 ) = (−0.26, 0.39, 0.88 ) c) a unit vector directed from Q toward P: a QP = 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 = (24 xy, 12 (x2 + 2 ), 18 z2 ) , or 10 = (4 xy, 2 x2 + 4, 3 z2 ) , so the equation is 100 = 16 x2 y2 + 4 x4 + 16 x2 + 16 + 9 z4 2. 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 ) = (24 x, 12 x2 + 24, 18 ), from which Gx = 24 x, Gy = 12 x2 + 24, Gz = 18, and G = 6 √ 4 x4 + 32 x2 + 25.

Plots are shown below. Given the vector field E = 4 zy2 cos 2 x a x + 2 zy sin 2 x a y + y2 sin 2 x a z for the region x y , and z less than 2, find: a) the surfaces on which Ey = 0. With Ey = 2 zy sin 2 x = 0, the surfaces are 1) the plane z = 0, with x. A field is given as G = 25 (x2 + y2 ) (x a x + y a y) Find: a) a unit vector in the direction of G at P(3, 4,−2 ): Have G p = 25 /(9 + 16 )× (3, 4, 0 ) = 3 a x + 4 a y, and G p = 5. Thus a G = (0.6, 0.8, 0 ). B) the angle between G and a x at P: The angle is found through a G a x = 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 a ydzdx ∫ 4 0 ∫ 2 0 25 x2 + y2 (x a x + y a y) a ydzdx = ∫ 4 0 ∫ 2 0 25 x2 + 49 × 7 dzdx = ∫ 4 0 350 x2 + 49 dx = 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 R AB = (−3, 1, 7 ) and R AC = (−1,−5, 3 ) to form R AB R AC = R AB R AC cos θA. Obtain 3 + 5 + 21 = √59√35 cos θA.

Solve to find θA = 65.3◦. B) Use R BA = (3,−1,−7 ) and R BC = (2,−6,−4 ) to form R BA R BC = R BA R BC cos θB. Obtain 6 + 6 + 28 = √59√56 cos θB. Solve to find θB = 45.9◦. 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 R MN: R MN = (−0.2, 0.1, 0.3 )− (0.1,−0.2,−0.1 ) = (−0.3, 0.3, 0.4 ). B) the dot product R MN R MP: R MP = (0.4, 0, 0.1 ) − (0.1,−0.2,−0.1 ) = (0.3, 0.2, 0.2 ).

R MN R MP = (−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 R MN on R MP: R MN a RMP = (−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 R MN and R MP: θM = cos−1 ( R MN R MP R MN R MP ) = cos−1 ( 0.05√ 0.34 √ 0.17 ) = 78◦ 4. 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 R AB + R BC on R AD: R AB + R BC = R AC = (8, 1, 4 ) − (10, 12,−6 ) = (−2,−11, 10 ) Then R AD = (−2,−5, 8 ) − (10, 12,−6 ) = (−12,−17, 14 ).

So the projection will be: ( R AC a RAD) a RAD = (−2,−11, 10 ) (−12,−17, 14 )√ 629 (−12,−17, 14 )√ 629 = (−6.7,−9.5, 7.8 ) b) the vector projection of R AB + R BC on R DC: R DC = (8,−1, 4 )− (−2,−5, 8 ) = (10, 6,−4 ). The projection is: ( R AC a RDC) a RDC = (−2,−11, 10 ) (10, 6,−4 )√ 152 (10, 6,−4 )√ 152 = (−8.3,−5.0, 3.3 ) c) the angle between R DA and R DC: Use R DA = − R AD = (12, 17,−14 ) and R DC = (10, 6,−4 ). The angle is found through the dot product of the associated unit vectors, or: θD = cos−1 ( a RDA a RDC) = 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: F pG = 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: G pF = G− G F = G− G 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 R BA × R BC = (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: R BA × R BC R BA × R BC = (0.47, 0.82, 0.33 ) b) Find the area of the face ABC: Area = 1 2 R BA × R BC = 0.43 5. Three vectors extending from the origin are given as r1 = (7, 3,−2 ), r2 = (−2, 7,−3 ), and r3 = (0, 2, 3 ).

