Hmwk_6_Solutions

Hmwk_6_Solutions - (4»: J04 FHN “(a bw Let V = 10(p +...

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Unformatted text preview: (4»: J04 FHN “(a bw Let V = 10(p + 1).:2 cos¢V in free space. a) Let the equipotential surface V = 20 V define a conductor surface. Find the equation of the conductor surface: Set the given potential function equal to 20, to find: (p+ 1).:2 c0545 = 2 b) Find p and E at that point on the conductor surface where 915 = 0.271- and z = 1.5: At the given values of {15 and 2, we solve the equation of the surface found in part a. for p, obtaining p = ._1_Q. Then 8V 1 8V 8V E=—VV=—— _-_ ___, 6p ap p 345 alt 62 a' 2 ‘ p+1 9 . = ~102 cos¢ap+10 z“ smgba¢—2O(p+1)zcos¢az p Then E(.10, .27r, 1.5) = —18.2 ap + 145 21¢ — 26.7az'V/m c) Find Ipsl at that point: Since E is at the perfectly—conducting surface, it will be normal to the surface, so we may write: E ' E ... - p5 = EOE-n = so =em/E-E : coy/(18.2? + (145)2 + 26.7 2 = 1.32 110 1112 IE! , - —— surface (2) Given the potential field V = 100:1:z/ (:1:2 44 4) V. in free space: a) FindD at the-surface z = 0: Use ‘ . 8 :1; V 100:1: At 2 = 0', we use this to find D(z ; 0) = eoE(z = 0) = —1006c;a:/(a:2 +4) a: C/mz. b) Show that the'z '= 0 surface is an equipotential surface: There are two reasons for this: 1) E at z = O is everywhere z-directed, and so moving a charge around on the surface involves doing no "work; 2) When eValuatiug the given potential function at z = 0, the resultisOforallxandy._' ' c) Assume that the z = Ousurface is a conductor and find the total charge on that portion of the conductor defined by 0 < a: < 2, -3—< y < 0: We have ' _100€oa: - z=0 $2 + 4 C/m2 7 so: §‘;V/°I/22.'15°€5%a ¥‘§('3)(i00)e lilin.A(>2-+4i)j2-_I;15061112—4092110 7-30 z?+_4 1" of ° *"—-, . - Atomic hydrogen contains 5.5 X 1025 atoms/m3 at a certain temperature and pressure. When an electric field of 4 kV/m is applied, each dipole formed by the electron and positive nucleus has an efi'ective length of 7.1 X 10‘19 m. :1) Find P: With all identical dipoles, we have P = qu = (5.5 x 1025x1602 >< 10-19)(7.1 x 10-19) = 6.26 x 10—12 C/rn3 = 6.26 pC/m’-’ '0) Find 51.: We use P = eoxgE, and so P 6.26 x 10-13 -4 = ._ = —_—.—__, = 1.76 x 10 X5 60E (8.85 x 10-12)(4 x 10°) Then 5, = 1+ Xe = 1.000176. A coazdal conductor has radii a. = 0.8 mm and b = 3 mm and a. polystyrene dielectric for . which a, = 2.56.,If P = (2/p)ap nC/m2 in the dielectric, find: a) D and E as fimctions of p: Use _ P __ -(2/p) x 10-9ap 144.9 _ eo(e, - 1) _ (8.85 x 10—12)(1.56) _ p a” Wm Then —9 -9 r) D=60E+P=2><10 ap 1 +1 =3.28x10 a,J C/mz=3..8ap nC/m2 p 1.56 p p b) Find Vab and Xe: Use 0.8 I 144.9 ‘ 3 V;=— —-——d =144.91n —— =12V b A p p (0.8) 9 Xe = e, — 1 = 1.56, as found in part a. c) If there are 4 x 1019 molecules per cubic meter in the dielectric, find p(p): Use (2 x 10-9/p) = 5.0 x 10-29 4x1019 a” Tape“ The surface 2: = separates two perfect dieleCtrics. For J: > 0, let 6,. = an = 3, While 61-3 = 5 where r < 0. If E; = 80ac — 6‘an ~ 30a; V/m, find: a) EN-L: This will be E1 - az = 80 V/m. b) Em. This