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Elements of electromagnetics 6th edition pdf solutions sadiku
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ELECTROMAGNETICS6 EditionISBN: 8220101369376Elements of Electromagnetics (The Oxford Series in Electrical and Computer Engineering)6 EditionISBN: 9780199321384Elements of Electromagnetics6 EditionISBN: 9780190213879 1. Solutions Manual Accompanying Elements of Electromagnetics, Third Edition Matthew Sadiku, Jerry
Sagliocca, and Oladega Soriyan 2. Chapter l Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter ll Chapter 12 Chapter 13 Chapter 14 Chapter 15 TABLE OF CONTENTS 40 l 13 129 164 193 221 240 280 328 350 384 389 3. CHAPTER 1 P. E. 1.1 i A + B = (1,o.3)+ (5,2,—6) = (6.2,—3) i iA+B!
=/36+4+9=Z I (8) (b) SA — B = (5,0.15)— (5,2,—6) = (0,—2,21) ’_‘ (C) The component ofA along a, . is A_, . = Q l (d) 3A + B = (3,o,9)+ (5,2,—6) = (s,2,3) I A unit vector parallel to this vector is I (s,2,3) . *1 “H = i x/64+4+9 1, = :(o.9117a_, + 0.2279(1), + 0.341911: ) 1 P. E. 1.2 (a) The distance vector r9, = r, — re = (o,3,s)— (2,4,6) — Zax —ay + 2a,
(b) The distance between Q and R is J4 +1 + 4 = g r9’ = (1,—3,5)— (2,4,6) = (—1,—7,—1) : rQR-re, = __7_ PQK I; -Qk“I'Q, .| 3s/ gi i i i 5 i’¢2Ri= (0) Vector re, = r, — cos 6 QPQR = (d) Area = y2|rQ, , x rQ, ,| = %| (15,—4,13) = W-_12 4. P. E. 1.3 Consider the figure shoxm below: A‘ 40 ‘ UZ = U,, -1-U”. =—350a4, +j3—(—a". +a, .) E ' = -3 78¢‘. +
28.28a, . OI‘ u = 379.341 75.72” 1 1 1 ‘y N i u :5 u X E . ~ 11 "‘ 1 S r [i P. E. 1.4 i ‘I -I At point ( 1,0), G = a, ; '1' at point (O, l), G = -a, ; at point (2,0), G = a, ; p in , , = -——-—-—; n . at o't(1l)Gt —a"+a’ adsoon ~/5 It is evident that G is a unit vector at each point. Thus the vector field G is as sketched in Fig. l.8. i P. E. 1.5 Using the dot product, .I A a B
— I3 13 - I; 9 = =————— , °°S "’ AB J15./6? 50 9 —— or using the cross product, r | Ax B! 481 V E sine-‘”= AB 650 5. . ( (E-F)F 2 — 1o(4.—1o.5) E . = E. . . = , (a) . a. )a1 ‘F W I‘ = — o.2837a, + 0.709211, — 0.3S46a: §; ' ax a), a: (b)ExF = 0 3 4 = (55,l6,—l2) ,1 4 -10 5 aE_, . = :(O.9398,0.2734,—0.205) P. E. 1.7 a + b + c = 0 showing that a. b. and c
form the sides ofa triangle. (1 - b = 0, hence it is a right angle triangle. Area = -1-| axb| = llbxci = l| cxai 2 2 2 40-1 134 Area: -3-s/9+289+l44=l .51 1 1 1 1 —2—| axb -5‘ l_—2—| (3,-17,12) P. E. 1.8 (3)P1P2= (x2'x1)2'*'(. V2". V1)2+(32‘z1)2 = ./ K472171574" = 94931 (b) r, , = rn +/ i.(r, .1 -r, ') = (1,2,—3)+ 2(- 5,—2,8) = (1 —5,1,2—.2.1,—3 +82) (c) The
shortest distance is d= l’, P,sin9=iP, P,xa, .|, .,i , . 6 -3 5 -5 — 2 8 1 1 T5 1 7753 — 1(—14.—73.—27) = 8.2 6. ‘ Prob. l.1 r - (-3.2.2) — (24.4) = (-5,-2,-2) (-5 r —2.——2) = ,— = ——-——-— = —0.8703 , - 0.3482 , - 0.3482 . 1ri . /25+4+4 a a’ a‘ 1 Prob. 1 2 (a) A + 23 = (2.5,-3)+ (6,-—8,0) = sax - 3a, - 3a: (b) A - 5c = (2.5,-3_) - (5,s,5) = (- 3,0,-8) 1A-sq =
/9+0+64 = s.544 (c) 13 = 314:, - 41m, [k3] = 1/912 +16k2 = i5k = 2 = > k_= _i: C-4. (d) A - 3 = (2,5,-3)-(3,-4,0): 6- 20+0 =14 2 5 "3’= (-12,-9,—23) 3 -4 0 AxB (12 9 23) AxB= ’ 5 7 = 0.8571ax + 0.6428ay + 1.64211: Prob. 1.3 I (a) A — 23 = (2.1,-3)— (0.2-2) = (2.-1.—1) V A-23+c=5a, +4a, -+611, I (b) A + 3 = (22-4) C - 4(A + 3) = (3,5,7)- (8,8,-16) = 5a, - 3a, , + 23a) (c) 2A - 33 = (4,2,-6)- (0.3,-3) = (4,-1,-3) 1c| =./9+25+49 -911 2A - 33 I I ICI = 0.43%‘ — O. l1ay — 0.3293a: 7. (d)A-C=6+5—21=—l0. 131 = .5 IE A-C—]B[l= —l0+2=:8_ " (e)1A + 1c -6.1,-1j+[3 3 Z) = (1 .4167.1.5833.0.75) . ' 3 4 3 4‘4’4 1 1 1 I 1 0 1 "1 1 —Bx -A+-C = - =1.1667a_. -0.7084a, ,-0.7084a, 1‘ 2 3 4 21.4167 1.5833 075 ___.