Engineering Electromagnetics Solutions Manual

Find: a) a unit vector perpendicular to both r1 and r2: a p12 = 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 a p = (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 a x = 2, where r = (x, y, z): This will be the plane x = 2. B) r × a x = 2: r × a x = (0, z,− y), and r × a x = √ z2 + y2 = 2. This is the equation of a cylinder, centered on the x axis, and of radius 2.

Point A(−4, 2, 5 ) and the two vectors, R AM = (20, 18,−10 ) and R AN = (−10, 8, 15 ), define a triangle. A) Find a unit vector perpendicular to the triangle: Use a p = R AM × R AN R AM × R AN = (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 R AN: a AN = (−10, 8, 15 )√ 389 = (−0.507, 0.406, 0.761 ) Then a pAN = a p × a AN = (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 )( a AM + a AN), where a AM = (20, 18,−10 ) (20, 18,−10 ) = (0.697, 0.627,−0.348 ) 6. (continued) Now 1 2 ( a AM + a AN) = 1 2 (0.697, 0.627,−0.348 )+ (−0.507, 0.406, 0.761 ) = (0.095, 0.516, 0.207 ) Finally, a bis = (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 R AB = (1.73,−1, 1 ) − (1.71, 4.70,−3 ) = (0.02,−5.70, 4 ) and therefore a AB = (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: a AB a ρ = 0.003 cos 70◦ − 0.82 sin 70◦ = −0.77. A AB a φ = −0.003 sin 70◦ − 0.82 cos 70◦ = −0.28. Thus a AB = −0.77 a ρ − 0.28 a φ + 0.57 a z. C) a unit vector in cylindrical coordinates at B directed toward A: Use a BA = (−0, 003, 0.82,−0.57 ). Then a BA a ρ = −0.003 cos (−30◦ )+0.82 sin (−30◦ ) = −0.43, and a BA a φ = 0.003 sin (−30◦ )+ 0.82 cos (−30◦ ) = 0.71. Finally, a BA = −0.43 a ρ + 0.71 a φ − 0.57 a z 1.19 a) Express the field D = (x2 + y2 )−1 (x a x + y a y) in cylindrical components and cylindrical variables: Have x = ρ cos φ, y = ρ sin φ, and x2 + y2 = ρ2. Therefore D = 1 ρ (cos φ a x + sin φ a y) Then Dρ = D a ρ = 1 ρ cos φ( a x a ρ)+ sin φ( a y a ρ) = 1 ρ cos2 φ + sin2 φ = 1 ρ and Dφ = D a φ = 1 ρ cos φ( a x a φ)+ sin φ( a y a φ) = 1 ρ cos φ(− sin φ)+ sin φ cos φ = 0 Therefore D = 1 ρ a ρ 7.

Evaluate D at the point where ρ = 2, φ = 0.2 π, and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5 a ρ. To express this in cartesian, we use D = 0.5 ( a ρ a x) a x + 0.5 ( a ρ a y) a y = 0.5 cos 36◦ a x + 0.5 sin 36◦ a y = 0.41 a x + 0.29 a y 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 R AB = (−4.77,−7.30, 4 ). B) a unit vector at B directed toward A: Have R BA = (4.77, 7.30,−4 ), and so a BA = (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 r B = (−1.71,−4.70, 2 ), and so − r B = (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 R CD = (−4,−6, 9 ) and Rρ = R CD a ρ = −4 cos (33.7 ) − 6 sin (33.7 ) = −6.66. Then Rφ = R CD a φ = 4 sin (33.7 )− 6 cos (33.7 ) = −2.77. So R CD = −6.66 a ρ − 2.77 a φ + 9 a z b) a unit vector at D directed toward C: R CD = (4, 6,−9 ) and Rρ = R DC a ρ = 4 cos (−104.0 ) + 6 sin (−104.0 ) = −6.79. Then Rφ = R DC a φ = 4− sin (−104.0 ) + 6 cos (−104.0 ) = 2.43.