has components of IE; not normal to the surface, or Em = —60a4, — 30a: V/m. “m c) En = V’(60)2 + (30‘2 = 67.1V/m_. d) E1 = «/(80)2 + (60)2 + (30)2 = 104.4V/m. e) The angle 01 between E1 and a normal to the surface: Use El-ax 80 m5 ‘91 = E1 104.4 ‘ f) DNg = DNI = ErléoENl = X 10~12)(80) = 2.12nC/m2. g) 1312 = ergeoETl = 5(8.85 x 10‘12)(67.1) = 9.97nC/m2. h) D3 = erleoEmax + ErgfoETl = 2.123a — 2.66% — 1.33az nC/m2. P2 = D2 — EQEQ = D2 [1 — (ll/67.2)] = = — —1.068.z nC/mz. j) the angle 02 between E2 and a normal to the surface: Use Eg-ax=D2-az E2 D2 COS 62 = = Thus 92 = cos-1(.581) = 54.5°. Two perfect dielectrics have relative permittivities 5,.1 = 2 and 67-2 = 8. The planar interface between them is the surface r—y+22 = 5. The origin lies in region 1. If E1 = 100a,, +200ay —— 50az V/m, find E2: We need to find the components of E1 that are normal and tangent to the boundary, and then apply the appropriate boundary conditions. The normal component will be E’Nl = E1 - n. Taking f = :1: — y + 22, the unit vector that is normal to the surface is This normal will point in the direction of. increasing f, which will be away from the origin, or. into region 2 (you can visualize a portion of the surface as a. triangle whose vertices are on the three coordinate axes at :1: = 5, y = —5, and z = 2.5). So Em = [100 — 200 — 100] = —81.7V/m. Since the magnitude is negative, the normal component points into region 1 from the surface. Then EM = —81.65 (%),[a¢ — a1, + 2az] = —33.33a¢ + 33.33ay — 66.67az V/m Now, the tangential component will be En = E1 — ENl = 133.3az + 166.7ay + 16.67az. Our boundary conditions state that En = Em and EN2 = (erl/e,2)EN1 = (1/4)EN1. Thus 1 E2 = ETz + EN2 = ET1 + zENl = 133.3% + 166.7% +16.67az — 8.3% + 8.3% — 16.673z = 125ac + 175a.y'V/m ‘ \ Let V(2:, y) = 462‘” + f(a:) — By2 in a region of free space where pu = 0. It is lmown that both Ex and V are zero at the origin. Find f (1:) and V(x, y): Since p1, = 0, we know that V2V = 0, and 5° 32V 62V 6’ f 2 __ __ = 22 _:_ _ = V V — —a$2 + ayz 166 + dmz 6 0 Therefore d2 d Eg=—1652”+6 ¢ 3%: 8622+6$+01 Now 6V d Ex = — :: 8522 + —f- 3 d2; and at the origin, this becomes E40) =s+ d—f dz; i=0 = 0(as given) Thus df/da: [3:0 = ~8, and so it follows that 01 = 0. Integrating again, we find f(a:, y) =' —4e2m + 3:1;2 + Cg which at the origin becomes f(0,0) = —4+ C2. However,_ V(0,0) = 0 = 4 + f(0,0). So f(O, 0) = —4 and 02 = 0. Finally, f(:27,y) = 4162” + 32:2, and V(m, y) = 452” — 4432‘” + 3:1;2 — 3312 = 3(22 — ' Given the potential field V = (Ap4 + B p“) sin 4gb: a.) Show that VQV = O: In cylindrical coordinates, q 1 3 8V) 1 62V V‘V = —— — + ——,— p3p< 39 P? 3¢2 > 1 8 _ . 