._; ___-__. __—. ._; _.. -_---___. __~ 1' is Prob.1.4 '-‘ (a) T= ( , -2,1)ands= (4,6,2) 1 (c) distance = lrrsl = 4/1+64+ = 8.124m Prob. 1.5 LetD= aA+, BB+C i: I = (5a-3+8)a, +(3a+43+2)ay+(—2a+63)a. Dx=0-—>5a—, B+8=0 (1) D: =O—>—2a+6,B=0—>a=3,B (2) ; I Substituting (2) into (1), - | v 8 4 15 — 8:0 = ——-= —— /3 '34’ "W 14 7 i | Thus I a: -—-, fl=
—— 8. 6 [ Prob. 1.6 _— A-B=0—~>0=3a+,8—24 (1) ‘ A-C=0—>O=5a-2+4y (2) ’ B-C=0—>0=l5—2fl—6y (3) In matrix form. 24 3 1 0 a . ' 2 = 5 o 4 6 -1 15 0 2 6 7 ’ 1 3 1 0 . A=5 0 4=3(0—8)—1(30—0)+0(l0—0)= -24-30=-54 1,- 0 2 6 : 24 1 0 : ‘ A, = 2 o 4=—24x8—(l2——60)= —144 15 2 6 3 24 o A, =5 2 4=3(12—6o)—24x3o= —s64 0 15 6 3 1 24 I A, =5
0 2-—12—75+240=153 3_' 0 2 15 ‘E A, — 4 a= Z—= :§—=2_. _6__gz 1 A2 -864 r”= z'= ‘3=2 : y= —AA—’= —:%= —2.s33 , . Prob. l.7 ‘ (a)A-B= ABcos19A, ’ Ax B = ABsin0A, a,, (A - 3)’ +1A x 312 = (AB)3(cos’ 6,4,, +s1n’ 0.4,, )= (AB)’ 9. a_xa, a 1 J " : —“-: ((I 1 a, ,~a‘, xa 1 1 ' 1 a_>. D6 = (x+ : )cos6 sinttt = r(sin0 sint) + cos6)cos6 sin¢. D, = (x+
z)cos(I> . D = 1'(sin0 cost) + cos0)[sin0 sinttt £1. + cost) sin¢ 21., + cos¢ 51,]. = r(sin6 cost) + cos9)cos¢. (b) Cylindrical: P d E, = — (yz -— x2)sin¢ + xyzcos¢ = p2 cos2¢sin¢ + p2 cos2¢sin¢ + p2zsin¢cos2 ¢. E__= x2—z2 = p2 cosz ¢— 22. E = p2 cos¢(zsin’¢- cos2¢)a, , + p2 sin In spherical: 5. E0 E4 E, = (y" — x")sin0 COSCI) + xyzsin0 sinttt + (x2 —
z")cos9; but x = rsin9 cost), E9 = (y’ — x’)cos9 cos¢ + xyzcose sin¢ — (x’ — z’)sin6; — r’ sin"9 cos2¢ cost) cost + r’ sin’ 0 cos’9 sin°¢ cost) - r’(sin’ 6 cos’¢ — cos’ 6)sin0; E, = (x: —y")sin¢ + xyzcost) = r’ sin’ 9 cos2tt sins) + r’ sin’ 0 cos’0 sintt cost); cos¢ sing} O y’ - x2 — sin¢ cos¢ 0 xyz (yz — x2)cos¢+ xyzsinqfi p2(sin2 05- cos2 (15) cos¢ +
pzzcosgésinz (0 - p2 cos2¢cos¢ + pzzsinz ¢cos¢. / sine cost: sin6sin¢ cost) cos0 sintt - sine cost cost) cost) — simt ‘ x’ — 2’ y: rsin9 sin¢, z= rcos0; = r’ sin’6(sin’¢ — COS2¢)COS¢ + r3 sin’ 6 cos0 sin‘, (1 cost) + r"(sin" 0 cos’ «I: )cosL= ;‘: 24. —r’ sin’ 0 cos2 cose cos + r3 sin’e cos’e sin’ ¢ cos¢ — r: sine(sin"‘ e cos3¢ - cosle) a0 + I 22 + [r" sin’ e cos2¢
sin¢ + r’ sin’ 0 cos’ 0 sins) cose]a. , Prob. 2.5 (a) 5» 1: F; F: 1 . F, = [pcos2¢+ ps1n2¢] = 1 . . F, = —’0-2-/ ——7[—pcos¢s1n¢+pcos¢s1n¢] = 0; _ 1 _ _ F = (pap+ 40;). Z In Spherical: F. I F. F. cos¢ sin¢ 0 — sin¢ sin¢ 0 O O 1 sine cos¢ sine sine cose cost cose sin¢ — sine — sine I cost» X Ip2+Z2 J’ Ip2+z2 4 cose 0 Ip2+z2 /0 Ip2+z2’ ‘: |-l‘: I"< E = [—r’
sin’ecos2d)+ 1-’ sin’ecose sin’¢tcos¢ + r'°(sin’ecos’¢ — cos"e)cose]c-1, + 25. II II II II I. -I 7 '7 . 7 . 7 . 3 4 , : : sin'ecos‘e + : sin‘esin‘e + —cose = sin‘e + —cose; I 7' I‘ I‘ I’ . . . . . 4 . . 4 . F; = smecose cos'e + sine cose sin'e - -sine = s1necose— -sine; I I‘ I’ F, = — sine cose sine + sine sinecose = 0; I . , . 1 _ - . -I . I F = (sin' 0 + -sine)a, + sine(cose — -)ao.
__ 1. (b) I xp’ ' 2 ,2 I G, cos¢ sin¢ 0 V” *, “ 1 . . )’P G, = —sin¢ sin¢ 0 —-; —-; 1 G, 0 0 1 Vflpfiz = ————_r~—, ,. , I 2 P3 _ 2 ' 2 . GP — P2 + 22 [pcos ¢+ psin ¢] — P2 + Z2 , G, = 0; G - Z”: - : p’ + 2’ ’ 2 G = f’ , (p2t, ,+za. ). 1' p + Z Spherical : I , _ _ xrsine I G, sine cose sine sine cose r I G, = cose cose cose sine — sine ysine I G, — sine cose 0 zsine El) ‘U;
26. 24 G, = rsin‘° e cos"e + rsin‘, 0 sin"e + rcos‘° e sine = rsin’e + rcos’ sine = rsine. G = rsinze cose cos’ + rsin’e cossin’ — r sin’ e cose 0 :1‘ sin: e cose — rsin‘, 0 cose = 0. G = — rsinze sine cose + rsin’e cose sine = 0. G = rsine 61,. I Prob. 2.6 (a) A, cose — sine 0 p(z’ + 1)I A, = sine cose 0 —pzcose A, 0 0 I 0 A, = p(z‘ + l)cose + pzsine cose =