So R DC = −6.79 a ρ + 2.43 a φ − 9 a z Thus a DC = −0.59 a ρ + 0.21 a φ − 0.78 a z c) a unit vector at D directed toward the origin: Start with r D = (−1,−4, 2 ), and so the vector toward the origin will be − r D = (1, 4,−2 ). Thus in cartesian the unit vector is a = (0.22, 0.87,−0.44 ). Convert to cylindrical: aρ = (0.22, 0.87,−0.44 ) a ρ = 0.22 cos (−104.0 )+ 0.87 sin (−104.0 ) = −0.90, and aφ = (0.22, 0.87,−0.44 ) a φ = 0.22− sin (−104.0 ) + 0.87 cos (−104.0 ) = 0, so that finally, a = −0.90 a ρ − 0.44 a z. A field is given in cylindrical coordinates as F = 40 ρ2 + 1 + 3 (cos φ + sin φ) a ρ + 3 (cos φ − sin φ) a φ − 2 a z where the magnitude of F is found to be: F = √ F F = 1600 (ρ2 + 1 )2 + 240 ρ2 + 1 (cos φ + sin φ)+ 22 1 /2 8. 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◦ × 2 π × 3 + 30 ◦ 360◦ × 2 π × 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. R PQ = Q − P = 5 a x − 4 a y − 6 a z Then R PQ = √ 25 + 16 + 36 = 8.8 b) cylindrical coordinates. At P, ρ = 5, φ = tan−1 (4 /− 3 ) = −53.1◦, and z = 5.

Now, R PQ a ρ = (5 a x − 4 a y − 6 a z) a ρ = 5 cos φ − 4 sin φ = 6.20 R PQ a φ = (5 a x − 4 a y − 6 a z) a φ = −5 sin φ − 4 cos φ = 1.60 Thus R PQ = 6.20 a ρ + 1.60 a φ − 6 a z and R PQ = √ 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◦. R PQ a r = (5 a x − 4 a y − 6 a z) a r = 5 sin θ cos φ − 4 sin θ sin φ − 6 cos θ = 0.14 R PQ a θ = (5 a x − 4 a y − 6 a z) a θ = 5 cos θ cos φ − 4 cos θ sin φ − (−6 ) sin θ = 8.62 R PQ a φ = (5 a x − 4 a y − 6 a z) a φ = −5 sin φ − 4 cos φ = 1.60 10. (continued) Thus R PQ = 0.14 a r + 8.62 a θ + 1.60 a φ and R PQ = √ 0.142 + 8.622 + 1.602 = 8.8 d) Show that each of these vectors has the same magnitude. Each does, as shown above. Given point P(r = 0.8, θ = 30◦, φ = 45◦ ), and E = 1 r2 ( cos φ a r + sin φ sin θ a φ ) a) Find E at P: E = 1.10 a ρ + 2.21 a φ. B) Find E at P: E = √1.102 + 2.212 = 2.47.

C) Find a unit vector in the direction of E at P: a E = E E = 0.45 a r + 0.89 a φ 1.26. A) Determine an expression for a y in spherical coordinates at P(r = 4, θ = 0.2 π, φ = 0.8 π): Use a y a r = sin θ sin φ = 0.35, a y a θ = cos θ sin φ = 0.48, and a y a φ = cos φ = −0.81 to obtain a y = 0.35 a r + 0.48 a θ − 0.81 a φ b) Express a r in cartesian components at P: Find x = r sin θ cos φ = −1.90, y = r sin θ sin φ = 1.38, and z = r cos θ = −3.24. Then use a r a x = sin θ cos φ = −0.48, a r a y = sin θ sin φ = 0.35, and a r a z = cos θ = 0.81 to obtain a r = −0.48 a x + 0.35 a y + 0.81 a z 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. (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 R AB = B − A = 5.62 ax + 4.90 a y + 7.77 a z b) Find the spherical components of the vector at P(2,−3, 4 ) extending to Q(−3, 2, 5 ): First, R PQ = 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 R PQ a r = −5 sin (42◦ ) cos (−56.3◦ )+ 5 sin (42◦ ) sin (−56.3◦ )+ 1 cos (42◦ ) = −3.90 R PQ a θ = −5 cos (42◦ ) cos (−56.3◦ )+ 5 cos (42◦ ) sin (−56.3◦ )− 1 sin (42◦ ) = −5.82 R PQ a φ = − (−5 ) sin (−56.3◦ )+ 5 cos (−56.3◦ ) = −1.39 So finally, R PQ = −3.90 a r − 5.82 a θ − 1.39 a φ c) If D = 5 a r − 3 a θ + 4 a φ, find D a ρ at M(1, 2, 3 ): First convert a ρ to cartesian coordinates at the specified point.

