ele401TermTest_1

ele401TermTest_1 - Version “A 1’ M49]: Field W: Tea-II...

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Unformatted text preview: Version “A 1’ M49]: Field W: Tea-II Tell #l' Fem-my I'd, ease I h r— "" j g at: Tn YEP—inmate! 34 % {fl j “i an r TH en] sedans RyEI'SflIl Fag “£123”: fl WIS V Department of Electrical and Computer Engineering ELE401: Field Theory Term Test #1 Examiners: Jill-i Sitnfloerg Sections: ti] - DB Paul Ksntorelt Time: -- 1 5‘5 hrs _—_—_—_.——_————- instructions: o Closed book test. Allis sueh as vector operators and inteng tables will he provided if needed. No questions are to he asked during the test. [t‘sorrue aspect oEn question is unclear make suitable mumpth Le. write down the assumptions. and oont'trrne 501mg the problem Calculators will not he required and are not permitted. o Answer all three {3} questions. Weight ofeaeh question is indicated. Part marks will be given. Remember to simplify all answers and indicate their units, and clearly indieate by underlining or “hosting” Il1e final answers. o Show all work in lhl: were presided. If: "fight" answer appears without any indication on how it “'83 derived that InsWer will be considered to be "Wrong". Use Ilse heel-t sides ofpeges it" more working space is needed. Note.- {WJ marks are available; {ill} marks is taken as a full paper. therefore the extra {Ell} marks are bonus marks. Tomi Mari: (as; no Marita (as) Tm rm r: em “A” Ethan-I M, m ELENI': F to“ Illa-my: Problem I: Three Short Questions [2" marks total] {a} Derivations: [I5 marile (5)5]1ow how the "Continuity ofCuznmt" Equation '5‘" J = - fi' 51.. I E! t can be derived from the concept that the total current 'I " flowing out from a ciosed surface is _1 equal. to the rate of decrease of the charges contained in the ctosed Voiorne. [2] m4“; f [Hatrng cry?" I = § F..ng :I_§§ Qfiwc some. t5 V = ‘gfrfid’nr Daft-9W dd: tr' warms“. 3" .'.- —- ,L. disarm = + f ado” HM 'T-r‘ W .E‘IGL'JHFJAJEE WE In! 7H? HAML‘J'S {ii}Using the Continuity of Current Equation as the starting point derive the sol-called “defusion of charge” aquation: 1:41“ng W L p» tlt= tutu) e '*”T" "fin-fa ~AP ,1 8m” if"? at £=£ 51-1 v .5:- s‘; v-E":ug . we Erna: are E cliff: E a: “C”; (flfifJ-Jfi-E fife:— AfiE Liv—J?“ {fie—flewoeu7 com eye‘an Vdflrflfifi-L‘S.‘ ——=l-._.II 34”” V“ D = fl Tans Emafctm Egr'flfl?df‘3 C7-— = fl; _ E- {E' Lad/S? drag-5 3y fl 4.5 mrmfrflyxdfi flax/5" a“; firumué any; awe—:3 say £th 3' Ere—auwflfiacsrdc‘r Wale ‘5. , fig H i5; = _ r a: flea; ,1? s _ 3: as: F'- I'I'EH P E xii E W Max? JMFEQC’HIY/Az’fi z 55m?! dimes—Fm {are} M, F g PM r —" fir - A -_— “FEE.- .fi ‘3- : j gar : E (2% fl g E” a? (etc-a} are.) It" fins) F“be c3 - I;ng Fruneey mgxua foET/Jrf‘flfizfi 5”” 3W7” ‘r Ti: weer: —v—E.:1L ,1“ e2 3 u r @Efl' MAR-'5 (‘9 1‘5} ,1" ¢ A3: A t “A #5“ -"'.I' "grow “1/5 . I“ .1" Tm Tn: nu Vernon A PM Iii. H0! ELEAIJI: PM Them}: Continuation: 1. Three Shun Questions: [h] Fieids , Voltages, and Energy Inside A Coaxial Conductor. [I5 mm'ksl A. thesis! infinitely long conductor centres! 