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Engineering Electromagnetics, 8th Edition Solution Manual

Engineering Electromagnetics, 8th Edition Solution Manual provides expert solutions to textbook questions, making complex problems easier to solve.

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    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. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2). a) Find the unit vector in the direction of ( A B ): First A B = ( a x + 2 a y + 3 a z ) (2 a x + 3 a y 2 a z ) = ( a x a y + 5 a z ) whose magnitude is | A B | = [( a x a y + 5 a z ) · ( a x a y + 5 a z )] 1 / 2 = p 1 + 1 + 25 = 3 p 3 = 5 . 20. The unit vector is therefore a AB = ( a x a y + 5 a z ) / 5 . 20 b) find the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of A and B : The midpoint is located at P mp = [1 + (2 1) / 2 , 2 + (3 2) / 2 , 3 + ( 2 3) / 2)] = (1 . 5 , 2 . 5 , 0 . 5) The unit vector is then a mp = (1 . 5 a x + 2 . 5 a y + 0 . 5 a z ) p (1 . 5) 2 + (2 . 5) 2 + (0 . 5) 2 = (1 . 5 a x + 2 . 5 a y + 0 . 5 a z ) / 2 . 96 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 = 1 3 B (2 , 2 , 1), we use the fact that | B A | = 10, or | (6 2 3 B ) a x (2 2 3 B ) a y (4 + 1 3 B ) a z | = 10 Expanding, obtain 36 8 B + 4 9 B 2 + 4 8 3 B + 4 9 B 2 + 16 + 8 3 B + 1 9 B 2 = 100 or B 2 8 B 44 = 0. Thus B = 8 ± p 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

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    1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector in rectangular components that lies in the xy plane, is tangent to the circle at ( p 3 , 1 , 0), and is in the general direction of increasing values of y : A unit vector tangent to this circle in the general increasing y direction is t = a . Its x and y components are t x = a · a x = sin , and t y = a · a y = cos . At the point ( p 3 , 1), = 150 , and so t = sin 150 a x cos 150 a y = 0 . 5( a x + p 3 a y ). 1.5. A vector field is specified as G = 24 xy a x + 12( x 2 + 2) a y + 18 z 2 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) p 26 = (0 . 59 , 0 . 20 , 0 . 78) d) the equation of the surface on which | G | = 60: We write 60 = | (24 xy, 12( x 2 + 2) , 18 z 2 ) | , or 10 = | (4 xy, 2 x 2 + 4 , 3 z 2 ) | , so the equation is 100 = 16 x 2 y 2 + 4 x 4 + 16 x 2 + 16 + 9 z 4 1.6. Find the acute angle between the two vectors A = 2 a x + a y + 3 a z and B = a x 3 a y + 2 a z by using the definition of: a) the dot product: First, A · B = 2 3 + 6 = 5 = AB cos ✓ , where A = p 2 2 + 1 2 + 3 2 = p 14, and where B = p 1 2 + 3 2 + 2 2 = p 14. Therefore cos ✓ = 5 / 14, so that ✓ = 69 . 1 . b) the cross product: Begin with A ⇥ B = a x a y a z 2 1 3 1 3 2 = 11 a x a y 7 a z and then | A ⇥ B | = p 11 2 + 1 2 + 7 2 = p 171. So now, with | A ⇥ B | = AB sin ✓ = p 171, find ✓ = sin 1 p 171 / 14 = 69 . 1 1.7. Given the vector field E = 4 zy 2 cos 2 x a x + 2 zy sin 2 x a y + y 2 sin 2 x a z for the region | x | , | y | , and | z | less than 2, find: a) the surfaces on which E y = 0. With E y = 2 zy sin 2 x = 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 = ⇡ / 2, with | y | < 2, | z | < 2. b) the region in which E y = E z : This occurs when 2 zy sin 2 x = y 2 sin 2 x , or on the plane 2 z = y , with | x | < 2, | y | < 2, | z | < 1. c) the region in which E = 0: We would have E x = E y = E z = 0, or zy 2 cos 2 x = zy sin 2 x = y 2 sin 2 x = 0. This condition is met on the plane y = 0, with | x | < 2, | z | < 2. 2

