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Fall_2007_HW_7_Solns

# Fall_2007_HW_7_Solns - Problem 3.32 Find the gradient of...

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Unformatted text preview: Problem 3.32 Find the gradient of the following scalar functions: (3)?" = 3&2? +22), ('0) V = 13/334, (c) Uzzcos¢f{l+r2), (a) W: E'Rsinﬁ, (e) 324%.?-2 +513, (f) Nzrzcoszq), (g) M=Rcosﬁsiﬂ¢. Solution: {11) From Eq. (3.?2), 61: 6 FT:—i——i Z {x3 + 2332 W - (b) From Eq. (3 .72], W’ : iy’z4 + £21322“ + 34%;. (c) From Eq. (3.32], Zrzcostj) n zs'mtj) A cosd) {1+r2]2 —¢r[1+r2)+z]+r2- 1"F'Uz—l' (a) From Eq. (3.33), FW 2 —ﬂe_R sin 3 + é[e_RfR] cos B. (e) From Eq. (3.72], S— — 43:21:77“ +1.23, 1VS: i gé+ 3,323+ 1 3+:— _ igxe'z +3.13}!2 — E4123”. {1) From Eq. (3.82], N : r23 C052 (1), n 1 EN 3N A VN— — r“Ell — N+¢;\$ +z—— — f2rco52¢—¢2rsin¢coa¢. (g) From Eq. (3.33], M : Roosﬂsintj), a 3M » 1 3M 1 3M « c054) VM—R—+H——+¢RsiﬂB—a¢ —Roosﬂsi11¢—Bsinﬂsm¢+¢taﬂ.ﬂ Problem 3.39 For the vector ﬁeld E : it: — \$3722 — 21y, verify the divergence theorem by computing: (a) the total outward ﬂux ﬂowing through the surface of a cube centered at the origin and with sides equal to 2 units each and parallel to the Cartesian axes, and (b) the integral of 1V -E over the cube‘s volume. Solution: (:1) For a cube, the closed surface integral has 6 sides: ngnds =Fmp +FLuuom+Fﬁgm +Fleﬁ +F£rmt+Fbacb 1 1 1 1 : —f xydydx = (Xi)| = '3': 12—1 _—1 4 y__1 I——l l l A A h A Fm : 1—4 y__1 (n3 — \$3122 — my] Iz——l [—zdy dx) 1 1 1 1 = xydy dx = (ﬂ) = D, x=—l y=—1 4 J'=—1 1'_—l l l n A A A Fright : 12—1 z:_1 [1122 — m2 — Ely) |y=1 [ydzdx] l 1 1 23 1 —4 :—f zzdzdxz— 1—)l 2—: 12—1 32—1 3 22—1 x:_l 3 l l A A A n Flag 2 ——1 —1 [m — 39:22 — zry) Iy——1 [—Vd'z dx) 1 1 1 l _ =—f zzdzdx=— E) =—4. I=—l z=—l 3 z=_1 1‘: l 3 Ed —1:u+11+_4+_"'l+11+11—_E 5g '5‘ 3 3 _ 3' Problem 3.41 A vector ﬁeld D : {'13 exists in the region between two concentric cylindrical surfaces deﬁned by r = ] and r = 2, with both cylinders extending between 2 = D and z = 5. Verify the divergence theorem by evaluating: (a) ﬁD-ds, (b) va-Dd-V. Solution: {3} f D'dSZFm+Fouter+FbottDm+Ftep1 FEE]: f; [ti-r3) (—frdszHr—l 2 [— —r Hdquyﬂr: 2—101: || || hahhé‘z” hgﬁﬁhhh II t: D m .‘-'l :D[[1'r3} (lwrdzd¢])|_2 II t: [r4 r.-a!.z-d.1n)|r=2 = lﬁﬂn, =0 231(ﬁr-3) - [_ird¢dr)}|z=u = [11 TL E. Fm}, ([rr3)- [swim] L:j = 0. 1 {1 Therefore, HD - ds = 15911:. (b) From the back cover, V-D = {1 fr)[3f3r)(rr3) = 412. Therefore, ff vndvz f:nf¢:f: 4r2rcfrd¢cfz= (({r r4)|i_l)|:ﬂ)|:=o=150m Problem 3.45 Verify Stukes‘s theorem for the VEClIDI' ﬁeld B = (i? (:05th + 1351314)) by evaluating: (3) % B 'tﬂ over the semieh'cular contour shown in Fig. P3.46(a], and C (h) f [v x B) -ds ever the surface of the semicircle. 