329sp10he1sol - ECE 329 Introduction to Electromagnetic...

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Unformatted text preview: ECE 329 Introduction to Electromagnetic Fields Spring 10 University of Illinois Roach, Kim, Goddard, Kim Exam 1 Thursday, Feb 18, 2010 — 7:00-8:15 PM l 1 Section: 9 AM 12 Noon 1 PM 2 PM I Please clearly PRINT your name in CAPITAL LETTERS and CIRCLE your section in the above boxes. This is a closed book exam and calculators are not allowed. You are allowed to bring notes on a 3x5 index card *- both sides of the card may be used. Please show all your work and make sure to include your reasoning for each answer. All answers should include units wherever appropriate. Problem 1 (25 points) Problem 2 (25 points) Problem 3 (25 points) Problem 4 (25 points) TOTAL (100 points) {-wrn (over) 1. girth S (25 pts) Consider a static charge distribution consisting of two infinitesimally thin, parallel sheets of charge in the z = 0 and z r: 4 planes. The bottom sheet (at z : 0) has an unknown surface charge density 933 C / m2, while the top sheet has a surface charge density psT of —8 C/m2. The displacement field in the region between the sheets is known to be D : 102 C/m2, which is a superposition of the fields generated by the two surface charges. a) (10 pts) Determine the unknown surface charge density 983 of the bottom sheet. L©s§£_@\@ ‘ .7. @97lZC/m” 5? b) (7 pts) Write an expression for the volumetric charge density 9(33, y, 2) C / m2 in terms of the top and bottom surface charge densities, psT and 983, and appropriately shifted delta functions. fitter) :- 65 Se) + 5; 39,43 C/M; c) (8 pts) Determine D for the region 2 > 4 m. @4271 D 4 N N Mew)», 1?: =+@=°2> e5 3 2 O m L A (\ 232% (over\ glu’efl‘fl S' 2. (25 pts) A cylindrical coaxial cable consists of two conducting regions — a central conductor of radius a and an outer conductor of radius b enclosing the central conductor. A dielectric of permittivity a fills in the space between the two coaxial conductors. While the conductor may be considered to be infinite in length, we would like to consider only a segment of length 2 (>> b). Keeping these in mind do the following: a) (10 pts) Making the observation that the electric field E'should be in the radial direction and a function of r, apply Gauss’s law in integral form to determine the radial field E(r) = Er(r)f._ (Hint: choose a cylindrical surface of radius r and length E as the surface over which the integra- tion of the displacement vector should be performed and assume that the total charge enclosed in the cylinder is +62.) 5le § 5 EU“)? mt Ertr) 7< ante/«Ad war awkward {We/OF what; r out WM 9~ . C? Q- 45 = Gamma = <9 § Answer: E,(r) = Q > ) wife/Q r b) (10 pts) By integrating the resulting Er(r determine the voltage difference V between the two conducting cylinders in terms of a, b, Q, and fl. (Hint: You may want to assume the outer conductor is grounded and find the voltage on the inner nductor, which is biased at voltage v.) a . ~ Q V: " fibre” : 21:99. in») Q 0"” (v) RGYGSL c) (5 pts) Determine the capacitance C of the coaxial cable of length Z. (If you don’t know the formula, capacitance represents the rate at which the charge stored in the coaxial cable increases as the voltage V increases.) Q ~ WEQIL C t T ” Mir/o gzrrrel ,QMU/a.) CF) Answer: C = Lower) gt? (M’ifomg 3. (25 pts) In a certain region of free space7 the Charge density is specified as 2x (a2 + $2)2 p(w, y, z) = Q C/m3. a) (4 pts) Considering the problem’s geometry, what is the appropriate form for Poisson’s equation? gMCQ )3 W5 @«vLy am )9 7+3 or (Cl PvaL%, 2 2 r __ dZV : _Jb(><) v v 4;: .9 3;; ‘ f) b) (4 pts) Give a brief physical explanation of Why is maximum at a: z 0. X C 0) all ‘HNQ 03159.14 aka/mgr, and WW 05AM 3 Wayk‘ +00 “(0 F a «geld M‘sz '3? dwecfiwgg I ”O W , j} aicvgo" . I: u, Q E K g Orqu/MC‘M (0% M 50 O. (064:4 m_ ‘ , o o. __ W SL091 cu} (4:0 Mb “NO-re ‘Haed‘ c) (3 pts) In what 22mm does E point at x 2 0?