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20091ee1_1_hw5-sol

20091ee1_1_hw5-sol - 5.1 Given the current density J =...

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Unformatted text preview: 5.1. Given the current density J = —10‘[sin(2z)e"’ag + coe(2c)e"'agl kA/m’: a) Findthetotalcurrent crosingtheplaney=l intheavdirectionintheregion0<z <1, 0<z<2z Thisisfoundthrough I=/LJ-nlsh=L2/olJ-avlu_1dzh=/:£1—10‘coe(2z)e4dzdz = -10‘(2)%ain(2s)|;e" = 4mm b) Findthe totalcurrent leavingthereg'ono <z,:r < 1,2 < z <3byintegrating J-dSover the surface of the cube: Note first that current through the top and bottom surfaces will not exit, since J has no 3 componem. Also note that there will be no current thread: the a: = 0 plane, dnce J. = 0 there. Current will pens through the three remaining airfacee, and will be found through dyds 1=/:/o1.1.(_a,)|'_odzds+/:£J-(a,)|'_1dzdz+/:/01.I.(a,)M = 101:]; [coe(2z)e‘°-coe(2c)e"] da-ds -10‘ [a a I sin(2)e"’dyds = 10‘ G) ein(2z)|;(3 — 2) [1 -e-’] + 10‘ G) sin(2)e-’V|;(3 — 2) = g c) Repeat part b, but use the divergence theorem: We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume. We have 8 8.1 J: 4- VJ = E 7: = -10“ [2 «amp-’1' — 2m(2c)e-’V] = g as expected 5.3. Let J=7+_4"A/'” a) Find the total current flowing through that portion of the spherical surface r = 0.8, boundedby0w1t<0<03ar 0<¢<2m Thiswillbe I: [/1 nl do: fat/1:r%an—O(B)9in0d0d¢=% :5...” do =34“ ;[1_ oos(29)]d0= —.77_4A .11! b) Find the mags value of J over the defined area. The area is Area: [0" [;:'(.8)nsin0d9d¢= 146.1:9 The average current density is thus J." = (77.4/1A6)a. = 53.0m- Alma. 5.5. Let 25 20 __ _— 2 J‘ p " p9+om"A/"' 3) Find thetotalcurrentcroaingtheplnnez =0.2inthen; direction florp<OAz Use ,= 1m =1f1‘-” =—(2) 20m. 01 + p2)|:(21r)= —201rln(l7)— _ —_178. 0A b) Calculate 8p./8t: This is found ning the equation of continuity: §&=_ ___ 81;— 18 8‘ VJ— Pimps: c) Find the outward current croea'ng thecloeedeurfece defined by p: 0.01, p=0A, 2 =0, andr=0.2: Th'nwillbe I= [ti/25 01a, (-ap)(01)d¢ds+//m4a, (a,)(4)d¢dz +11%]p’+018‘(-“)”d”“+/k/Pa+—Dasl (as)Pdpd¢=Q since the integrals will cancel end: other. d) Show that the divergence theorem'neet'ufied SorJ andthe eurfeceepecifiedin pertb. Inpertqthenetoutwurdfluxwae foundtobezero,endinpertb,thedivergenceof.l m found to be zero (as will be in volume integral). Therefore, the divergence theorem is eet'ufied. 5.8. The conductivity of carbon '3 about 3 x 10‘ S/m. a) What she and shape sample of carbon has a conductance of 3 x 10‘ S? We know that the conductance '3 G = aA/t, where A h the crosseactional area. and I "a the length. To malaG=a,wemay meanyregularehapewhoselength'sequaltoib area. Examples include a. square sheet of dimensions l x l, and of unit thicth (where conductance is meuured end-to-end , a block of square dos-section, having length l, and with cros- section dimensions t x s/i,orasolid cylinderoflength (and radiusa=\/l 1r. b) What is the conductance if every dimension of the sample found in part a is halved? In all three cues mentioned in part a, the conductance '3 one-half the original value if all dimens'ons are reduced by one-half. Th'n is easily shown using the given formula for conductance. 5.13. A hollow cylindrical tube with a rectangular cram-section has external dimensions of 0.5 in by l in and awall thicknasofODS in. Alums that the material is bras, for whidi a = 1.5 x107 S/m. Acurrentof2wAdcisfiowingdown thetube. a) What voltage drop is present acres a. 1m length of the tube? Converting all measurements tometers,thetuberes'uta.nceovera1 mlengthwillbe: l R‘ = (1,5 x 107) [(254)954/2) x 10-4 — 2454(1 — .1)(2.54/2)(1 — 2) x 10-41 = 7.33 x 10“ a ThevoltagedropisnowV =IR1 =200(7.38 x10"=m b) Find the voltage drop if the interior of the tube '3 filled