3 Solution: (3) j£B-dl: B-dl+]B-dl+ 1341, L] L1 L} B -d1 = [i‘reosd)+\$sin¢] - (i-dr+\$rd¢+2dz) = rms¢dr+r5m¢d¢, Liam: (fiﬁrcosder) 4:0. zﬂ+ (ﬁﬂrsqunj = Miami, 2:0 L2 2 L2 1‘ x 11])& x —2 L3 ﬂ L1 2 l 2 (a) (b) Figure P3 .46: Contom' paths for (a) Problem 3.45 and {1}] Problem 3.46. hill-cf]: (figreosﬁm'r) 220+ (A:Drsin¢d¢) = D + {—ZCOSqJJILU = 4, LiB-dl: (/:2reos¢dr) ¢=ﬁrﬂ+ ([41:1551'114352'43) z =(_lr2)|_ +2: 2 5gB-dl:2+4+2:8. r=2, z:{} (b) va— vx[i-rcos¢+\$sin¢) (:— 2.3M) (2m) — 2) +2; (\$251)?» — £99354”) : m+\$u+a§tsin¢+trsm¢n = ism): (1 +1), ffva .25. = féf; (hind) (1 + a) «{ira'rqu) :f;£:DsJ-ﬂ¢(r+l)drd¢= ({—eos¢(%rz+r))|:u)l IE ¢=n =3- Problem 3.49 Find the Laplacian of the following sealer ﬁmetiens: (a) I? = 413923, (b) V = 1? +yZ+zx, u)F=3ﬂﬁ+ﬂL (d) F = ﬁe'rees I1), (11-) F = lﬂe'RsirlE-l. Solution: (:1) From Eq. (3.110),?2ny223) = 3:23 + 2413122. (wwwew+m=ﬂ {e} From the inside back eever of the bunk, \$112112): v2{3r'2) = 121:4 = ﬂ. 0!) Wﬁe'rcosqa) = Se_"e05¢ (1 — é — g). {e} thlem 4.5 Find the total charge on a circular disk deﬁned by r 11'. a and z : E} if (a) pa = psﬂCDSq) (Cfmz), (b) p5 = p505iﬂ2¢ ((391112), (‘3) P5 = Pang—r (chug)! ('1) p5 = page“ sing \$110119), where p30 is a constant. Solution: (3} a 211 r2 :1 21'! szpsdszf f psgcesdJ rdrd¢=Psn — 5111(1) :13. r=ﬂ 41:0 2 a D (b) a 21: a 2n _ 9:] f [3505].“qu rdrdﬁmﬁ f (M) 614) 3:0 ¢=u 2 u. I} 2 _ panel _ 5in2¢ 2“ _ I'm2 w ma- {‘3} a 211 a 2:] f page'rrdrddbzlnpsuf re'rdr r=u 41:11} a : Expat} [—re'r —e'r]: = Empsuﬂ —e_”[1+a]]. {d} a 21-1 {32 [ page—rsinztjj rdra'tj) =0 H a 21: = pan f FEE—rdrf sin2 ¢d¢ r n ¢=u = Psﬂ[1—e_a(1+ﬂ}]'” = Ipsuﬂ - (”(1 “)1- Problem 4.9 A square with sides 2 In each has a charge of4ﬂ MC at each of its four corners. Determine the electric ﬁeld at a point 5 m above the center of the square. Figure P43: Square with charges at the corners. Solution: The distance lRl between an}; of the charges and point P is |a| = «12 +12 + 53 = 1/21 E Q [R1 R2 R3 R4 2 man IRP + IRP + |R|3 + IRI?‘ _ Q —i—j‘r+is+i—§r+is+—i+jr+i5+i+§r+is _ 41:89 {2a)3f2 {2am {2am (my: 5 5 40 c: 1.42 2% Q :i X ’“ = —>q1a-'5 me}:i51.2 (kwm). {2?}3f1 11:29 {ET} 31"? nag TIT-Eu. Problem 4.14 A line of charge with uniform density p; extends between 2 = —Lf2 and z = LIE along the z-axis. Apply Coulomb’s law to obtain an expression for the electric ﬁeld at any point Prlr, (1), 0] on the x—y plane. Show that your result reduces to the expression given by Eq. (4.33] as the length I. is extended to inﬁnity. Solution: Consider an element of charge of height dz at height 2. Call it element 1. The electric ﬁeld at P due to this element is dEl. Similarly, an element at —2 produces dEg. These two electric ﬁelds have equal z—components, but in opposite directions, and hence they will cancel. Their components along i- will add. Thus, the net ﬁeld due to both elements is A. 2p; cosﬂdz _ i'p; cos Eldz dE:dE dE : —_ 1+ 2 1 4142032 2mm where the cosﬁ factor provides the components of dEl and dE-z along i'. Our integration variable is 2, but it will be easier to integrate over the variable B from B : 0 to Figure P414: Linc charge of length L. Hence, with R : rfcosﬂ, and z : rtanﬂ and dz : :i'sm:2 3 dB, we have Lﬂ an an 33 E: dEzf 515:! E '3” “05 rscclﬂdﬂ 3' 3:0 0 a =n' P; fﬂcosﬁdﬂ 211280? In A P: A [3; Lil 21' 51113 21' —. new 0 2mm 1fr2+{Lf2)2 FarLEsr, sz H] r2+[Lf2)3 N ’ and P E=i‘~ I (inﬁnite line ofcharge). 