Use a ρ = ( a ρ a x) a x + ( a ρ a y) a y. At A(1, 2, 3 ), ρ = √ 5, φ = tan−1 (2 ) = 63.4◦, r = √14, and θ = cos−1 (3 /√14 ) = 36.7◦. So a ρ = cos (63.4◦ ) a x + sin (63.4◦ ) a y = 0.45 a x + 0.89 a y. Then (5 a r − 3 a θ + 4 a φ) (0.45 a x + 0.89 a y) = 5 (0.45 ) sin θ cos φ + 5 (0.89 ) sin θ sin φ − 3 (0.45 ) cos θ cos φ − 3 (0.89 ) cos θ sin φ + 4 (0.45 )(− sin φ) + 4 (0.89 ) cos φ = 0.59 1.29. Express the unit vector a x in spherical components at the point: a) r = 2, θ = 1 rad, φ = 0.8 rad: Use a x = ( a x a r ) a r + ( a x a θ ) a θ + ( a x a φ) a φ = sin (1 ) cos (0.8 ) a r + cos (1 ) cos (0.8 ) a θ + (− sin (0.8 )) a φ = 0.59 a r + 0.38 a θ − 0.72 a φ 12.

1.29 (continued) Express the unit vector a x 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 a x = sin (105.5◦ ) cos (33.7◦ ) a r + cos (105.5◦ ) cos (33.7◦ ) a θ + (− sin (33.7◦ )) a φ = 0.80 a r − 0.22 a θ − 0.55 a φ 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 a x = sin (59◦ ) cos (40.1◦ ) a r + cos (59◦ ) cos (40.1◦ ) a θ + (− sin (40.1◦ )) a φ = 0.66 a r + 0.39 a θ − 0.64 a φ 1.30.

Given A(r = 20, θ = 30◦, φ = 45◦ ) and B(r = 30, θ = 115◦, φ = 160◦ ), find: a) R AB : 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 R AB = R B − R A = (−32.6, 2.2,−30.0 ), and so R AB = 44.4.

B) R AC , 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 R AC = R C − R A = (7.07, 7.07,−17.3 ), and R AC = 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 ( 2 π 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 4 π0 d2 = 4√ 2 × (10 −8 )2 4 π(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 = x a x + y a y; the vector directed from the second charge to the third is R23 = (x − 0.8 ) a x + (y + 0.6 ) a y. The force on the third charge is now F3 = Q3 4 π0 Q1 R13 R13 3 + Q2 R23 R23 3 = Q3 × 10 −6 4 π0 0.1 (x a x + y a y) (x2 + y2 )1.5 + 0.2 (x − 0.8 ) a x + (y + 0.6 ) a y (x − 0.8 )2 + (y + 0.6 )21.5 We desire the x component to be zero. Thus, 0 = 0.1 x a x (x2 + y2 )1.5 + 0.2 (x − 0.8 ) a x (x − 0.8 )2 + (y + 0.6 )21.5 or x (x − 0.8 )2 + (y + 0.6 )21.5 = 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 4 π0 R CA R CA 3 + R DA R DA 3 + R BA R BA 3 where R CA = a x − a y, R DA = a x + a y, and R BA = 2 a x. The magnitudes are R CA = R DA = √ 2, and R BA = 2.