1 - . = (p(4Ap3 — 4Bp 5)) sm4¢ — F16(Ap4 + 3,0 4)Sm4¢ = 16(Ap3 + Bp's) sin4q§ -— -1—S(Ap4 + Bf“) sin4q5 = 0 b) Select A and B so that V = 100 V and El = 500 V/m at P(p = 1,¢ = 225°; 2 = 2): ’1 First, E=—VV=-2Kap—lfla¢ 6p . p 8¢ . = -4 [(Ap3 — Bp'°) sin 445 an + (Ap3 + Bp's) cos 4915 ad and at P, E; = —4(A —— B) a,,. Thus |Ep| = i4(A — B). Also, Vp = A+ B. Our two equations are: 1‘ 4(A _— B) = i500 and A+B =100 We thus have two pairs of values for A and B: A = 112.5, B = —-12.5' or A = -12.5, B = 112.5 a“ Let V = (cos 2gb) / p in fires space. V _ a) Find the volume charge density at point 'A(0.5, 60—",1): Use Poisson’s equation: _; 2 _:_ 13 Eli - i‘fl p”_ E°VV~ 60(pap pap +p23¢2 (1 8 <~cos2¢> 4 cosZcp) 360c0s2q’9 =—Eo — —— —— 53p p 92 9 113 So at A we find: 3 1900 m = M = —l2eo = —106pC/m3 0.53 -—-—' b) Find the surface charge density on a conductor surface passing through E (2, 30°, 1): First, we find E: 8V 1 6V E: —VV =—a—pap- _ cosZgb a + 2sin2¢ a, 92 p P2 4' At point B the field becomes cos 60° a + 25in 60° 4 p 4 The surface charge density will now be E3 = a4, = 0.125 a,."+ 0.433 as ' m = $11331 = WEB] = $045150 = $0.399pC/m2 The charge is positive or negative depending on which side of the surface we are consid- ering. The problem did not provide information necessary to determine this. Coaxial conducting cylinders are located at p 0.5 cm and p = 1.2 cm. The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at 100V and the outer at 0V, find: a) the location of the 20V equipotential surface: From Eq. (16) we have 1n(-012/p) _» ln(.012/.OO_5) We seek p at which V: 20 V, and thus we need to solve: V(p) = 100 V ln(.012/p) _ .012 ln(2.4) => p- 20 = 100 b) Ep mm: We have 3V W 100 Ep=-———-_——-= 8p dp p ln(2.4) « whose maximum value will occur at the inner cylinder, or at p = .5 cm: 4 . 100 =____=,2 04v =22.8kV Em“ .005m(2.4) 2_8x.1 /m J c) eT if the charge per meter length on the inner cylinder is 20 nC/m: The capacitance per meter length is 216061- Q C = m(2.4) ‘ vb We solve for 67-: - - (20 x 10-9)1n(2.4) = — = 3.15 21Ij_eo'(100) I 1y Concentric conducting spheres are located at 7' = 5 mm and r = 20 mm. The reoicn between D the s heres ' filled ‘ - - ‘ - . Spherlz at 0 ‘1: With a. perfect dielectric. If tne inner sphere is at 100 V and the outer 3.) Find the location of the 20 V equipotential Surface: Solving Laplace’s equation gives us V(r) = V0 where V0 = 100, a = 5 and b = 20. Setting V(7‘) = 20, and solving for 1' produces -r = 12.5 mm. b) Find Emma: Use dV VOar E = -VV = —— = —— W m- e V Em“ = E(r = a) = 0 — ——109——— = 26.7 V/nun = 26.7 kV/m a(1 - (0/5)) _ 5(1 — (5/20)) c) Find er if the surface charge density on the inner sphere is 1.0 pC/m2: p, will be equal in magnitude to the electric flux density at r = «1. So ,05 = (2.67 X 104 V/m)ereo = 10‘6 0/1112. Thus er = 11;; ' We want in design a spherical vacuum capacitor wiih a given radius a for the outer sphere. wnicn wul he able in sicre the greatest amount of electrical energy subieci in the constraint that the eiecingfieid strength at the surface of ihe inner sphere may not exceed Ea. What radius 33' ehouid be chosen icr iheinner spherical csnductcr. and? how mucfijfiggg can be stored? .. I __ _ ‘_ A __-._-—...- . 5 Pique" I»: seesaw:- swig“ _ = m" = 4Wé° “’5 ~_- “Wotan-L _,___jnenls___ 4' °~ ‘L-‘D cvka~ 1—11 a :3 -__,\_.._: _. “QM..- __ _._..__... -_ ._ ._..-__ _ __~ _ ...- --__- .— e = a . \3 e a e a. _ <<<< °~ _¥:T________.________ - _._ ____.-_____ _ if“: 2 Q (J'Reb - -<7 _ . _ -__ -___ ._ M _._.