with a conducting material for whicha=1.5x 10‘ S/m: 'I'heres'utanceofthefillingwillbe: 1 W =23: x 10-2 n Hg: The total ree'atanoe is now the parallel combination of R1 and Ra: R7- = R¢R¢/(R1 + R9) = 7.19x1040, andthevoltagedrop’nnowV = 2MB;- = .1 V. 6.1. Atomic hydrogen contains 5.5 x 10’“ atoms/n1a at a cousin tempuoture and presume. When an electric field of 4 kV/in 's applied, each dipole formed by the electron and podtive nucleus has an eflective length of7.1 x 10-19 in. c) Find P: With all identical dipoles, we hm P = qu = (5.6 x 109°)(1.602 x 10-19 )(7.1 x 10-“) = 6.26 x 10-11. cm2 = 6.26 pC/m9 b) Find er: WemP=¢ox.E, undue P 6.26 x 10-19 = _ = = 1.76 104 X‘ 603 (6.66 x 10-”)(4 x 16') x Thenc,=1+x. = 1.003176. 6.3. A coaxial conductor bu radii a = 0.8 mm and b = 3 mm and a polystyrene dielectric for which 6.- = 2.56. If p = (2/10)!» nC/m9 in the dielectric, find: a) DandEasfimctioneofp: Use E_ P _ gzmxlo-992 _144.9 v/ ——co(e,—l)_(8.85x10-19)(1.56)_ e “P "' Then —9 —9 D =eoE-l-P = —2 " I: " [&+1 = —3'28 "fi" " C/m’ = _3'28"' 110/!!!” A b) Find Va and x.: Use -' 144.9 3 x.=e,—1=l.56,esfoundinputa. c) Ifthere are 4 x10” molecule per cubic meter in the dielectric, find p(p): Use 6.11. Capacitors tend to be more expensive as their capacitance and maximum voltage, V“, increase. The voltage V"... '5 limited by the field strength at which the dielectric breaks down, EDD. Which of these dielectrics will give the largest CV“ product for equal plate areas: (a) air: 6, = 1, E30 = 3 MV/m; (b) barium titanate: e, = 1200, Esp = 3 MV/m; (c) silicon dioxide: 6.- = 3.78, Esp = 16 MV/m; (d) polyethylene: e.- = 2%, Esp = 4.7 MV/m? Note that V..." = EBDd, where d 'n the plate separation. Also, C = WA/d, and so VwC = 6.6041330, where A is the plate area. The maximum CV”... product is found through the maximum 9E3]; product. 'Ii'ying this with the given materiab yiel<h the winner, which 'n barium titanate. 6.13. A parallel plate capacitor '3 filled with a nonuniform dielectric characterized by e.- = 2 + 2 x 10°29,where z isthe distance from one plate. HS = 0.02m”, and d = 1 mm, find 0: Start by asumingcharge density p. on the top plate. Dwill, asumal, be o—directed, originatingatthe topplateandterminafing onthe bottom plate. Thebyhereisthat D m'llbe constantocer the distance Lawn plates. This an be understood by considering the :e—mying dielectric as constructed of many thin layers, cad: having constant permittivity. The permittivity changes from layer to layer to approximate the given function of z. The approximation becomes exact asthelayerthicknmes approachzero. We knowthat D,which'nnormaltothelaysrs, will be continuous acres each boundary, and so D '5 constant over the plate separation d'stance, and will be given in magnitude by p,. The electric field magnitude is now -£-_L soc, - 60(2 +2 x 10329) Thevoltagebeweenplata'sthen -a V°=/w L- &_ 2—W5‘ I”: p_. 1 (g) o 60(2+2X10‘2’)- Cox/4x13“.n 2 0 502x103 Now Q = p.(.02), and so 6.14. Repeat Problem 6.12 winning the battery is diconnected before the plate separation is increased: The ordering of parameters '3 changed over that in Problem 6.12, a the progesion of thought on the matter is different. a) Q: Remainstheflsincewiththebatteryd'nconneded,thedlargehunowheretogo. b) ps: AsQ'uunchanged,ps isako undlanged, since the plate area'uthe same. c) D:AsD=ps,itwillrelnaintheamabo. d) E: SinceE=D/eo,andasD'nnotrhanged,Ewillaboremainthe&. 0) V0: We require E x d =%, whereE has not changed. Therefore, V0 hm increued by a factor ofl_0. f) C: AsC=£oA/d,immuingdbyafactoroftendecmsquyafadmofIL—l. The same resuh. occurs because C = 0M, where V0 is increased by 10, whereas Q he not changed. 3) W5 Use W. = ”20%! =1/2QV0, to observe its increue by a factor offl. ...
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