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    1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between two vectors by finding the angle between A = 3 a x 2 a y + 4 a z and B = 2 a x + a y 2 a z . Does this ambiguity exist when the dot product is used? We use the relation A ⇥ B = | A || B | sin ✓ n . With the given vectors we find A ⇥ B = 14 a y + 7 a z = 7 p 5  2 a y + a z p 5 | {z } ± n = p 9 + 4 + 16 p 4 + 1 + 4 sin ✓ n where n is identified as shown; we see that n can be positive or negative, as sin ✓ can be positive or negative. This apparent sign ambiguity is not the real problem, however, as we really want the magnitude of the angle anyway. Choosing the positive sign, we are left with sin ✓ = 7 p 5 / ( p 29 p 9) = 0 . 969. Two values of ✓ (75 . 7 and 104 . 3 ) satisfy this equation, and hence the real ambiguity. In using the dot product, we find A · B = 6 2 8 = 4 = | A || B | cos ✓ = 3 p 29 cos ✓ , or cos ✓ = 4 / (3 p 29) = 0 . 248 ) ✓ = 75 . 7 . Again, the minus sign is not important, as we care only about the angle magnitude. The main point is that only one ✓ value results when using the dot product, so no ambiguity. 1.9. A field is given as G = 25 ( x 2 + y 2 ) ( 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 ✓ . So cos ✓ = (0 . 6 , 0 . 8 , 0) · (1 , 0 , 0) = 0 . 6. Thus ✓ = 53 . c) the value of the following double integral on the plane y = 7: Z 4 0 Z 2 0 G · a y dzdx Z 4 0 Z 2 0 25 x 2 + y 2 ( x a x + y a y ) · a y dzdx = Z 4 0 Z 2 0 25 x 2 + 49 ⇥ 7 dzdx = Z 4 0 350 x 2 + 49 dx = 350 ⇥ 1 7  tan 1 ✓ 4 7 ◆ 0 = 26 1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle between any two diagonals of a cube, where each diagonal connects diametrically opposite corners, and passes through the center of the cube: Assuming a side length, b , two diagonal vectors would be A = b ( a x + a y + a z ) and B = b ( a x a y + a z ). Now use A · B = | A || B | cos ✓ , or b 2 (1 1 + 1) = ( p 3 b )( p 3 b ) cos ✓ ) cos ✓ = 1 / 3 ) ✓ = 70 . 53 . This result (in magnitude) is the same for any two diagonal vectors. 3

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    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 R M N : R M N = ( 0 . 2 , 0 . 1 , 0 . 3) (0 . 1 , 0 . 2 , 0 . 1) = ( 0 . 3 , 0 . 3 , 0 . 4). b) the dot product R M N · R M P : R M P = (0 . 4 , 0 , 0 . 1) (0 . 1 , 0 . 2 , 0 . 1) = (0 . 3 , 0 . 2 , 0 . 2). R M N · R M P = ( 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 M N on R M P : R M N · a RM P = ( 0 . 3 , 0 . 3 , 0 . 4) · (0 . 3 , 0 . 2 , 0 . 2) p 0 . 09 + 0 . 04 + 0 . 04 = 0 . 05 p 0 . 17 = 0 . 12 d) the angle between R M N and R M P : ✓ M = cos 1 ✓ R M N · R M P | R M N || R M P | ◆ = cos 1 ✓ 0 . 05 p 0 . 34 p 0 . 17 ◆ = 78 1.12. Write an expression in rectangular components for the vector that extends from ( x 1 , y 1 , z 1 ) to ( x 2 , y 2 , z 2 ) and determine the magnitude of this vector. The two points can be written as vectors from the origin: A 1 = x 1 a x + y 1 a y + z 1 a z and A 2 = x 2 a x + y 2 a y + z 2 a z The desired vector will now be the di ↵ erence: A 12 = A 2 A 1 = ( x 2 x 1 ) a x + ( y 2 y 1 ) a y + ( z 2 z 1 ) a z whose magnitude is | A 12 | = p A 12 · A 12 = ⇥ ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 + ( z 2 z 1 ) 2 ⇤ 1 / 2 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) 4