IIx2+y2 (22+l)—-—-—-—I;2—):7y§- + I/ ?:? "_( zxy x’ + y’ xyz 2 = x(z’+1)+ Z. x+y A, = p(z’+ 1)sin e — pzcos’e . 1/.1)_-, .y_. - IIx‘ +y" ‘Ix‘ +y" l : y(z2+[)— - « x2+y2 ’ A_= 0; - 2 . ‘.A= 2+1 +—iy—Z——, + 2+1-————-xz '. lx(Z ) WI“ [)’(Z ) ay I-I‘ 27. B [ 2x’ x2 1 + = —‘——? " + , + '**3“‘7— ax ‘Ix’, + y2 + 22 (x2 + y2)‘Ix2 + y2 + 2‘ x I’ J’ . 25 B, sine cose
cose cose — sine 2x B, = sine sine cose sine cose rcose cose B__ cose - sine 0 —rsine B, = Zxsine cose + rcos‘e cos’e + rsin’e 2x3‘Ix" + y" ‘Ix’ + y‘? + 2’ ( xz ) y’, ) + -—--—: ,'——"-; — , + X’ + " 2" '5 4 ‘Ix-’+y"/ x-’+y°+ 2’ x2+y' + 2‘ x’+y‘ } X‘ + y‘ .1 2x3 2 .7 .7 .7 xz y x +y +2 "‘T_T"T ‘I’ , 1 ‘D ‘I I ‘L 2 .2 3 I ‘Ix‘+y‘+z‘ (x +y‘)‘Ix‘+y'+z' -“ ‘U’ I By = Zxsine
sine + rcosze sine cose — rsine cose 2xy‘Ix3+y" ‘Ix2+ yz + z‘°(xyz2) _ ( xy I = “+ 2 2 2 2 2) I ‘Ix‘1+y" x2+y’+z" 35+)’ +2 x+. V 2xy xyz’ xy‘Ix" + y’ + 2‘: I = ———-—-——— + — —--————-—-; ‘Ix2 + y‘7 + 2’ x2 + y"‘Ix’ + y2 + 22 ‘Ix’ + y’ I B, = 2x cos— rsine cose cose 2xz ‘Ix’ + y’ + z’ (xy)‘Ix’ + y’ ‘Ix’ + y’ + z’ (x’ + y’ + 2’) ‘Ix’ + y’ 2xz x2 x2 : -——-——————— — —-———————— : ———————-———; 5. ‘Ix’ + y‘° + 2*’ ‘Ixz + y’ + 2’ ‘Ixz + y’ + 2’ '- y"‘Ix’ + y’ + z’ _ f 2xy xy xy‘Ix2+y2+z2 _ [ 2 2 2+ 2 2 2 2 2- -’ 9 ]at”+ x+y+z (x+y)x+y+z x+y I xz - ' [ 3 1 1]a: I 1"l) 28. Prob 2.7 (a) (b) cos¢ - sin¢ O cos¢ 0 zsin¢ -pcos¢ 2p: sin¢ xyz xy x2 + y: + }—-—--1 2 2 x +y zsin¢cos¢ + psin¢cos¢ = t, + yz x2,/
x2 + y’ x2+y2 2 yz x2+y’ - zsinz ¢— pcosz ¢ = C__ = 2,02: 2z, [x2+, 1‘; Z yz x2+y’ X)/ Z '- C= (x2+y2 , /x2 + y E1, xy - _ + )ax+( )a, + 22 sine sin6 cos¢ cos6 cos¢ — simb cose sine sin¢ cose sinq5 cos¢ cost) 4 - sin¢ _—_. j._j. _. cos"0sin¢ + ————~——- - —— - - V2 — ‘[x”+y‘°(x’+y"+z")’ _. ._: ___. _;_. sine cos9 29. Prob. 2.8 (a) (b) [In &, , = (cos¢E1,, —
sin¢[2,)o K1,. = cos¢ 61,: £1, = (cos¢ci, , - sin¢c}, )o £2, = ’— sin¢ sin¢ 51,1 (-1,, = (sin¢c}, . + cos¢&, )o 51,, 21,- E1, = (sin¢E1,, + sin¢Z1,)~ 5:, = cos¢ Since [1, , E1. , and [1, are mutuallyonhogonal Also, 21,9 21; = 0; cos¢ we 51, = 1; 5;! ‘ E1, = 0; E1_. ¢ E1. = 0. 5,0 5; = 0. 0 éx. ép éx. é; a &: sin¢ cos¢ 0 = 51,0 21,, 51,6 51. 5,6 51.- 0 0 1 2m 5, a, s
a. an a, In spherical system: [1, = sine cose £1, + cose cost £19 — sine 51.. &, = sine sin 21, + cose sin¢ 59 - cost (3.. 3 E1, = cose E1, - sine 219. Hence, E1» E1, = sine cos¢; &, e 519 = cose cos¢; [rye E1, = sine sin¢: ; (-1,: [19 = cose sinda; an a, = cose; a_. e an = ~ sine; —, _ 30. ‘, ‘ ' (x2 * 2: (p2+z2. -j. '. ' 1. 28 or I p = ,/ x’ + y’ = ,/ r’ sin’ ¢9cos’ ¢+
r’ sin’ 6?sin2 ¢. = rsinfl; 95. z = rcosél; (25 r: +y‘+z 0: tan"£; ¢= ¢. I: (a) From the figures below, Hence, From the figures below, a, , = cose E19 + sine 51,; E1, t V: ___j _. .. __ _. .. ___. L., ___: _ ___= __. ___. __ ____ . _. __% . - Z I I’ [1, = sine Ev; + cose E19; (ire = cose 51,, — sine ('1,-; E1. = 51. cose 51,- sine E19; 21. = (-1.. 31. P 6-7;» sim9 cos6l 0 63,
6.1‘; = 0 0 l £19 0“: cos0 — sin 6? 0 a'__ Prob. 2.10 (a) Hp cos¢ sin¢ 0 xy’z H. = — sin¢ cose 0 xzyz H__ 0 0 I xyz" H‘, = xy’zcos¢ + x’yzsin¢ = p’zcos’¢sin’¢ +p’zcos’¢sin’¢. 2 H, = — xy’zsin¢ + x’yzcos¢ = - p’zcos¢ sin’ ¢ + p’zcos¢ sin¢ p’zsin’2¢ = p’zcos¢ sin¢cos2¢. H_. = xyz’ = p’z’ sin¢ cos¢o. - I - I — I _ H: -—p’zsin’2¢ ap + —p’zsin2¢ cos2¢»a. +
3p’zsin2¢ a: . 2 2 [HA sine cos¢ sine sin¢ cose xy’z I H0 I = cose cost» cose sin¢ - sine x’yz 3 H, ; E — sin¢ cos¢ 0 _. xyzzd x= rsinecosd>. _v= rsinesin¢>. := rcose. 32. 30 H, = xy: [ysine cose + xsine sine + . -cose = r’ sin’ 9 cose sine[rsin’ e sine cose + rsin’ e sine cose + rcos’ e] He = xyz[ycose cose + xcose sine — zsine] = r’ sin’ e cose sine cose [r
sine cose sine cose + rsine cose sine cose — rcose sine] H, = x_v: [— ysine + xcose] = r" sin’ e cose sine cose[—rsine sin: e + rsine cos’e] I = r’sin’ecose sinecos2e. H: r’ sin’ e cose sine cose[(sin’e sin 2e + cos’ e)(1, + (sine cose sin 2e — cose sine)Ez, , + sine cos2e 13,]. (b) At (3— 45), H(x, y,z) = -60(--1,3,5) @mya= uu I This will help check H(p, e,z)