011 I112 zraxis consists ufa solid ehndueling, eyl'Lhder with radius "a" [rn]I and a thin cylindrical conductiun uuter shell with a rsflius “2:” [m]. The spec; bclwuen um conducting surl‘accs is filled with a dieleuu-ic material with permiuivity "s". i .-"'F I {'1} 1r Lhe millage 31 any poinl beiween the eunuueimg surfaces is given by: __ d. , _.' _ Vipi=+(3anln{2Hl-1(23fpl [V1 I s. and the surfsee charge density um the inner muduemr is found in NJ [as =+35Vflf(a1n[2}} [Cfmz] Determine the energy stored in an " L ” [m] length hflhe coaxial mnduemr using We = 5'3 f. p, vies [J] [fljmees WM I- “ wig): + 3% _,§,_£_s) Ev] .6311 (2) F Page 3 1, H J} ELEHI: Field Tm: Thu-n1 Tan #1 Var-5m A Feta-um? M, 200‘ Can-m: 1. Three Shari flute-Liens: Cant: {h} Fields. Voltage and Energy mside A Cumin] Cunductur. {iifll‘ instead Inowng due mlmge, the the electric field E is kmwn at any palm between me eendueling Surfimw and in given by: EIP} = +{3Vufflnli21fll}an[Wm] delemfine the energy stored in an “L” [m] length 01‘ me cunducler 'LL'iing the relationship: WE=V1IvD'EdH-' [41mages rarer. where the field axial. only in the space between the concluet'mg surfiea. » 4‘. m Am] .21:— Be’ Va Eff-“d Page 4 . H 51‘ Tenn Tear ll! Vernon A February 15, mi ELEJ-fll': Fin-H Tit-coy.- Cent: L Three fibut‘t Questing: [e] Boundary ConditioosJE met-lute] Two different dielectric material not on the z = t: plane. Region #1 is considered to be z 1- {l : and Region #2 is r. < I]. The expressions as: the electr‘te field intensity vectors EL and the eluttrie flux density vector field D. in their respective regions are given by: E1 = 31" +31 : and D] = En. 9.1: 'l" 5 3],} Determine the values of the two relative pennittivllles em and en t and soive for the electric field E; and the electric flux density field D1 iflhe surface charge density on the boundary between the - l . |._ - —. '— —- 3 l—a LWIJ tlIElLELt'lCS ts p5 — 2 ED [Ctm‘]. Q T? E. , 33 1.. [2.13.532] —-'—1- fl'h' 3):": J' 42:. =En€rt rag :lflé 0 ti .4 ,_._'~. fir. = EMT 232‘!— : Enérz £2: ' 't/EEéar'C-AJ {’80 fiA./}j5 ,4; ’54 K. W ll dbl I P“ Tl? cl” Tn QE} Mi!” LIKE writ". = at L ’F‘\ LU oil 1. in all \J '- 1? 6-“ 145.1 per”: mfioflfi THE MtETAEE-g. mm! Hmte w [3- r: ' Ties—1.5 r reiterate e13.” cure 'FWE gab“ 4; Foe. creel: redeem Tc“ .- v“ ’5 E15101! Film flurry: Tm Tut #J' an A FMM, 3m Problem 2. A Non—Qniform Egg-fore flhflflt Eigihulinu, [29 marks totall In line spam: :1 nonuniform smfanc charge density [is = up [Um’] (when: at is a constant} is found to exist on Line surFELoe defined by: namagpihnflflsosn I?" «:5 {a} Determim: Ulelotalchargc worm um mini-1|; m] _ a 5 ¢=F '53:“ T ¢zzgf=w 2 JL 3 '— 0‘“? *1 [E] a: in? of) [611' {b} Develop an expression [or the voltage "V" at the originr assuming that IJic ironan atinfiniiyJfimnrfl] 2 =[A m4- 7 , Ufa) ='- g £5 .r' ‘fi‘xfxE'fiE I=fi; Elsa-(36?: JET-f“ = i’F‘Ffiiz/e Page 6 Fusion “A ’3 Elm}: Field My: Tam Tea: #1 PM M, M! Cam..- 2. A Nun-Unit" Ch ' hu {1:} Calculate the electric [Md E at the urigin dut- ID the surface charge tlensitj.r . Express lhE final answer in three fnrn‘als: {t} in Cartesian unit vectors. {if} in Cylindrical unit vectors andfiiflinSpherrical unit vet—trims. I11} marks] = f2; J55“ dag“. My mm M-wacflflpctfidflfi Ami/w“ A _— Meat/o W /'”‘”’"L—-. :35”: xi”:t:fl fl “a [ PH] rah]. r“-—l| $3759 W I Iii:- Fffffl “1x 5.:ng g E: if A 1 w; :4 H a rig at W ~ cry: ,3“ r W1”? at T a < titan F, a} t a ts 1% 1t: / a r t t A. __1 "- ....H -"" a”; Pfirmefl w I ,; Ego} = 2% Matted/“Mo fl Ii Tc?