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    1.14. Given that A + B + C = 0, where the three vectors represent line segments and extend from a common origin, a) must the three vectors be coplanar? In terms of the components, the vector sum will be A + B + C = ( A x + B x + C x ) a x + ( A y + B y + C y ) a y + ( A z + B z + C z ) a z which we require to be zero. Suppose the coordinate system is configured so that vectors A and B lie in the x - y plane; in this case A z = B z = 0. Then C z has to be zero in order for the three vectors to sum to zero. Therefore, the three vectors must be coplanar. b) If A + B + C + D = 0, are the four vectors coplanar? The vector sum is now A + B + C + D = ( A x + B x + C x + D x ) a x + ( A y + B y + C y + D y ) a y + ( A z + B z + C z + D z ) a z Now, for example, if A and B lie in the x - y plane, C and D need not, as long as C z + D z = 0. So the four vectors need not be coplanar to have a zero sum. 1.15. Three vectors extending from the origin are given as r 1 = (7 , 3 , 2), r 2 = ( 2 , 7 , 3), and r 3 = (0 , 2 , 3). Find: a) a unit vector perpendicular to both r 1 and r 2 : a p 12 = r 1 ⇥ r 2 | r 1 ⇥ r 2 | = (5 , 25 , 55) 60 . 6 = (0 . 08 , 0 . 41 , 0 . 91) b) a unit vector perpendicular to the vectors r 1 r 2 and r 2 r 3 : r 1 r 2 = (9 , 4 , 1) and r 2 r 3 = ( 2 , 5 , 6). So r 1 r 2 ⇥ r 2 r 3 = (19 , 52 , 37). Then a p = (19 , 52 , 37) | (19 , 52 , 37) | = (19 , 52 , 37) 66 . 6 = (0 . 29 , 0 . 78 , 0 . 56) c) the area of the triangle defined by r 1 and r 2 : Area = 1 2 | r 1 ⇥ r 2 | = 30 . 3 d) the area of the triangle defined by the heads of r 1 , r 2 , and r 3 : Area = 1 2 | ( r 2 r 1 ) ⇥ ( r 2 r 3 ) | = 1 2 | ( 9 , 4 , 1) ⇥ ( 2 , 5 , 6) | = 33 . 3 5

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    1.16. If A represents a vector one unit in length directed due east, B represents a vector three units in length directed due north, and A + B = 2 C D and 2 A B = C + 2 D , determine the length and direction of C . (di culty 1) Take north as the positive y direction, and then east as the positive x direction. Then we may write A + B = a x + 3 a y = 2 C D and 2 A B = 2 a x 3 a y = C + 2 D Multiplying the first equation by 2, and then adding the result to the second equation eliminates D , and we get 4 a x + 3 a y = 5 C ) C = 4 5 a x + 3 5 a y The length of C is | C | = ⇥ (4 / 5) 2 + (3 / 5) 2 ⇤ 1 / 2 = 1 C lies in the x - y plane at angle from due north (the y axis) given by ↵ = tan 1 (4 / 3) = 53 . 1 (or 36 . 9 from the x axis). For those having nautical leanings, this is very close to the compass point NE 3 4 E (not required). 1.17. 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) p 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) 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) 6

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    1.18. A certain vector field is given as G = ( y + 1) a x + x a y . a) Determine G at the point (3,-2,4): G (3 , 2 , 4) = a x + 3 a y . b) obtain a unit vector defining the direction of G at (3,-2,4). | G (3 , 2 , 4) | = [1 + 3 2 ] 1 / 2 = p 10. So the unit vector is a G (3 , 2 , 4) = a x + 3 a y p 10 1.19. a) Express the field D = ( x 2 + y 2 ) 1 ( x a x + y a y ) in cylindrical components and cylindrical variables: Have x = ⇢ cos , y = ⇢ sin , and x 2 + y 2 = ⇢ 2 . Therefore D = 1 ⇢ (cos a x + sin a y ) Then D ⇢ = D · a ⇢ = 1 ⇢ [cos ( a x · a ⇢ ) + sin ( a y · a ⇢ )] = 1 ⇢ ⇥ cos 2 + sin 2 ⇤ = 1 ⇢ and D = D · a = 1 ⇢ [cos ( a x · a ) + sin ( a y · a )] = 1 ⇢ [cos ( sin ) + sin cos ] = 0 Therefore D = 1 ⇢ a ⇢ b) Evaluate D at the point where ⇢ = 2, = 0 . 2 ⇡ , and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0 . 5 a ⇢ . To express this in cartesian, we use D = 0 . 5( a ⇢ · a x ) a x + 0 . 5( a ⇢ · 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. If the three sides of a triangle are represented by the vectors A , B , and C , all directed counter- clockwise, show that | C | 2 = ( A + B ) · ( A + B ) and expand the product to obtain the law of cosines. With the vectors drawn as described above, we find that C = ( A + B ) and so | C | 2 = C 2 = C · C = ( A + B ) · ( A + B ) So far so good. Now if we expand the product, obtain ( A + B ) · ( A + B ) = A 2 + B 2 + 2 A · B where A · B = AB cos(180 ↵ ) = AB cos ↵ where ↵ is the interior angle at the junction of A and B . Using this, we have C 2 = A 2 + B 2 2 AB cos ↵ , which is the law of cosines. 7