and H(r, e,e) ‘ 4 p = 5, z: 5, ¢= 360°- tan"3= 306.87° - 1 - 1 . 1 _ i H: 3(125)(5)(—0.96)a, ,+ 3(125)(5)(—0.90)(-0.277))a. + 3(25)(5)(—0.96)a, II =2w¢+wA—um¢ I ' Sphfllcala T—__—-——-j- 5 l 5 l l = 50:5 2;‘ ’6=——-= --; 6: —-= ——. r x/ _ / — sm 5‘/ E ‘/ E cos 5‘/5 ‘/ E I , 4 3 I & s1n¢= —§, cos¢= -5-. I _ I 1 I2 1 12 l _ 1 12 1 - 1 9 _ H = 2500(§)(f)(- 2—5)
[{§*2(- §§)+ §}a, + {§*2(- 2—5)- §}ao+ 35- ; —:}a, ] f_ = -8435&, +4158&, +s4&, Prob 2.11 (a) 33. (b) 31 g :7 7 ‘I 7 x x- A_, =pcos'e= x‘+y' _, _, =—-, —-j I- x w I I A, = psine cose = ,Ix’+y3 , xy _. = ‘I X + .1’ ‘Ix’ + y’ 1 , - - - ' A__ = ———. ,—-—‘, [," a. + xya, + y: :a_. ]. I ‘Ix +y : At (3,-4,0) x=3, y= -4, z=0; T - 1 _ , A = §[9a, ~ l2a_, .]. II I I/ ’1I=3 I 12-- A,
sint9cos¢ sin0sin¢ cos0 /7 x A, = cos¢cos¢ cos6sin¢ —sin9 -1 A, —sin¢ cos¢ 0 pz XE. . P x= rsinecose, y= rsinesine, z rcose, p= rsine. r’ sin’ e cos2e r’ sin’ e cose sine = -——. *--~ sine cos + . sine sin + ’ rsine ¢ rsine ‘I’ I r’ sine cos2e , —-—-, --—s1necose rsine = rsin’ e cose + r’ cos’ e sine A0 = rsine cos2e cose cose + rsine cose sine cose sine — r‘,
cos’ e sine sine = rsine cose cose — r’ sine cos’ sine = rsine cose[cose — rcose sine] - A, = — rsine cos"e sine + rsine cose sine cose = 0. - . - - I! A = 1-[sin‘ e cose +1-cos’ e sine]a, + 1' sine cose[cose - rcose sine ]ao. , 34. 32 AI (3— 4,0), 1‘: 5,e = n /2, e = 306.83 cose = 3/5, sine: -4/5. 21 = 5[1~’ *-fi—+ 5(0)(—1/5)]£z, + 5(1)(0)a, , = 361,. I21I= 3 I
Prob 2.12 A, cose —sine 0 A9 A, = sine cose 0 A, A _ 0 0 1 A__ x y 0 ‘Ix’+ y’ ‘Ix’ + y’ A P = =—I~2-”—2 —'—I—2”-2 0 A. x +. y x +y A 0 0 I ' A, sin0cos¢ cos0cos¢— sin¢ A, A, = sin0sin¢ cos0sin¢ cos¢ A, A_, C050 -sine 0 A, I ‘Ix’ + y’ ‘Ix’+y’+z’ ‘Ix’+y’+z’ -y x xz II ‘Ix’ + y’ + z’ = ‘Ix’ + y’ ‘Ix’ + y’ + 2’ ‘Ix’ + y’ I z x §~. ‘ix _. __3_’_ . _______«_"______ = ‘Ix’ +
y’ + z’ ‘Ix’ + y’ ‘Ix’ + y’ + z’ ‘Ix’ + y’ z - : .- . -_-. ==. -._: _—-—_—__ ——: ___%= =_ _ _ ____ 35. 33 Prob 2.13 (a) lfsing the results in Prob.2.9, A, ,= ,0:sin¢ = r’sinI9cos0sin¢ A, = 3pcos¢= 3rsin0cos¢ A__ = pcos¢sin¢ = rsin6cos¢sin¢ Hence. A, sine 0 cose r’ sine cose sine A6 = cose 0 — sine 3r sine cose A 0 I 0 r sine cose sine O A(r, e,e) = rsineIsine
coseIr sine + cose)a, + sine(rcos’e - sine coseIa, , + 3cosea, I At (10,1t/2,311/-I), r=10,e = II /2,e = 31:/4 —- . 3 A = 10(0a, + 0.5a, ‘ — —‘/3-a. ) = 5a, — 21.21a, P (b) B, =r’= (p’+z’), Be=0, B, =sine= —-——- 2 p+z I B, sine cose 0 B, I B, = 0 O 1 B, I B_, cose - sine 0 . ) I p I B(p, e,z)= ‘Ip’ + z’[pap + p, + Z, a, + zazj _At (2,1r/6,1), p = 2,e = 1:/6,z= 1 3 =
‘/3(2a, + 0.4a, + a_, ) = 4.47211, + 0.89-Ma, + 2.236a, I Prob 2.14 I‘ (a) .1: ‘I(6— 2)= +(—1-1)2+(2-5 2 = ‘/ E = 5.385 I I I (b) '31 I 36. I I d’ =1o= + 53 — 2(I0)(5)cos£cos~’5 4 6 I 71 7r 317 I I if = (10)(5)sin§sin-6-cos7Z- -4- I d = ‘/9—9.I’2‘ = ‘_). _9__5___6__. II Prob 2.15 I' (a) An infinite line parallel to the z-axis. I II (b) Point (2,-1,10). I II (c) A circle of
radius rsinI9=5 , i. e. the intersection of a cone and asphere. (d) An infinite line parallel to the z-axis. I (e) Asemi-infinite line parallel to the x-y plane. (1) Asemi-circle of radius 5 in the x-y plane. ' Prob.2.l6 ’ At T(2,3,-4) I -, ‘/73’ -4 I - « 13 cose = —: = = -0 7428,sine = 7%: 06695 " _ —IZ= -I: ’.= e—tan x tan 2 56.31 2 _ 3 cose sine: -J7 I 6:, = cose &, —
sine a‘, = -0.7428&, - 0.6695139. ' II (1, = sine cose d, + sine sine ¢1_, + cose cf. I = 0.3"]-la}, + 0.55 715:, - 0.7-1280,. ll __________. _.. _ II 37. A1 P(0,2.—5). e = 90°; B cose —sine 0 B I B : H, sine cose 0 B, 13, I0 0 1 B, 0 -1 () -5 = I 0 0 I » 0 0 I -3 ,1; = 2.1.4.1, -31}, (a) 21+ 13 = (2,4,10)+(—l, -5,—3) = a, —c1,+7c1:. (b) 9 _ AOB _ -52 ‘’°‘‘ 1” ‘ IIAII IIBII