Substituting these leads to F = (50 × 10 −9 )2 4 π0 1 2 √ 2 + 1 2 √ 2 + 2 8 a x = 21.5 a x µN where distances are in meters. Let Q1 = 8 µC be located at P1 (2, 5, 8 ) while Q2 = −5 µC is at P2 (6, 15, 8 ). A) Find F2, the force on Q2: This force will be F2 = Q1 Q2 4 π0 R12 R12 3 = (8 × 10−6 )(−5 × 10−6 ) 4 π0 (4 a x + 10 a y) (116 )1.5 = (−1.15 a x − 2.88 a y)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 4 π0 Q1 R13 R13 3 + Q2 R23 R23 3 where R13 = (x − 2 ) a x + (y − 5 ) a y and R23 = (x − 6 ) a x + (y − 15 ) a y. 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, 2.5 x, 8 ). With this restriction, the force becomes: F3 = Q3 4 π0 8 (x − 2 ) a x + 2.5 (x − 2 ) a y (x − 2 )2 + (2.5 )2 (x − 2 )21.5 − 5 (x − 6 ) a x + 2.5 (x − 6 ) a y (x − 6 )2 + (2.5 )2 (x − 6 )21.5 where we require the term in large brackets to be zero.

This leads to 8 (x − 2 ) ((2.5 )2 + 1 )(x − 6 )21.5 − 5 (x − 6 ) ((2.5 )2 + 1 )(x − 2 )21.5 = 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 4 π0 25 R13 R13 3 + 60 R23 R23 3 where R13 = −3 a x+4 a y−4 a z and R23 = 4 a x−2 a y+5 a z. AlsoR13 = √ 41 and R23 = √ 45. So E = 10 −9 4 π0 25 × (−3 a x + 4 a y − 4 a z) (41 )1.5 + 60 × (4 a x − 2 a y + 5 a z) (45 )1.5 = 4.58 a x − 0.15 a y + 5.51 a z b) At what point on the y axis is Ex = 0?

P3 is now at (0, y, 0 ), so R13 = −4 a x + (y + 2 ) a y − 7 a z and R23 = 3 a x + (y − 4 ) a y + 2 a z. AlsoR13 = √ 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 4 π0 25 × (−4 ) 65 + (y + 2 )21.5 + 60 × 3 13 + (y − 4 )21.5 To obtain Ex = 0, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic: 0.48 y2 + 13.92 y + 73.10 = 0 which yields the two values: y = −6.89,−22.11 15.

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 E P = 120 × 10 −9 4 π0 R AP R AP 3 + R BP R BP 3 where R AP = 0.5 a x − a z and R BP = 0.5 a x + a z.

AlsoR AP = R BP = √ 1.25. Thus: E P = 120 × 10 −9 a x 4 π(1.25 )1.5 0 = 772 V /m b) What single charge at the origin would provide the identical field strength? We require Q0 4 π0 (0.5 )2 = 772 from which we find Q0 = 21.5 nC. A 2 µC point charge is located at A(4, 3, 5 ) in free space. Find Eρ, Eφ, and Ez at P(8, 12, 2 ). Have E P = 2 × 10 −6 4 π0 R AP R AP 3 = 2 × 10−6 4 π0 4 a x + 9 a y − 3 a z (106 )1.5 = 65.9 a x + 148.3 a y − 49.4 a z Then, at point P, ρ = √82 + 122 = 14.4, φ = tan−1 (12 /8 ) = 56.3◦, and z = z.

Now, Eρ = E p a ρ = 65.9 ( a x a ρ)+ 148.3 ( a y a ρ) = 65.9 cos (56.3◦ )+ 148.3 sin (56.3◦ ) = 159.7 and Eφ = E p a φ = 65.9 ( a x a φ)+ 148.3 ( a y a φ) = −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: E P = 10 −6 4 π0 − R1 R1 3 + 2 R2 R2 3 − R3 R3 3 where R1, R2, R3 are the vectors to P 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 = 2.06R2 = 2.24, and R3 = 2.50.

Thus E P = 10 −6 4 π0 − (0, 2, 0.5 ) (2.06 )3 + 2 (0, 2, 1 ) (2.24 )3 − (0, 2, 1.5 ) (2.50 )3 = 89.9 a y + 179.8 a z Now, at P, r = √5, θ = cos−1 (1 /√5 ) = 63.4◦, and φ = 90◦. So Er = E P a r = 89.9 ( a y a r )+ 179.8 ( a z a r ) = 89.9 sin θ sin φ + 179.8 cos θ = 160.9 Eθ = E P a θ = 89.9 ( a y a θ )+ 179.8 ( a z a θ ) = 89.9 cos θ sin φ + 179.8 (− sin θ) = −120.5 Eφ = E P a φ = 89.9 ( a y a φ)+ 179.8 ( a z a φ) = 89.9 cos φ = 0 16. 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: E P = 100 × 10 −9 4 π0 R AP R AP 3 where R AP = (x + 1 ) a x + (y − 1 ) a y + (z − 3 ) a z, and where R AP = (x + 1 )2 + (y − 1 )2 + (z− 3 )21 /2.