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    1.21. Express in cylindrical components: a) the vector from C (3 , 2 , 7) to D ( 1 , 4 , 2): C (3 , 2 , 7) ! C ( ⇢ = 3 . 61 , = 33 . 7 , z = 7) and D ( 1 , 4 , 2) ! D ( ⇢ = 4 . 12 , = 104 . 0 , z = 2). Now R CD = ( 4 , 6 , 9) and R ⇢ = R CD · a ⇢ = 4 cos(33 . 7) 6 sin(33 . 7) = 6 . 66. Then R = R CD · a = 4 sin(33 . 7) 6 cos(33 . 7) = 2 . 77. So R CD = 6 . 66 a ⇢ 2 . 77 a + 9 a z b) a unit vector at D directed toward C : R CD = (4 , 6 , 9) and R ⇢ = R DC · a ⇢ = 4 cos( 104 . 0) + 6 sin( 104 . 0) = 6 . 79. Then R = R DC · a = 4[ sin( 104 . 0)] + 6 cos( 104 . 0) = 2 . 43. So R DC = 6 . 79 a ⇢ + 2 . 43 a 9 a z Thus a DC = 0 . 59 a ⇢ + 0 . 21 a 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: a ⇢ = (0 . 22 , 0 . 87 , 0 . 44) · a ⇢ = 0 . 22 cos( 104 . 0) + 0 . 87 sin( 104 . 0) = 0 . 90, and a = (0 . 22 , 0 . 87 , 0 . 44) · a = 0 . 22[ sin( 104 . 0)] + 0 . 87 cos( 104 . 0) = 0, so that finally, a = 0 . 90 a ⇢ 0 . 44 a z . 1.22. A sphere of radius a , centered at the origin, rotates about the z axis at angular velocity ⌦ rad/s. The rotation direction is clockwise when one is looking in the positive z direction. a) Using spherical components, write an expression for the velocity field, v , which gives the tan- gential velocity at any point within the sphere: As in problem 1.20, we find the tangential velocity as the product of the angular velocity and the perperdicular distance from the rotation axis. With clockwise rotation, we obtain v ( r, ✓ ) = ⌦ r sin ✓ a ( r < a ) b) Convert to rectangular components: From here, the problem is the same as part c in Problem 1.20, except the rotation direction is reversed. The answer is v ( x, y ) = ⌦ [ y a x + x a y ], where ( x 2 + y 2 + z 2 ) 1 / 2 < a . 1.23. The surfaces ⇢ = 3, ⇢ = 5, = 100 , = 130 , z = 3, and z = 4 . 5 define a closed surface. a) Find the enclosed volume: Vol = Z 4 . 5 3 Z 130 100 Z 5 3 ⇢ d ⇢ d dz = 6 . 28 NOTE: The limits on the integration must be converted to radians (as was done here, but not shown). b) Find the total area of the enclosing surface: Area = 2 Z 130 100 Z 5 3 ⇢ d ⇢ d + Z 4 . 5 3 Z 130 100 3 d dz + Z 4 . 5 3 Z 130 100 5 d dz + 2 Z 4 . 5 3 Z 5 3 d ⇢ dz = 20 . 7 8