‘ \ ‘-~~\\ 515% dag Obi/vi ha I :1— ” J Wat/W14 \ + d) (4 pts) Give a brief physical explanation of why E = 0 at :r 2 —00. M ream WSTS‘B 0&3 cc POSF+>Ja wot mega/W Slab wl/xage (law; can al. A (Whetstw 4M2 CW Clint-.454,me resembsz a. at: «Km Add ms 40 am , 57W? Wfirf’cfie 1665;: ((flflnwe‘t‘ W??? Alternative We W QW’W“ (Mid 1'“ theha‘bove 0%servat1 ms to solve o ‘ e) (10 pts) Use r the potenti V(x,y, 2) if the zero of potential 9 S is at :c : ~00. Hint: 2mm 1 _ E {1125:er CMW /(a2 +$2)2 : ‘a2 +3122 +0 .I—UmMcfiQ 61A 44w. ‘Mfiw dx 1 m (3ng W ame /a2+m2=Earctan(;l—)+c Lew, +0 96) 541948 4. (25 pts) Consider the following spherically symmetric configuration of composite materials in steady- state equilibrium (the figure belows details a spherical cross section of the geometry): (i) Region 1: r < a, where r = V162 + 3/2 + 22 is the radial distance from the origin is a perfect conductor (01 z 00) and holds a net charge of Q = —5 C. (ii) Region 2: a < r < b, composes of a perfect dielectric shell with e r: 460. (iii) Region 3: b < r < c, has 6 = 260, (73 = 107 S/m, and holds a net charge of Q = 3 C. (iv) Region 4: 7° > c, is occupied by free space (6 = 60). Note in all four regions ,u = no. Utilize the integral form of Gauss’s law, 3% D-dS = fV pdV, when applicable and what you know about steady—state fields within conducting materials to deter- mine the electric field vector, displacement vector, polarization vector in each region and also the surface charge densities at each interface. a) (7 pts) Determine D, E, and P in region 1. What is the surface charge density, p, in C / m2units, at r = a? 6 glud; 0’: O”, a ye/feef (cMIMCJ-W, all charges UUIHHU MIDMH ’l‘v HM Sodom~ le; ypvsicto :0 Giausgb Lou) gndg: 85w :0 2 650 0 l3: {27: E30 I [W 44443 Can (’WWL MUM? V0 a (cow/umLof //U 9&4ch 3%a2‘e. W SUle‘CQ/ ,4; /5‘ c —Q: v/f/ firz‘! ’71: 5 WT” m “77"; ML (9 KW clwys W w WA (over) fl aArAb b) (5 pts) Find D, E, andPin region 2. Us"; Aqausgrs éfi’dé 3 Zena : ‘gc- CW5? Gaussiafi $144902 05 a 3:510? (a a? é WIH‘ (aohus r m FM» “you, Sum é 1) I6 {aphid/2 lifecfed 0M fa/al/c/fi) 7% JIP/t/enfié? éi/fiééf/ii ‘ ’.-- 7- 9 Li {Z:'§C 37)) =15: :11; :c 7 r V r W}, .. __ .. AV ‘ 3‘; _> _ " o A 5%“ f: 22%: " (gr 9: c) (8 pts) Find D, E, and P in region 3. What are the surface charge densities at 69 JM flay“) 3 (bum) , r¢03am w/a My /0?5/M 2) Val? (5M Melt» ciw - We WSW/u beam; Vail 019 Elm/34 _ a m swig Shh, as w Ma) Eso/ P/ o (‘5? On 7% r: b Surfiu on mdm‘a Q: +§¢mv and Ho «€Cclwgufi r200 'A 1 (>9 0n Hm rsc mace, aracwm 62M: (+54) : +3 a /_ 0L9 MM) ’2 9995b3Q Ci; ‘525C:Q :f—ljfiL W flab Wlml ) a (:0 We USE (paugé's Law gird; : Zena Mae w pOrHQ é: '5 - 229% e0 5,, ) Wearz M Sfu‘fl 1:3 One (av/J clmlc bwwlura FY CMIHWS 0+ “HQ (:01, b,c surfing J70 va/lfia H42 @st . , ...
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This note was uploaded on 04/20/2010 for the course ECE ECE 329 taught by Professor Goddart during the Spring '10 term at University of Illinois at Urbana–Champaign.

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329sp10he1sol - ECE 329 Introduction to Electromagnetic...

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