The x component of the field will be Ex = 100 × 10 −9 4 π0 (x + 1 ) (x + 1 )2 + (y − 1 )2 + (z− 3 )21.5 = 500 V /m And so our condition becomes: (x + 1 ) = 0.56 (x + 1 )2 + (y − 1 )2 + (z− 3 )21.5 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. Determine E at P(0, y, 0 ): The field will be E P = 20 × 10 −9 4 π0 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 E P produces E P = 20 × 10 −9 4 π0 −6 a x (9 + y2 )1.5 from which E P = 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 E P = Q0 4 π0 −2 a x + a y − a z 61.5 Since the z component is of value 1 kV/m, we find Q0 = −4 π061.5 × 103 = −1.63 µC. (continued) b) Find E at M(1, 6, 5 ) in cartesian coordinates: This field will be: E M = −1.63 × 10 −6 4 π0 a x + 6 a y + 5 a z 1 + 36 + 251.5 or E M = −30.11 a x − 180.63 a y − 150.53 a z.

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 Eρ = E M a ρ = −30.11 cos φ − 180.63 sin φ = −183.12 Eφ = E M a φ = −30.11 (− sin φ)− 180.63 cos φ = 0 (as expected ) so that E M = −183.12 a ρ − 150.53 a z. 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 E M. This will be: Er = E M a r = −30.11 sin θ cos φ − 180.63 sin θ sin φ − 150.53 cos θ = −237.1 2.12. The volume charge density ρv = ρ0 e− 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 ρ0 e − x− y− z dx dy dz = 8 ρ0 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 = ∫ 2 π 0 ∫ π 0 ∫.05.03 0.2 r2 sin θ dr dθ dφ = 4 π(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.

Let ρv = 5 e−0.1 ρ (π − φ ) 1 z2 + 10 µC /m 3 in the region 0 ≤ ρ ≤ 10, − π. (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 (2 θ) 25◦ 0 −2 cos ( θ 2 )1.1 π.9 π = 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 E P = ρl 2 π0 R P R P 2 where R P = (1, 2, 3 )− (1,−2, 5 ) = (0, 4,−2 ), and R P 2 = 20.

So E P = 16 × 10 −9 2 π0 4 a y − 2 a z 20 = 57.5 a y − 28.8 a z V /m b) Find E at that point in the z = 0 plane where the direction of E is given by (1 /3 ) a y − (2 /3 ) a z: With z = 0, the general field will be E z=0 = ρl 2 π0 (y + 2 ) a y − 5 a z (y + 2 )2 + 25 We require Ez = − 2 Ey , so 2 (y + 2 ) = 5. Thus y = 1 /2, and the field becomes: E z=0 = ρl 2 π0 2.5 a y − 5 a z (2.5 )2 + 25 = 23 a y − 46 a z 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. A) Find E at P(x, 0, z): In general, we have E P = ρl 2 π0 R+ P R+ P − R− P R− P where R+ P and R− P 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 R− P = (x, 0, z)− (0,.6, z) = (x,−.6, 0 ). So E P = ρl 2 π0 x a x + 0.6 a y x2 + (0.6 )2 − x a x − 0.6 a y x2 + (0.6 )2 = 0.4 × 10 −6 2 π0 1.2 a y x2 + 0.36 = 8.63 a y x2 + 0.36 kV /m 20. (continued) b) Find E at Q(2, 3, 4 ): This field will in general be: E Q = ρl 2 π0 R+ Q R+ Q − R− Q R− Q where R+ Q = (2, 3, 4 )− (0,−.6, 4 ) = (2, 3.6, 0 ), and R− Q = (2, 3, 4 )− (0,.6, 4 ) = (2, 2.4, 0 ). Thus E Q = ρl 2 π0 2 a x + 3.6 a y 22 + (3.6 )2 − 2 a x + 2.4 a y 22 + (2.4 )2 = −625.8 a x − 241.6 a y 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) −∞. 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/(2 π0 (0.8 )) a y. The force on a differential length of the line at the positive y location is d F = dq E = ρldz E.