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    1.23 c) Find the total length of the twelve edges of the surfaces: Length = 4 ⇥ 1 . 5 + 4 ⇥ 2 + 2 ⇥  30 360 ⇥ 2 ⇡ ⇥ 3 + 30 360 ⇥ 2 ⇡ ⇥ 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( ⇢ = 3, = 100 , z = 3) and B( ⇢ = 5, = 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. Two unit vectors, a 1 and a 2 lie in the xy plane and pass through the origin. They make angles 1 and 2 with the x axis respectively. a) Express each vector in rectangular components; Have a 1 = A x 1 a x + A y 1 a y , so that A x 1 = a 1 · a x = cos 1 . Then, A y 1 = a 1 · a y = cos(90 1 ) = sin 1 . Therefore, a 1 = cos 1 a x + sin 1 a y and similarly , a 2 = cos 2 a x + sin 2 a y b) take the dot product and verify the trigonometric identity, cos( 1 2 ) = cos 1 cos 2 + sin 1 sin 2 : From the definition of the dot product, a 1 · a 2 = (1)(1) cos( 1 2 ) = (cos 1 a x + sin 1 a y ) · (cos 2 a x + sin 2 a y ) = cos 1 cos 2 + sin 1 sin 2 c) take the cross product and verify the trigonometric identity sin( 2 1 ) = sin 2 cos 1 cos 2 sin 1 : From the definition of the cross product, and since a 1 and a 2 both lie in the x - y plane, a 1 ⇥ a 2 = (1)(1) sin( 1 2 ) a z = a x a y a z cos 1 sin 1 0 cos 2 sin 2 0 = [sin 2 cos 1 cos 2 sin 1 ] a z thus verified. 1.25. Given point P ( r = 0 . 8 , ✓ = 30 , = 45 ), and E = 1 r 2 ✓ cos a r + sin sin ✓ a ◆ a) Find E at P : E = 1 . 10 a ⇢ + 2 . 21 a . b) Find | E | at P : | E | = p 1 . 10 2 + 2 . 21 2 = 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 9

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    1.26. Express the uniform vector field, F = 5 a x in a) cylindrical components: F ⇢ = 5 a x · a ⇢ = 5 cos , and F = 5 a x · a = 5 sin . Combining, we obtain F ( ⇢ , ) = 5(cos a ⇢ sin a ). b) spherical components: F r = 5 a x · a r = 5 sin ✓ cos ; F ✓ = 5 a x · a ✓ = 5 cos ✓ cos ; F = 5 a x · a = 5 sin . Combining, we obtain F ( r, ✓ , ) = 5 [sin ✓ cos a r + cos ✓ cos a ✓ sin a ]. 1.27. The surfaces r = 2 and 4, ✓ = 30 and 50 , and = 20 and 60 identify a closed surface. a) Find the enclosed volume: This will be Vol = Z 60 20 Z 50 30 Z 4 2 r 2 sin ✓ drd ✓ d = 2 . 91 where degrees have been converted to radians. b) Find the total area of the enclosing surface: Area = Z 60 20 Z 50 30 (4 2 + 2 2 ) sin ✓ d ✓ d + Z 4 2 Z 60 20 r (sin 30 + sin 50 ) drd + 2 Z 50 30 Z 4 2 rdrd ✓ = 12 . 61 c) Find the total length of the twelve edges of the surface: Length = 4 Z 4 2 dr + 2 Z 50 30 (4 + 2) d ✓ + Z 60 20 (4 sin 50 + 4 sin 30 + 2 sin 50 + 2 sin 30 ) d = 17 . 49 d) Find the length of the longest straight line that lies entirely within the surface: This will be from A ( r = 2 , ✓ = 50 , = 20 ) to B ( r = 4 , ✓ = 30 , = 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. State whether or not A = B and, if not, what conditions are imposed on A and B when a) A · a x = B · a x : For this to be true, both A and B must be oriented at the same angle, ✓ , from the x axis. But this would allow either vector to lie anywhere along a conical surface of angle ✓ about the x axis. Therefore, A can be equal to B , but not necessarily. b) A ⇥ a x = B ⇥ a x : This is a more restrictive condition because the cross product gives a vector. For both cross products to lie in the same direction, A , B , and a x must be coplanar. But if A lies at angle ✓ to the x axis, B could lie at ✓ or at 180 ✓ to give the same cross product. So again, A can be equal to B , but not necessarily. 10

