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Unformatted text preview: Version “A 1’ M49]: Field W: TeaII Tell #l'
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RyEI'SﬂIl Fag “£123”: ﬂ WIS V Department of Electrical and Computer Engineering ELE401: Field Theory Term Test #1 Examiners: Jilli 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 heelt 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” EthanI M, m ELENI': F to“ Illamy: 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 Daft9W dd: tr'
warms“. 3"
.'. — ,L.
disarm = + f ado” HM 'Tr‘ W .E‘IGL'JHFJAJEE WE In! 7H? HAML‘J'S {ii}Using the Continuity of Current Equation as the starting point derive the solcalled
“defusion of charge” aquation: 1:41“ng W L p» tlt= tutu) e '*”T" "ﬁnfa ~AP ,1 8m” if"? at £=£
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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 dieleuuic material with permiuivity "s". i ."'F I {'1} 1r Lhe millage 31 any poinl beiween the eunuueimg surfaces is given by: __ d. , _.' _
Vipi=+(3anln{2Hl1(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: Thun1 Tan #1 Var5m A Fetaum? M, 200‘ Canm: 1. Three Shari ﬂuteLiens:
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 ELEJﬂl': FinH Titcoy. Cent: L Three ﬁbut‘t Questing: [e] Boundary ConditioosJE metlute] 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
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E15101! Film ﬂurry: Tm Tut #J' an A FMM, 3m Problem 2. A Non—Qniform Eggfore 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 mini1; 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?:
JETf“ = i’F‘Fﬁiz/e Page 6 Fusion “A ’3 Elm}: Field My: Tam Tea: #1
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Cam.. 2. A NunUnit" 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 Mwacﬂﬂpctﬁdﬂﬁ
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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 ﬂieitcuic 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" [:_al 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: ‘ fciZFdzz ,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: — = adse EMJ as ea As Win
were): oars (ea2e) _— :3 cu;
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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”?
Esmes 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
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Team #J Vetawn A February Id. It'lli HEN}. P't‘aH 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 elemrt} 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] VFWum a “I?
‘r/Eﬁ'EZeﬂ 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 min9‘5 Page h—x _ __,.RH_\__ _f“—__,—/— “‘“\ J I: ﬁﬂﬁ‘fﬁlé [ﬁij . .‘U
ELEdM: F‘Ieﬂﬂmy: Tenn ten ii: Vemart “A
February 15. m Cent: 3. Cuueenjﬂ; ﬁghterteal Shells.t
{c} A Linear. ﬁguHuygeneoﬁ Esauepic. NDB—Cnnducﬁng Dialects1:13 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. speciﬁer11:.. 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'{?a3r}) 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: —? rag1'3“ If]? ﬂy t_._—t.._..d
ﬁrsHM” —‘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: italtrial. mew9' '—""'_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 ”
Febmm} 15. M4 Cam.. ﬁ. Qungntﬂg ﬁuhedtzﬂ Sahel;I
Can: {u} A Linuu', Nun—Homugcggng; Isutmpit. NunCunducting 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_
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This note was uploaded on 04/17/2008 for the course ELE 401 taught by Professor Forgot during the Spring '08 term at Ryerson.
 Spring '08
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