‘ 4—42oo -52 = " = 14 .26°. 0,, cos ( 4200) 3 : O : : ——j: — A 21 " ”°B 52 3739 (C) H an B ‘fig . . Prob. 2.18 At I>(8, 30°, 60°)‘ = 1>(r, e,¢), x = r sine cose = 8sin 30° cos60° = 2. y = rsine sine = 8sin30° sin60° = 2‘/3 z= rcose = 8(§-x/3): 4‘/3. - /3- 1- a, = — sine c1,+ cose E1, = — —2—a, + 30,; I I I I I G = 14£i, + 843 &, .+ (-18+ 24)&, = (14,13.86,72);
I 1 I ‘. = (Go c1.>&. = d: :p’ITIcos3¢d¢: IIc1: ¢=0 1-0 I= , = (4)’11(1)=6411 1’ =0+ 0+ 641 = g_4_£ By the divergence theorem, I211; = Iv. A111 8' V - I18 16 (M = _____ 3 2 _____z - VoA pap(p cos¢)+p6¢ s1nI>+ d 1 = 3p cos’¢ + 'p‘COS¢. I - 1 I I1’: IVA Adv: I (3pcos’¢ + ; cos¢)pdI>dzdp I V V 2: I = 3Ip’dp Icos"¢1d¢ Idz + Idp IcosI>d¢ Id: 0 I} 0 0 43
= 3(7)11(1) = 6411. P. E. 3.8 ' (a) v x 21- E1.(1-0)+£1,(y-0)+ E1,(4y-z) &, + yE1,+ (4y- z)Zz, At (1,-2,3), v x A = 21,- 251,-1111, (b) ~- - - . - 1 1 V x B = a, ,(0— 6pzcos¢)+ a, (p sm¢ — 0)+ a; ;(6pz‘ cose - pzcos¢) = —6p: cos¢ 11,». p sir—1I>a. + (6:- 1): cos¢ E1, I At 15.41;. -1). V1 13- 511,. 46. I II 44 - . 1 . E1. C 1 . ' r sine 2 9 ' ' 3 , ' _‘ (r “cose - 0) + -(—————r°°_‘°‘ s'"I - -1-'1-)+ 5'i(0- 2rSln0 cos¢1) r sine 2 r = r"" cote 61,- (2cote sin¢ + -2-r""I‘I)c—ze— 2sine cotI1 &. At (Lag), v x C: 1.732£1,- 4.551.-0.551, P. E. 3 9 zC; ISI[1,, and d5‘: pdtI)dp 5, I(V x }l)od§= I_Ipsin¢dI1dp S But (V x / '1): sin¢ci: + V2U= -3-(2xy+yz)+ 2-(xz + xz)+ —0‘I-(xy) 036 03: fit = 2y. 47. 45 (b) . ' l é’ 1 6? 6° V2V= —— 2' 2 . —
_' -223-—' — ' 22 2 p0_}0p( s1n¢+ p)+p_( p'sin¢ apsm¢cos¢)+ d_(psin¢+ cos 05) I 1 . I . 1 1 I = —(: sin¢+ 4p)— 3 (: psin¢+ 2:‘cos2¢)+ 2cos' ¢. I ,0 /7 72 = 4+ 2cos3¢— "2 cos2¢. ' P 1 I (C) , 1 6 1 1 _ , 1 , _. , V'f = 75-r—[r—2;cose sin¢ + 21-‘¢1] + ’__, Sine 56-[— sin esin¢1lnr] 1 . + r_, sin, 0 [- cose sine lnr] 1 = ;, —cose sinI1(1— 2lr. .'- cscze lnr) + 6e ‘I
II P. E. 3.12 ' If B is conservative, V x B = 0 must be satisfied. I; .5. 3. 3. 1 V x B: ax ay 62 " y+zcosxz x xcosxz = 0c1,, + (cosxz- xzsinxz— cosxz+ xzsinxz)E1,+(1—1)&_. = 0 1 Hence B is a conservative field. ll 48. 46 Prob.3l (3) d/ = pd¢; p= 3 g If 37: L: [d1 : 3[d¢: 3(§—Z): —4—= 2. 56 I (b) dl= rsin0d¢; r: 1, 0: 30°; % 7!" L= [dl = rsm0_[ d¢ : (l)sin3O°
[(3)-0] : 0.5236. 0 (C) ‘ dl= rd¢ g ,7: 71' 47: L— [d1:r[d6v: 42--E): -3- :4.19 3 Prob.3.2 (a) I ds: pd¢dz 5 g It l07r S: [dS= p[[d¢dz: 2 [dz [d¢ = 2(5)[3—3} : —6—: 5. 6 0 5 "= ""‘ 3 (b) In cylindrical, dS = p dp dy , 3 n P s: Ids: [pap M5 = ;i= 114 I 0 (c) In spherical, d S = r’ sine dd; dB 21: s: [dS= 100 sin6a9 [d¢= 100(2n)(—cos6 1.}: -.: ,u. [§" 21! )-[ = 2001:
(0.5—o.7o71): 7 7.58 .4 —. —__ 49. 47 Pr0b.3.3 (a) 61'! ’ = drdyd: V: [dxdy 2: [dx[dy[d; : (1) (2- 1)(3— —3): g (b) dV= pd¢dpdz I 5 4 V: [pa'p[dz[d¢ : (4— -1)(n — g): %(25— 4)(5)(2T“)= 35:: :1 2 I 5 3 - 0 H (c) dV : r’ sin9 dm9d¢ Prob 3.4 x: I y=3, [p, dS: [ [(x-’+xy)dydx =0 y=0 ’§- N) “I 50. 48 Prob. 3.5 [H-3] = [(x3dx+y" dy) I. But on L, y: x’ dy= Zxdr I 3 6
"— —-’ 4 -5-: ’‘—'—i 1-06667 [H. d1-0[(5 +x .2x)dx— 3 +26 ()-3+3-_. ______ Prob.3.6 In a a 2 . Bnaj V: [ [ [r sm6a9drdd)= (1—cosa) 6:0 9:0 r= u 3 n 21ta3 1 naj V(a= ;) , = 3 - 1: na’ 21ra3 V(a=3)= 3 (1-0: 3 F Prob.37 (3) _ - I =0 ‘:2 = / . -=3 , (:0 [ . d1: [(x"—z2)dy[_:0 + [(2xydx[:0+ [(—3x2‘)dz[M i=0 . ’= ) . ‘=/ ) ' 0 2 1 xzfi 3 2 Z3 3 — + ()2 — a°’ -rag I rV9
_ (39 sin9 _ sine I N9 1 , - V x (Sing) = ;s1n0 ao rV0 Thus, VI» = V x (sine) Prob 3.24 (a) VV= (6xy+z)Z1,+ 3x c-1,+xc_1: VeVV= 6y 2 _6_ 2 I VxVV= ax ay 02‘: _0__ II 6xy+z 3x2 x I (b) VV= zcos¢ c-1,, -zsin¢ E1.+pcos¢ 01, I O : —-— zc "' = — -" = V VV I a( os¢)+zcos04~0 zcoso zcosd) 0 060 p p 9 9 = I VxVV= 0 . 62. 60 — I 4 , (c) V V = :2—