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    1.28 c) A · a x = B · a x and A ⇥ a x = B ⇥ a x : In this case, we need to satisfy both requirements in parts a and b – that is, A , B , and a x must be coplanar, and A and B must lie at the same angle, ✓ , to a x . With coplanar vectors, this latter condition might imply that both + ✓ and ✓ would therefore work. But the negative angle reverses the direction of the cross product direction. Therefore both vectors must lie in the same plane and lie at the same angle to x ; i.e., A must be equal to B . d) A · C = B · C and A ⇥ C = B ⇥ C where C is any vector except C = 0: This is just the general case of part c . Since we can orient our coordinate system in any manner we choose, we can arrange it so that the x axis coincides with the direction of vector C . Thus all the arguments of part c apply, and again we conclude that A must be equal to B . 1.29. Express the unit vector a x in spherical components at the point: a) r = 2, ✓ = 1 rad, = 0 . 8 rad: Use a x = ( a x · a r ) a r + ( a x · a ✓ ) a ✓ + ( a x · a ) a = sin(1) cos(0 . 8) a r + cos(1) cos(0 . 8) a ✓ + ( sin(0 . 8)) a = 0 . 59 a r + 0 . 38 a ✓ 0 . 72 a b) x = 3, y = 2, z = 1: First, transform the point to spherical coordinates. Have r = p 14, ✓ = cos 1 ( 1 / p 14) = 105 . 5 , and = tan 1 (2 / 3) = 33 . 7 . Then a x = sin(105 . 5 ) cos(33 . 7 ) a r + cos(105 . 5 ) cos(33 . 7 ) a ✓ + ( sin(33 . 7 )) a = 0 . 80 a r 0 . 22 a ✓ 0 . 55 a c) ⇢ = 2 . 5, = 0 . 7 rad, z = 1 . 5: Again, convert the point to spherical coordinates. r = p ⇢ 2 + z 2 = p 8 . 5, ✓ = cos 1 ( z/r ) = cos 1 (1 . 5 / p 8 . 5) = 59 . 0 , and = 0 . 7 rad = 40 . 1 . Now a x = sin(59 ) cos(40 . 1 ) a r + cos(59 ) cos(40 . 1 ) a ✓ + ( sin(40 . 1 )) a = 0 . 66 a r + 0 . 39 a ✓ 0 . 64 a 1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft. We assume a constant altitude, a plane earth, a flight along the x axis from 0 to 10 units, no vertical velocity component, and no change in wind velocity with time. Assume a x to be directed to the east and a y to the north. The wind velocity at the operating altitude is assumed to be: v ( x, y ) = (0 . 01 x 2 0 . 08 x + 0 . 66) a x (0 . 05 x 0 . 4) a y 1 + 0 . 5 y 2 a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind would be x -directed, and so we look at the x component only. Over the flight range, this function maximizes at a value of 0 . 86 / (1 + 0 . 5 y 2 ) at x = 10 (at the end of the trip). It reaches a local minimum of 0 . 50 / (1 + 0 . 5 y 2 ) at x = 4, and has another local maximum of 0 . 66 / (1 + 0 . 5 y 2 ) at the trip start, x = 0. b) Repeat for headwind: The x component is always positive, and so therefore no headwind exists over the travel range. c) Repeat for crosswind: Crosswind will be found from the y component, which is seen to maximize over the flight range at a value of 0 . 4 / (1 + 0 . 5 y 2 ) at the trip start ( x = 0). d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizing the denominator accomplishes this; in particular, the lattitude associated with y = 0 gives the strongest tailwind. 11

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    CHAPTER 2 2.1. Three point charges are positioned in the x - y plane as follows: 5nC at y = 5 cm, -10 nC at y = 5 cm, 15 nC at x = 5 cm. Find the required x - y coordinates of a 20-nC fourth charge that will produce a zero electric field at the origin. With the charges thus configured, the electric field at the origin will be the superposition of the individual charge fields: E 0 = 1 4 ⇡✏ 0  15 (5) 2 a x 5 (5) 2 a y 10 (5) 2 a y = 1 4 ⇡✏ 0 ✓ 3 5 ◆ [ a x a y ] nC / m The field, E 20 , associated with the 20-nC charge (evaluated at the origin) must exactly cancel this field, so we write: E 20 = 1 4 ⇡✏ 0 ✓ 3 5 ◆ [ a x a y ] = 20 4 ⇡✏ 0 ⇢ 2 ✓ 1 p 2 ◆ [ a x a y ] From this, we identify the distance from the origin: ⇢ = q 100 / (3 p 2) = 4 . 85. The x and y coordinates of the 20-nC charge will both be equal in magnitude to 4 . 85 / p 2 = 3 . 43. The coodinates of the 20-nC charge are then (3 . 43 , 3 . 43). 2.2. Point charges of 1nC and -2nC are located at (0,0,0) and (1,1,1), respectively, in free space. Determine the vector force acting on each charge. First, the electric field intensity associated with the 1nC charge, evalutated at the -2nC charge location is: E 12 = 1 4 ⇡✏ 0 (3) ✓ 1 p 3 ◆ ( a x + a y + a z ) nC / m in which the distance between charges is p 3 m. The force on the -2nC charge is then F 12 = q 2 E 12 = 2 12 p 3 ⇡✏ 0 ( a x + a y + a z ) = 1 10 . 4 ⇡✏ 0 ( a x + a y + a z ) nN The force on the 1nC charge at the origin is just the opposite of this result, or F 21 = +1 10 . 4 ⇡✏ 0 ( a x + a y + a z ) nN 12