2113811!" x—y plane Problem 4.20 Given the electric ﬂux density D = i29%)?) + \$1316 - 2y} (Cfmz), determine (a) pv by applying Eq- (4-26), (h) the total charge Q enclosed in a cube 2 n1 on a side, located in the ﬁrst octant with three of its sides coincident 1with the 1-, y—, and Z-axes and one of its corners at the wig-in, and (e) the total charge Q in the cube, obtained by applying Eq. {4.29:}- Solution: {:1} By applying Eq- (4-26) —v D—me 2 ]+3(3 —2 )—o Pv— _ 3x + y 3y I y — i (b) Integrate the charge density,r over the volume as in Eq. (4.2T): 2 2 2 Q:/V’-de=f f dedydz=u 'V 1:0 y=D 2:0 (c) Apply Gauss." law to calculate the total charge from Eq. (4.29) Q = jgn'dﬁ :Ffront +Fback+Fﬁgm +F1eﬁ +Fmp+Fbatbom1 . (i dz dy) x=2 2 2 Fm = f f {i2(x+y3+a3x—2y)) y=D 2:1] 2 2 2 2. 1 :1 f 2{x+y)i dzdy = 22 (2y+ —J12) l I = 24, y=o 2:11 F2 2 L0 y=0 2 2 ﬂack: [ f (iztwywm—zyn «(—idzdy} 12:0 2:1] x:0 2 2 2 2 = —f 2(x+y) dzdy = — 2312 = —8, F0 2:“ 1:0 2::- y:ﬂ 2 2 Fright = f f (mm + 343:: — 2y» *(S’dzdx) 1:0 2:1] 33:2 2 2 3 2 2 = f f [31: — 2y) dz dx = z (—x2 — 4x) = —4., 1:0 3:0 2 y:2 2:0 1:0 2 2 A A A Fm = [—9 [4020 +2!) + 2(3x - 220) {-2 dde} I— “ F0 2 2 3 2 2 = —f [3x—2y) dzdx : — z (—12) = —12: 1:0 2:0 2 y:0 2:0 x=0 2 2 Fmp= [ 0f nti2tx+y3+ir£3x—2y)) -{idydx) I: 2: 2:2 2 2 2/ f 0 dydxzﬂ, 1:0 2:0 2:2 2 2 FM: 1 0f “(22(x+y)+3:[3x—2y)) -{idydx) I: 2: 2:0 dydx: 0. 2:0 2 2 =/ f 0 1:0 2:0 ThusQ: 3€Dvd5224—8—4—12+0+0:0. Problem 4.22 Charge Q1 is uniformly distributed. over a thin spherical shell of radius a, and charge Q; is uniformly distributed over a second spherical shell of radius 5, with b :5 :11. Apply Gauss’s law to ﬁnd E in the regions R < a, a {i R <1 I}, and R I} 5. Solution: Using symmetry considerations, we know D : ﬁDR. From Table 3.], d5 : ﬁRZ sinﬂdﬂ dd) for an element of a spherical surface. Using Gauss’s law in integral form (Eq. (4.29)}, %D'd5 : glut: S where QM is the total charge enclosed in 5'. For a spherical surface of radius R, 2.31. 1: . . {RIDE} . {1:32 SinEldEl (111)) : QM, 41:11 3:11 Dﬂazmnn— cos 13E,L = QM, _ Q1131 DR_4ER2' From Eq. (4.15), we know a linear, isotropic material has the constitutive relationship D 2 BE. Thus, we ﬁndE from D. {a} In the region R a: a, Q1201: n, E : REE : I D {Vim}. (b)Intheregiona-{R{b, n R as = a, E = HER = 4,338 (Wm) {c} In the region R 3: b, A ﬁt + as = 91+ 92, E = HER = {9‘ 92} (Wm). Problem 4.24 In a ccrtain region of space, the charge density is given in cylindrical coordinates by the function: pv 2 50m" (emf). Apply Gauss’s law to ﬁnd D. Solution: tel—II— —' r Figure P424: Gaussian surface. Method 1: Integral Form of Gauss’s Lari-Ir Since p1,. varies as a function of r only, so will D. Hence, we construct a cylinder of radius r and length L, coincident with the z-axis. Symmetry suggests that D has the functional form D = fl). Hence, jgﬂ-ds = Q, S [fD-dszﬂﬂm-LL Q=2mLf SDrE-J—r-rdr D : 1001:L[—rze'r+2[1 —e-*(1 +r})]., sz'sz'Sﬂ [Eﬂ —e'r{1+r)) —re"' . r Method 2: Differential Method v'szv-J DZE'DH with 1),. being a function of r. 1 a ; E (rDr) 2 50m“, a —r E (rm) 2 50:33 , fr % {rﬂrlldr 2 fr Sﬂrleﬂdr, 0 ﬂ rDr = 5:3[2{1— 3-11 +r}) — rig-r], 2 Dzi‘rDr = f'ﬁﬂl:—{1—3'r{1+ I'D —re"' . r ...
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Fall_2007_HW_7_Solns - Problem 3.32 Find the gradient of...

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