(24r’)cos0sin¢+ rrsjzsg (cos‘ 0sin’ 0) 4 — rl sin, 0cos0sin¢ _ 4cos¢ , 4cos0sin¢ , I = 24rcos0sm¢+ Sing - 8cos¢s1n0- I V x V V = 0 ' Prob. 3.25 V" ' (a) ' (Va17)T= 3T: 6yzc1,+ 3xy"01,+ 3x"yzc1_. I (b) ' 0T 07 07-" _ - _ _ _ ax + _; 67+ 25-; = x (y a, + 2xyza_. )+ y(2za. + 2xya, .+ x"za, ,) X + z(2y01,, + x y0_. ) = 4yzc_1,, + 3xy c_1,, + 4x2yzZ1, V o F6» 1) = 3
(2xyz+ xy3 + x’yz’) = 6xyz+ 3xy3 + 3x2yz2 (d) I = 2(x2 + y’ + z’): 2r2 Prob.3.26 : ——x —-— —- ()V"_ (")+a(")+a(") arroxr oyyr ozzr where r" = (x3 + y3 + z: )"’2 63. 61 I1 11‘ 2” 2 2 27*’ -"7 * : ~": -/ Vr r=2x (§)(x +y +2 )‘ +2y (3) (x‘+y +2‘)- «I: 3 n 1 1 ‘I + 2z’(-2-) (x‘ + y‘ + : :3)- + r"+ r" + r" N = n(x'1 + y2 + 22) (x2 + y’, + 23)? ’ + 3r 7! = nr” + 3r" =
(n+ 3)r" - 8 6 ‘ (b)Vxr”r= —— —— 0 ax 0y 6: FX I‘y 1'2 413 -1 = [2y(5;-)z(x’ + y-’ + z~‘)- - 2z(§)y(x-’ + y-’ + z-’)3"]a, +.. . =0 Prob.3.27 I (a) Let V= lnr=1n‘Ix"+y“’+z" I 0V 1 1 1 x 3;= :3‘2*’’3=7 I 0V. V. V- I VV= —a. + a—a, + 9--a, = -4 I y(: ’+ N’ = [7 OX Oy 02 I’ = _ V 1-. . . (b) Let V V = = -r7 = : a, in spherical coordinates. V’ (lnr)= V V (I11r)= V A=
ri_, %(r’Ar)= —r1,—; d;(r) 1 -12 Prob3.28 (8) - V, =x3+y‘+z3 ‘IV ‘JV V3I’, =0‘*2'+0¢2'+§2I2/' dc c3» 0'2 6" 5 6° : _ 3 2 __ 3 ,2 j :3 d_, F_. = : “— I F¢ = —x‘° sin¢ + y"eos¢ = —p3cos'7¢ sing) + p‘° sin3d)cos¢: , 1 * , V oF= E-a%(p’cos3¢ + p’sin’¢)+ 22- pcos’¢ — 2p cosd; sin’¢ + 2p sind>cos"11)+ p sin3¢ = 2p cos3¢ + 4p sinjd) — 2p cosdw sin‘7¢a + 2p
cos2¢ simia + 22 [F-21$: [V-‘Fay Due to the fact that we are integrating sin¢ and cos¢ over 0 < 4) < Zn, all terms involving cos¢ and sin¢ will vanish. Hence, jFdS= [fizz pdpd¢dz= 22:l‘d¢: l2dz7lpdp 2‘ 2 = 2(21r)(? 0) = 162! = 50.26 Prob 3.34 (8) AodS= J‘V0_. l/—1dv, Vo_A= y+z+x , . I (x + y+ z)cixdydz ) 2 = 3lxdyldyldz= 3 VI _ ’ 1 —-”"= 75-p1, ' 213
E = . 0 1 ' 15s1,r or E _ 290 — 1580 r’ r — — 2p 2p V: E = - 0 _2d : 0 I ' 152,, I’ ' 15s, ,r + C’ Since V(r—>0)=0, C, =0; 290 -' V: l5e, ,r 1 (b) ' For rs a, _ azrs rs , 2 3 5 1 Q. ,.. = p. 1 = 4np. (2, + pzcos¢ 12. + p"sin¢ci_. ) 1 ~ 1 . . , W=3s0 I| E|‘dv = ”J(4p’z’ s1n’¢+p"2"cos’¢+ p‘sin‘¢)pd¢d2dp {*1 50 1:13 ‘3ut I cos‘, tbdd). 0 W 1/2 I si 0 2W TI + : [p ’dp
idzijijsinz ¢ do I -2 0 1:/3 1: 1 [1+ cos2¢: ]d¢= —-+ —si J 6 4 II I 2 ll 64 3 22” "“( 2! 1 2 n’¢d¢: 3I(1—cos2¢)d¢= 0 22” p‘ ' 4 — (0.3071)+— 3 , ! 4 I 0 Q21‘ -1‘. _4p ll 16 255 16 255(—; )(0.3071) + —; —(—3—)(0.7041) + : 417.67+ 239.394+ 838.59 : 1495.6 _ 1495.6 10'“ 2 (361:) . = 6.612 nJ 21: —-— = 0.7401 3 4096 6 J 2 ill} 1 2 it/3 : Ip’dzIz2dzIsin’¢d¢+
[p’dp[/ dz icos2¢dq5 I -2 0 I -2 0 sin’ 1 : 0.3071 p6 4 0.7401) + 7 | (4)(0.a071) 0 (0.30 71) 115. 113 CHAPTER 5 P. E. 5.1 as = 6.11.4.-up 1 1: 11.215: j[10:sin2¢pd. ~d¢1p: _. = 10(2)§ 21: 5 E11 —cos2¢)d¢ : 2401: ¢:0_-: / " I 0 if P. E. 5.2 ‘ 1: p, wu : 0.5x10"’x0.1x10 : 0.51; A V : IR : 10"x0.5x10"’ : 50 MV P. E. 5.3 o : 58x10’ S/ m : ~—-—-—) : -— : 7 : . .1 E
E J ‘W06 0138 v/ ° 0 5.8xl0 W‘ J 8x10‘ _4 . ' J= pvu —-—) u= ?= =4.42x10 m/ s -' P. E. 5.4 The composite bar can be modeled as a parallel combination of resistors as 71 shown below. RL ' Re For the lead, R, ‘ = I , S, ‘ = d" —7tr2 = 9-3 cm2 H o, lS, ‘ 4 | Ii R1‘ = 0.974 m. (2 f For copper. R. = ——l , S. = nr’ = 5 cm2 i‘ ‘ o(. S‘. ‘ 4 116. l14 4| 4 R = --————-—= 0.8781 m.0 5.8x10" x 3x10" R _ R, _RL. _ 0.97-. tx0.8781 “ ’ R, _ + R ' 0.97./ +0.8781 C = 461.7p.0 P. E. 5.5 p, ,,= l’oa, =ax‘7+b : Po(—gx)x=0:: -_b. I x=0 —- pp; Q, = Ipmds = -—bA +(aL2 +5)/4 = AaL" pm = —. v.P= —(—i—(ax’+b)= -zax Q, = _[p, ,,dv = f[(-2ax)A¢x = —AaL’ 0 Hence, Q, =Q, +Q, =—AaL’+,4aL’= o «g P. E. 5.6 ‘ i. to’ 2x10" -9
a, = 500a, , kV/ m N f‘ I I 1 P: xeaaE = (2.25— 1)x 36” xo.5x10“a, = 6.853a, pC/ m-’ ; pm. = Pea, = 6.853pC/ m2 P. E. 5.7 (a) Since P = s, ,xeE, P, =e, ,x, E, P‘ M091 361: 10" 21 ‘ : : "' x : _ ’-“ a, ,E. ion 5”‘ H. 117. 115 (b) E= P = -3-6-K-)i0—9~l—(3—14)10'°-5a -167 +667 v/ =i x, s,, 2.16 [On ‘ ’ ' I ' “~14 ' "= m 1 (C) ! ' D- E _ 2). - _l_ 2 i‘ -e, ,e, — X — 2.16