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    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 ⇡✏ 0  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 | = p 2, and | R BA | = 2. Substituting these leads to F = (50 ⇥ 10 9 ) 2 4 ⇡✏ 0  1 2 p 2 + 1 2 p 2 + 2 8 a x = 21 . 5 a x μ N where distances are in meters. 2.4. Eight identical point charges of Q C each are located at the corners of a cube of side length a , with one charge at the origin, and with the three nearest charges at ( a, 0 , 0), (0 , a, 0), and (0 , 0 , a ). Find an expression for the total vector force on the charge at P ( a, a, a ), assuming free space: The total electric field at P ( a, a, a ) that produces a force on the charge there will be the sum of the fields from the other seven charges. This is written below, where the charge locations associated with each term are indicated: E net ( a, a, a ) = q 4 ⇡✏ 0 a 2 2 6 6 6 4 a x + a y + a z 3 p 3 | {z } (0 , 0 , 0) + a y + a z 2 p 2 | {z } ( a, 0 , 0) + a x + a z 2 p 2 | {z } (0 ,a, 0) + a x + a y 2 p 2 | {z } (0 , 0 ,a ) + a x |{z} (0 ,a,a ) + a y |{z} ( a, 0 ,a ) + a z |{z} ( a,a, 0) 3 7 7 7 5 The force is now the product of this field and the charge at ( a, a, a ). Simplifying, we obtain F ( a, a, a ) = q E net ( a, a, a ) = q 2 4 ⇡✏ 0 a 2  1 3 p 3 + 1 p 2 + 1 ( a x + a y + a z ) = 1 . 90 q 2 4 ⇡✏ 0 a 2 ( a x + a y + a z ) in which the magnitude is | F | = 3 . 29 q 2 / (4 ⇡✏ 0 a 2 ). 2.5. Let a point charge Q 1 = 25 nC be located at P 1 (4 , 2 , 7) and a charge Q 2 = 60 nC be at P 2 ( 3 , 4 , 2). a) If ✏ = ✏ 0 , find E at P 3 (1 , 2 , 3): This field will be E = 10 9 4 ⇡✏ 0  25 R 13 | R 13 | 3 + 60 R 23 | R 23 | 3 where R 13 = 3 a x + 4 a y 4 a z and R 23 = 4 a x 2 a y + 5 a z . Also, | R 13 | = p 41 and | R 23 | = p 45. So E = 10 9 4 ⇡✏ 0  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 E x = 0? P 3 is now at (0 , y, 0), so R 13 = 4 a x + ( y + 2) a y 7 a z and R 23 = 3 a x + ( y 4) a y + 2 a z . Also, | R 13 | = p 65 + ( y + 2) 2 and | R 23 | = p 13 + ( y 4) 2 . Now the x component of E at the new P 3 will be: E x = 10 9 4 ⇡✏ 0  25 ⇥ ( 4) [65 + ( y + 2) 2 ] 1 . 5 + 60 ⇥ 3 [13 + ( y 4) 2 ] 1 . 5 13