Thus the force per unit length acting on the line at postive y arising from the charge at negative y is F = ∫ 1 0 ρ2 l dz 2 π0 (0.8 ) a y = 1.26 × 10−4 a y N /m = 126 a y µN /m The force on the line at negative y is of course the same, but with − a y. A uniform surface charge density of 5 nC /m2 is present in the region x = 0, −2. Given the surface charge density, ρs = 2 µC /m2, in the region ρ.

Given the electric field E = (4 x − 2 y) a x − (2 x + 4 y) a y, find: a) the equation of the streamline that passes through the point P(2, 3,−4 ): We write dy dx = Ey Ex = − (2 x + 4 y) (4 x − 2 y) Thus 2 (x dy + y dx) = y dy − x dx or 2 d(xy) = 1 2 d(y2 )− 1 2 d(x2 ) So C1 + 2 xy = 1 2 y2 − 1 2 x2 or y2 − x2 = 4 xy + C2 Evaluating at P(2, 3,−4 ), obtain: 9 − 4 = 24 + C2, or C2 = −19 Finally, at P, the requested equation is y2 − x2 = 4 xy − 19 b) a unit vector specifying the direction of E at Q(3,−2, 5 ): Have E Q = 4 (3 )+ 2 (2 ) a x − 2 (3 )− 4 (2 ) a y = 16 a x + 2 a y. Then E = √ 162 + 4 = 16.12 So a Q = 16 a x + 2 a y 16.12 = 0.99 a x + 0.12 a y 2.28. Let E = 5 x3 a x − 15 x2 y a y, and find: a) the equation of the streamline that passes through P(4, 2, 1 ): Write dy dx = Ey Ex = −15 x 2 y 5 x3 = −3 y x So dy y = −3 dx x ⇒ ln y = −3 ln x + ln C Thus y = e−3 ln xeln C = C x3 At P, have 2 = C/(4 )3 ⇒ C = 128. Finally, at P, y = 128 x3 24. (continued) b) a unit vector a E specifying the direction of E at Q(3,−2, 5 ): At Q, E Q = 135 a x + 270 a y, and E Q = 301.9. Thus a E = 0.45 a x + 0.89 a y. C) a unit vector a N = (l, m, 0 ) that is perpendicular to a E at Q: Since this vector is to have no z compo- nent, we can find it through a N = ± ( a E× a z).

Performing this, we find a N = ± (0.89 a x − 0.45 a y). If E = 20 e−5 y (cos 5 x a x − sin 5 x a y), find: a) E at P(π/6, 0.1, 2 ): Substituting this point, we obtain E P = −10.6 a x − 6.1 a y, and so E P = 12.2. B) a unit vector in the direction of E P: The unit vector associated with E is just ( cos 5 x a x − sin 5 x a y ), which evaluated at P becomes a E = −0.87 a x − 0.50 a y. C) the equation of the direction line passing through P: Use dy dx = − sin 5 x cos 5 x = − tan 5 x ⇒ dy = − tan 5 x dx Thus y = 15 ln cos 5 x + C. Evaluating at P, we find C = 0.13, and so y = 1 5 ln cos 5 x + 0.13 2.30. Given the electric field intensity E = 400 y a x + 400 x a y 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.

I have the book you are looking for First published just over 50 years ago and now in its Eighth Edition, Bill Hayt and John Buck’s Engineering Electromagnetics is a classic text that has been updated for electromagnetics education today. This widely-respected book stresses fundamental concepts and problem solving, and discusses the material in an understandable and readable way. Numerous illustrations and analogies are provided to aid the reader in grasping the difficult concepts. In addition, independent learning is facilitated by the presence of many examples and problems.

Important updates and revisions have been included in this edition. One of the most significant is a new chapter on electromagnetic radiation and antennas. This chapter covers the basic principles of radiation, wire antennas, simple arrays, and transmit-receive systems.