10“ (3,—14) nC/ m = l39.7a, —46.6ay+I86.3a_. pC/ ml F= —-LHS —-) p‘, =————2 F ' 25,, S V4 But p, = s, ,E = 3,, Hence 2 F ’V ’ -’ p‘2 = ____=5o ,4 ____> Vd2=2Fd 5' a’ SOS i. e. fwd’ " V. .1=V1‘V2= Leas as required. ', P. E. 5.9 (a) Since . .,, =. .,, L D, n = 1261,, D, , = _max +443 02" = D, " =12ax is D I E21 = E11 “" D21 = 828 I’ = 3§(-10¢. ’ +40,-)
= -411), +l.6a_- ‘ I u D, = D_, ,, + 1), , = —12a, — 461}, + 1.66, nc/ m’. D , / -4 2 1.6 2 (b) tan92= D” = -(-)—1§£—)—= 0.359 __—» 6.: 19.75" 2!! (C) 5,, = E_, , = E2 sine, = l2sin60° = 10.392 118. 116 Er] tan6,= 2.5 tans, = -. -I-tan60" = . 4.33 ——» 9, = 77" _ H r. ’ Note that 9, >92. P. E. 5.10 i, l -9 ‘ V D = gag = 36“ (6n,20,- 30)x10" = 0.531a, + 0.1
77ay - 0.265a_. pC/ m2 ll l -9 p5 = D” = |Dl= '3; (10). /36+ 4+ 9(10")= 0.619 pC/ m2 Prob. SJ 1 = Pods. dS = rsined¢dra, , H 2 2x 4 _ 1= — I Ir’sin29d¢dr W, = —(sin30")" r— ’(2n)= -21: = -6.283A r= o¢= o ‘ 4 0 'j—' I N! Prob. 5.2 L '. a 2x -3 1: [J-ds = I J -p—pd¢dp= 500(2na) = ]0001tx1.6x10 =1.61r = 5.026A p:0b=0 I Prob. < 3 a 2x a I = _[JodS' = 10
l J'e""""”pd¢dp = 2011 I pe'”"’/ "’dp p=0Q:0 p=0 llul J‘xe‘”d' = ed; (ar— I) U 119. 117 2011a" I’ruh. 5.4 " I 1: "9 : —3x10"e"' I’ d! ' In: -3(1) ~().3 m. =. I(1>=3.5) -0.3 c’ -160 n. « Prob. 5.5 (:1) VJV = --p, /a V’ : (—: ;(2xy2z)+ %(2x2yz) : 6xy: : pv = —8xyz(2e, ,) = —l6xyze, , «I (b; 1 : p, u = —l6xy‘7zs0(l0")a, , I ’0_, o.5 0.5 _ 5 x, 9 . _ __ 4 j_ __ — 0.5 1-11.43l6(I0) 36‘ ‘! xdx0zdz- l6(361:)(10 )7L 1 : —4(36n)(10")(0.5)’ : -1.131 m/ '. 1 8x10" 8 ' l. .. , 1 = := '-""-—-—-—-—--= ——= . P'"’ ‘(’ ‘” as 3xl0"x1tx25xI0'6 751: 3———-—-——-3 95”” 75 ' (b) 1: V/ R= 9317": 265.1 A ((-1 P= lV=2.38(> kw 1| _ pl 4.04 11d’ _ ' l’roh. 3.7 1.11 R: -3- ——» p : RS/ l= -I-5-; --7-=2.855xl0 ” ii 0 = I/ p = 3.5xl07 S’m (Aluminum) 40 _ 6 ,
113» .1: HS: —1———: .).66x10 A/ m“ I‘-¢’x90xI("‘ ill’ 1 J : GE : 3.5x0.1616.. -10' : 5.66.: -10" A 1 n1: . 1 1 : 207te"a’(%— 1)e°"'1,, " : (1+ 0) : 23.11a-’ . 1 e 120. 118 : —~ 5: 111-’ : 1:11’/11.11: 0.-(mm. l= N27rR = M10, D : 6.5 mm 150x71 (6.5)x10" R : ————-————, ——— : .4212 (0.4)- ‘I 5.8x10'"x7r -I” x10‘/ ’ ‘ | l° pcl S 2 - 3 -4 - -4 S , ,= 111; = n(1..
>) x10 = 2.2.>1tx10 , Prob. 5.9 (L1) R: 5,, = n(r, ,-’ -1,9) : nu-2.25)x10“ : 1.7510110" 1 = I 1.77x11.8x10"" .5 P_«Eg RR 5 S 225nx175nx10"‘ 3 . _. / ._. __. ..L: ‘ " _: _. ___. .'____. j'_. ..j = . 7 . R R” R° R, +Rn p_, +p, , I 10 1.77x10'“ + 11.8x10'“ 02 mg 3, 5,, 1.751tx10" 2.251rx10" 1, _£, ,_= 0.3219 = ' (b) V—1,R, —1oRo -—-—> 70-— R, -——-L669 0.1929 f 1,
+ 10 =1.19291,, :60 A I) l I = 50.3 A (Copper), 1 = 9.7 A (steel) 1‘ _ 10x1.77x10‘*' (c1 1- 1,5” 10-. ill Prob. S. l0 '1 , I, l h 2 5' R - — - ——-—-— --—-—-———- = 4m. () 1 ’ 0S “ 6n(b’ - a2) = 10’n(25- 9)x10"' , Prob. 5.ll | Pl= n| p|= 121211 = 2ned = x. e.. E (Q = 2e) _ 2nea' _ 2x5x10”x1.602x10"’x10"” E -9 5" «'-3% x10‘ TI = 0.000182 XL’ 2, : 1+ 7”, :
1.000182 , - 121. 119 11>1:%'| p1: 2x10"’x1.8x10‘-” : 3.6xI0'" I 1>:11>1a, : 3.6x10""a, C/ ml But P= xee, ,E or Xe = L= =0.0407 1:05 105 a, : 1+ ,1? : 1._04fl l'rob. 5.l3 (:1) E = —VV = -—5:£a_. =600:a_. ' I -9 3611 D : e, ,s, E : (2.4)600za, : 12.73za, nc/ m’ 1. pv= VeD= -6-P-‘-=12.73nC/ m3 I 11.0 6 I‘ 1.4 P : x, eaE : : -2-4-‘(12.732)a, : 7.42721, nC/ m2
pm: —Ve P= —7.427nC/ ml Prob. 5.14 1 ii : —V®P: Q, Pp» p, ,3=Pva, ,=5sinay l Prob. S. l5 (LII . - Hlxinv('ou| o1nh'sI;1. I . C . I D’: Q , b E = Q 2 a, , 4nr 41:s, ,r I or 21 I b . ‘ 2 -hrr 41ree, r I For 1' > h =4nQ2a, ——> E=4Q2a, , r near l‘hu. ~“ D = a r > 0 -lnsr’ " Q , a, . a < r < b E _ 4nss, r‘ Q a otherwise 0 a, = 107.1a, V/ m ll . _. A. .. . _.‘ . -_- . .—. -.. .-.