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    2.5b (continued) To obtain E x = 0, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic: 0 . 48 y 2 + 13 . 92 y + 73 . 10 = 0 which yields the two values: y = 6 . 89 , 22 . 11 2.6. Two point charges of equal magnitude q are positioned at z = ± d/ 2. a) find the electric field everywhere on the z axis: For a point charge at any location, we have E = q ( r r 0 ) 4 ⇡✏ 0 | r r 0 | 3 In the case of two charges, we would therefore have E T = q 1 ( r r 0 1 ) 4 ⇡✏ 0 | r r 0 1 | 3 + q 2 ( r r 0 2 ) 4 ⇡✏ 0 | r r 0 2 | 3 (1) In the present case, we assign q 1 = q 2 = q , the observation point position vector as r = z a z , and the charge position vectors as r 0 1 = ( d/ 2) a z , and r 0 2 = ( d/ 2) a z Therefore r r 0 1 = [ z ( d/ 2)] a z , r r 0 2 = [ z + ( d/ 2)] a z , then | r r 1 | 3 = [ z ( d/ 2)] 3 and | r r 2 | 3 = [ z + ( d/ 2)] 3 Substitute these results into (1) to obtain: E T ( z ) = q 4 ⇡✏ 0  1 [ z ( d/ 2)] 2 + 1 [ z + ( d/ 2)] 2 a z V / m (2) b) find the electric field everywhere on the x axis: We proceed as in part a , except that now r = x a x . Eq. (1) becomes E T ( x ) = q 4 ⇡✏ 0  x a x ( d/ 2) a z | x a x ( d/ 2) a z | 3 + x a x + ( d/ 2) a z | x a x + ( d/ 2) a z | 3 (3) where | x a x ( d/ 2) a z | = | x a x + ( d/ 2) a z | = ⇥ x 2 + ( d/ 2) 2 ⇤ 1 / 2 Therefore (3) becomes E T ( x ) = 2 qx a x 4 ⇡✏ 0 [ x 2 + ( d/ 2) 2 ] 3 / 2 c) repeat parts a and b if the charge at z = d/ 2 is q instead of + q : The field along the z axis is quickly found by changing the sign of the second term in (2): E T ( z ) = q 4 ⇡✏ 0  1 [ z ( d/ 2)] 2 1 [ z + ( d/ 2)] 2 a z V / m In like manner, the field along the x axis is found from (3) by again changing the sign of the second term. The result is 2 qd a z 4 ⇡✏ 0 [ x 2 + ( d/ 2) 2 ] 3 / 2 14

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    2.7. A 2 μ C point charge is located at A (4 , 3 , 5) in free space. Find E ⇢ , E , and E z at P (8 , 12 , 2). Have E P = 2 ⇥ 10 6 4 ⇡✏ 0 R AP | R AP | 3 = 2 ⇥ 10 6 4 ⇡✏ 0  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 , ⇢ = p 8 2 + 12 2 = 14 . 4, = tan 1 (12 / 8) = 56 . 3 , and z = z . Now, E ⇢ = E p · a ⇢ = 65 . 9( a x · a ⇢ ) + 148 . 3( a y · a ⇢ ) = 65 . 9 cos(56 . 3 ) + 148 . 3 sin(56 . 3 ) = 159 . 7 and E = E p · a = 65 . 9( a x · a ) + 148 . 3( a y · a ) = 65 . 9 sin(56 . 3 ) + 148 . 3 cos(56 . 3 ) = 27 . 4 Finally, E z = 49 . 4 V / m 2.8. A crude device for measuring charge consists of two small insulating spheres of radius a , one of which is fixed in position. The other is movable along the x axis, and is subject to a restraining force kx , where k is a spring constant. The uncharged spheres are centered at x = 0 and x = d , the latter fixed. If the spheres are given equal and opposite charges of Q coulombs: a) Obtain the expression by which Q may be found as a function of x : The spheres will attract, and so the movable sphere at x = 0 will move toward the other until the spring and Coulomb forces balance. This will occur at location x for the movable sphere. With equal and opposite forces, we have Q 2 4 ⇡✏ 0 ( d x ) 2 = kx from which Q = 2( d x ) p ⇡✏ 0 kx . b) Determine the maximum charge that can be measured in terms of ✏ 0 , k , and d , and state the separation of the spheres then: With increasing charge, the spheres move toward each other until they just touch at x max = d 2 a . Using the part a result, we find the maximum measurable charge: Q max = 4 a p ⇡✏ 0 k ( d 2 a ). Presumably some form of stop mechanism is placed at x = x max to prevent the spheres from actually touching. c) What happens if a larger charge is applied? No further motion is possible, so nothing happens. 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 E x = 500 V/m: The total field at P will be: E P = 100 ⇥ 10 9 4 ⇡✏ 0 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) 2 ] 1 / 2 . The x component of the field will be E x = 100 ⇥ 10 9 4 ⇡✏ 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 15

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