“ 1;?! d _..:' “h ‘n L'—'--l qg—F F-Efl Half—335:? a?“ ? fix '—-—.J 2- r- C? :Cmfidr | 6E can. CL A“ I; l\ {0 Pa" '19; will 1U .15 G .—. — 0"“ Pl ‘5???” +gm¢¢at MAKER“! {rt—ft; K v fig % N? ‘1‘ M K c: i / “i— “I' we “a. @‘H >25 I X + ’“2‘ , rl 'fi‘rga fife» ‘* a PE fl fl' L ‘ ,_; ‘\ ‘f‘F'EaH “7' .2 . {£141}! HEN}: Field Theory: Tm Test #1 Vernon Fm M. 2951' Pmblm 3: Concentric soheficfl Shell; [211 marks tow] it spherical structure consists ot‘two concentric oondueting shelLs eeoLred at the origin. The inner oonducting shell {#1} has a nadius = “e” [m] white the outer shell {#2} has a radius = “2s” Int}. The space between The conducting surfaces is tilted at various times wifl'l different dielectric andfot ooruzlucLhtg materials. [:1] Linear, Homogeneous, Isotropic flieitcu-ic Matefiel. [6 marks] when the dielectric malen'a! between the EUIIstCIing shells is homogeneous it is found to have a relative dielectric constant En = 3.121. The electric l'ttet densityr vector Dm becomes r D{1'}=+[ota2fr2}a, [Cfrn’]; wimreoisaconstant. ('1) Determine the sortaee charge demities pm and [352 on the inner and outer conductor P! [2:17 surfaces respectivehr. 2 mm A. A : see _ a: r: 1 I; = D m A; u if“ use—rm J—= 41- »! =-'1. fl'fifiJfH—H .—._L. J '1 {9 :. D" [:_a-l rd’q H -__——-0(q’2(= 3 2- sz.1. ma- jz=aau A: in” {iijfletermjne the total charge "Q J " on the surface of the outer {#2} conductor. cg; fll=g€6€5: + hgdi$= H5 #5- [1f- 5 s $2. = "i« 5‘27qu 5: ‘- fciZF-dzz ,w «H J — {iiijIl'me voltage in any poinl the capacitor is given by the exprmion.‘ [2? V[r}=oa{Za—r}a’[ficur} m wflffi; determine the voltage between the two conducting. surfaces, Le. find "Va: “ . the voltage on surface #1 with respect to surface #2. m \- .-'--..-. _—\ V02) = ease (53:4: — = ads-e EMJ as ea As Win were): oars (ea-2e) _— :3 cu; (Eric (-34?) r» a: 2 4x) “ \42 1" 6C: LU] All: W Peas ' 6; £5. . fl ,1 seem.- Field 1m.- s'mn m: w Fm A PM hi. WM If I r? 5' M Cum: 3: Concen'ltjg Eghefiea] Shelia (a C . - - e . .n' . '_ 2' I one. {a} Linea:-1 Humngeneous. lamp]: chlectnt: Material. ?‘r' (‘50 qty/6- 1—153 ,r “ij0193 for “C” the eapneitanoe ofthe eonoentrie sphe'gal shell structure. W—j i I Pu _ ._r--—'”— v M 25.... -_F:‘__ _,_,_._-e_—- r13 5:: __ 33“] E911 means mini. rent {a]. '3' “'4 TH'E’ fiufififlfi. it”? Esme-s NJ “*0 ‘ DR V r, [b] Conducting Material. [5 merits] ElmLfl BE JtAFPiE—D K._,_-W“—'——"_‘~—._ The dielectric n'atterial used in part {a} is replaoe wilh a weakly conducting material 1with permittivity e = En and oonduetivity 5 [Sim]. The voltages on the oomiuottng shells are now changed so that the outer shell {#2} is at - 2% [V] and the inner one is at 4 3V, [‘9']; where ‘9'“ is a constant. [1} Use Lapam's Equation. 1? = v = e , in man: for m}, the expression which gives the voltage at an}r point in speoe between the t‘Wfl oonfluet'tng SUI‘EIEBS. Evaluate the oonstants ofintegmtion [into the known boundary vaiues. iii—av = Er; gflffllgrkV>zg m §_L(¢2%Ly) :0, x; .ngéd , “"1 mm 915 ffirgfigAf/Alcf GATE ' "-2." I._ {229V : C " Ifl‘n i = Q LEVI 5H; 31-h it: _ Xe jLPnE—G£fl?