—_. a- : i._ ; . 124. 122 '8 4nQr_, a, , a< r< b r I] ll 0, otherwise Prob. 5.19 (a) ' pa, 0 a, eUE, (-lnrz) = p” 3 = "JE‘dl= pea +C2 3E0r AS r———>oo, V=0 and c_, =0 i At t = a, V(a+) = V(a') 2 , ‘ +c, = 2:“ : ——-> c, = 2:1‘, (28, +1) p0(28r + 2 Vd¢dzlp= ,=7Id¢Idz=501t= I 7.1 A , 0 0 Prob. 5.23 (a) 1! a ‘[04 4 4 I = JJodS = ”l5—€—; r2 sin6d6a’¢| ,=2 = (2)
(5)e"0 ' Jsin6d6 [dip = 40ne"0 ' I‘ (I 0 At t= O.1 ms, 1: 40ne" = 46.23 A U —8?"= VoJ —» pv= —JVoJc'3t | C‘! I V-J = -—]: :—1(r: J,)= i,e""l' . I‘- (‘V I‘- | I 126. At t=0.l ms and r = 2m, p = 5 e": 4598 pc/ m’ « 95 v ‘ 10-’ " ; 5- 3.Ix 36“ _ J ! Prob 524 (a) ——---— 2741 I0 ' ' 0 _ [045 T I1 I 10-’ 3 _ 6’ 361: _ 4 ( ) 6 - 10_, , —5.305xI0 s 11/'9 80 I 5.- . .._)_c_-3.§‘_7L_
707 ! 6 — I0--I — ' 1 . — . I‘ } Prob. 5.25 (a) Q: Qae"”' —'-V---> '0;-Q, -== . Q, e""r' e"”’ = 3 I i 1 2011s _ T’ 2 1,-‘ ln3 = 18”‘ j in l. 8,80 oT, I0"xI8.2 10'‘ (b) But Z: 6 , e, = ——= ———l5-_;1—= 20.58 f (c ) Qg = e"-»"> = e"‘”'“ = 0.1923 i. e. 19.23% I Prob. 5.26 p‘ = pme" ’> = 0.2984e‘-‘ "-‘ = 0.1898 cm‘ -9 T, =£= =4.42us 0' 5x10 x361: 10 10'“ Pw. ‘IQ;
=%~=0.298~IC/ m3 H. 3' . 'l0- [x8 ‘I 127. 125 Prob. 5.27 (a) E3, = E, , = —300aV, , +500.. . E, ,, = 700: D211: Din —_—+ 8.’E. ’n: £IEIn 7 15., , = 8-’ E, " = $1706.) = 43.75a_. . 4 E3 = -3001 + 500.‘, + 43.750: -9 D_, =e, ,e, E_. =-Ix ( 3611 10 [)2 = €nxe2E2 = 3x 361: (C) E, oa_. = E, cos6,, 70 cose , , = Prob. 5.28 I (3') Pl = EoxeIEI = 2x 30-’ + 50-’ +
702 —30,50,43.75) = —1.06Iax + 1.768ay +1.547a_, nC/ ml -9 (—30,50,-13.75) = 0.7958ax + 1.326ay + 1.I61a: nC/ m2 = 0.7683 ——> 9” = 39.79" -9 (I0,- 6,12) = 0.1 7680, — 0.1061ay + 0.2122a, nC / ml 361: _= _=__ E, ,,= -60,, E2,= E,, =10aX+12a_. D2» = D1” "“"“* 5252» = 515/1. 35 or 132,, = E: E, ,, = T5—€°: (—6a, ) = -46, E2 =10ax— 4ay+
I2a__ V/ m t 9 E-" 102+ 122 3905 e 7564" an : ‘j: : _ j—-) : _ 2 E2" 4 _2______ (c) w, 17005:; -elE|7 1 . 1 10"’ . . , . . .,, = 35,151 = 3x3x 363 (10- + 6- + 12-) = 0.2219 nJ/ m‘ 1 I . 1 _ 11', _. = : E_~? E:l‘ = ‘; .‘-I. .). ‘ 10"’ 367: (10-‘ + 4-‘ +12‘): 0.3208 n1 / m‘ . .t9__; _.: .$' 128. 126 Prob. 5.29 (a) D_. , - 12.1,, - o, ,, 15,, - -6.. . -9a_. D, , D. , E]! = E31 """ 7: : —: "
e, 53 _z»: , ‘3.5z: ,, 9 _ I 7] D, , —; :—D_. , — L580 (-641, + a_. )— — 40, +. a_. D, = 12a, , -14a, +2Ia_. nC/ m’ (12,- 14,21)x_10'9 5, = D, /5, = 10,, = 387.8a, , - 452.421, + 678.6a_. V/ m 3.3): 36“ D. 0.53,, (b) P_, = s, x,, ,5_, = 0.52,, E; = 155 (12,-6,9) = 4a, , - 2a, + 3a: nC/ m2 2 ‘ 0 I 1 pp‘, = V» P, :0 1 1 1 D D 1 122+ 142+ 21’) 10"” 1 (c) w, , = 313,5 5, =
3 8’; ’= 3(————1—0:, —"———= 12.62 mum’ 0 rl : 3.5x 36” 1 . 1 1 1 D, D, 1 122 6’ 9’ 10"" , w, ,., = 3 8-: = 3‘ ‘L +104)’ = 9.839 mJ/ m‘ 5x 361: Prob. 5.30 (a) 1 10" , 2 1 P, = s, ,xe, E, = 1.5x 36“ (2,5,- 4)xI0 = 26.53a_, + 66.31a, - 53.05a: nC/ m 1 0 26.53 I = —V P= ——— .5 = - 3 1 pm, 0 , p ap (26 3p) p nC/ m (b) 521 = 511 = 501-40.- 2.5 5.». =€
—’E1. = fi=5 2 D211 = Din _"—’ | E = aa+5a. -4a_. kV/ m 11 '~ 10" 361 (5.5.-4)x10-' = 110.521, + 110.56, - 88.-I2a__ nc/ m3 D_. = 2,2,5 = 2.5):
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