/AJ£; DNCE £13.91}. -' _ I I ——~- “EFF EVHLHAEMJG C. ETC—2- ( I VIE):— — C; E_ _ ' J L TL— + a @A=fl; Ufa) :+3L"0 K ml 3 'C' + C: a? __ 2 EQL=JQ; —;LéF :3 +C2n-—?£PJ} {a— go} £ IQ. 55L *2 I "M SLLETPFET Eu: (2) Farm flmfr] + J ff - C: 1 +51"; : 1C3: Pris—1'" = Ef(fli‘fit = '—_- H ,1, e; : — WAVE s: ’3 r- g :4, . fl ’1‘ Team #J Vet-awn A February Id. It'll-i HEN}.- P't‘a-H 13w.- Cartt: 3. Eggneentrlr: Enhgfinel Shel; Carer..- (b) Conducting Meterini. 1:21 "‘ FjIl‘the electric: field intensity vector Eir} heide the spherical structure is given by the expreeeien: Em = +[1fl a V“! r :I! If [Wm] determine the current elem-rt}- Vector Jtr} and Earn it suive For the total current “ I " flowing between the eendueiing surfaees. “F; 3;: chLUHVe a: [fl/mg] ’12 p: E =— fi_.-I- :M j?" A = Mae 0"”; a; . flaASthflafifim fl ¢=Cfl 9:3 _ a? .= fflfiefw [3?] Z: @345 fiery—5"“ a”: +2 fieflE—ff‘” I we; 07% [A] VFW-um a “I? ‘r/Efi'EZ-efl U ref—H 1._._._.——~-’"'M..___,d.rr‘~__._._.h..,_|__._r L l] [iiflDetermine the expression fix the resistance “12” between the two conducting surfaces. I _—Er' at hanw AM“? 599025 W “W Megan. “I '1 ITS" BE: “CA I J min-9‘5 Page h—x _ __,.--RH_\__ _f“—__,—/— “‘“\ J I: fiflfi‘ffilé [fiij . .‘U ELEd-M: F‘Ieflflmy: Tenn ten ii: Vemart “A February 15. m Cent: 3. Cuueenjfl; fighter-teal Shells.t {c} A Linear. figu-Huygeneofi Esau-epic. NDB—Cnnducfing Dialects-1:13 marks] The dielectric material mtween the mndueting spheres is changed to a linear. mu—curuiueiiun. nuna hnmgeneuus, isumeit: material. The relative pemflttivitjr, en. is new a fisruzu‘nn ut‘spaee wfiebles. specifier-11:..- e; = e: {r j. The expressicm litr pfllar'tzaflun vectur P [Ural] Lnside the dielectric is fiJIJJ'Id Lu be given by '. I'{I]= +{aa2frl}{( fia—3r}i'{?a--3r}) a, [Ctm'] while the electric flux density field [Hr] is the same as it was in part [a] Le. Dfrj=+{aa2iir3] a, {D'm’} :11 {i} Determine the {bouts}; polarization smfatsi: charge densities pm and [3951 en the dielectn'ii: “MM medals inner and outer surfaces ntspectively. 2 a; a; t; a. — 3s: \ (it: Fiffimfl : r 43 Eta-'34 32:41 A—‘i‘J— r P :,_gg4i4a—3a =Ha{_‘3_h e "gettfl P5: —? rag-1'3“ If]? fly t_._—t.._..d firs-HM” —‘L 2 F31 HIP if ' Z __ {fa 6Q _._r = 13,1 (ii}Eveiuale the {bound} pnlefizmien vnlumc charge izlisitsit}r psi.- inside the diEiEEfi'il: ital-trial. mew-9' '—""'_1_i 7"}: = _V.P __ 2%{A A: (?fl_3-’z-)Q .F—r' _ flats“? *3¢ ' = 1“ 395d -' f1 (0* F 2 ALE {rem F 5A) /e2(-’§7‘3 ‘34-) m1.- Field mm.- Tc:- Tut H: Vmian “A ” Feb-mm}- 15. M4 Cam..- fi. Qungntflg fiuhedtzfl Sahel;I Can: {u} A Linuu', Nun—Homugcggng; Isutmpit. Nun-Cunducting Dielectric. $.51 [iiijEince the maturiaj is linear, ajmnugh mn—hormgenmus, the the basic relationships still apphr: Le. 93’ an“? D=EGE+P;P=EDX¢E;andEg=l+Xc Us: Lhe given I? and D fi£1fi5 ID solve for the expression 1%): an {r}. _1_ mflc‘ x6: €F__‘ if” P: 6", Page 12 ...
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ele401TermTest_1 - Version “A 1’ M49]: Field W: Tea-II...

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