__._ ..- _€____.___,E=E-. .L.Hfi_e6¥’:_ezlfij_ _ flF—flwu ' MN""_~—fiE§'Z:\—}e5— swk.) k5 _ u a I _. _ Q ~ “I j k —. ‘IIH film—1 I 01H 01H ' I ' The potential difi‘erence AV bemeenfthem plates of a spherical capacitor iser conchan’r~ E Show that then the elecen'e field'at'the surface of ‘. the inner sphere will be a minimum if a = 1515. I _ Find this minimum value of E. b a Q dz: Q [l_}_] O-V(r) " LMEO? 41mm?» 1: V l 1-} = 4! —-- Q L!“ 43L; I V . a. - h I ; Smeea<0 for a<-i-, andgg>o fora); I' L‘K—E/iindfifcaleaoa—citor has outer and inner conduc- tors whoseradii are in the ratio of b/a = 4/1; The? inner conductor is to be replaced by a wire whosezi radius is one-half the radius of the original inneri; conductor. By what factor should the length be creased in order to obtain a eapacitance equal to that ' of the original capacitor? ' V3.0: lb wherea:=% Ili=_b_=§ , 2kln(;) a: ¢1/2 6: __. -———-—l—;'”' “ " ““--l-2_ ,.__ “Hm—~N‘ 2kC = = —— ' 1n(;‘; 111%) . 1 1, _ meg) _ ln(2 5;)-_ 1n2+1n(;¢;) _ 1.1 _ ‘ Emu"; “1-75? he,» ' 1am ‘- _ -.——___._"=_'1'.WL. ._._.. . i. s :-. i I. i . E . E l ! an » USS @ Current-carrying components in high-voltage power equipment must be cooled to carry away the heat caused by ohmic losses. A means of pumping is based on the force n‘ansmit- ‘ ted to the cooling fluid by charges in an electric field. The electrohydrodynamic (El-1D) pumping is modeled in Figure 6. The region between the electrodes contains a uniform charge p0, which is generated at the left electrode and collected at the right electrode. Cal- culate the pressure of the pump if p0 = 25 rnC/m3 and V0 = 22 kV. Figure 6 An electrohydrodynarnic pump; Area S Since pv -.—’- O, we apply Poisson’s equation '~ ' ' Viv = —— 8 The boundary conditions V(., = 0) = V0 and V(z = d) = 0 show that Vdepends only on 3 (there is no p or (b dependence). Hence sz _ -p° d:2 8 Integrating once gives dV -poz _ = ' A dz 8 T Integrating again yields , ,2 = —£9”— + A: + B a 28 where A and B are integration constants to be determined by applying the boundary condi— tions. When z = O, V = V0, Vo=—O+O+B—>B=Vo Whenz=d,V=0, 2 0——p°d +Ad+vo 28 ' 01‘ _&é_fi A— s d The electric field is given by The net force is d . F=fvadv=p°JdSJ Edz _ VOZLE 2— :ld . —poS[d :28(z dz) Dag F=PaSVoaz The force per unit area or pressure is p = g = pOVO = 25 x 10-3 x 22 x 103 =_ 550 N/m2 _ I _ . ,. ‘h 4””. - '5.“ ' ‘1 ‘ ‘ _ V ._ _ M W“ Whig“? l .. . ‘ .. . .~ ., ._. _. _.._ __ - _- ___. .:’N_'.'.___ em / Farce reguired +a move +his‘ Q T@ charge upward =35Q2/(2xf ' ‘r (’3 '* . h l T The second studen‘i‘ WW7 77777777777777? CGICUICH'ES C15 ‘FOHOWS: I . i I 00 1' . z I , _ _ KG a -3., l H...) I] WOFK —f(2x)2 - Inmal state ." h ' .a This is the correct answer. No‘te-tho'i' iF two real charges Q and -Q were being pulled apart symmetrically, the tow work done " would beKQZ/Zh, but the agency moving Q wo'ulci Supply only halF 0F i+-. , Given the current density J = —10“[si.n(2z:)e‘f"yam + cos(2:t)e"2yay] kA/m2: 3.) Find the total current crossing the plane y = 1 in the a.” direction in the region 0 < :L' < 1, 0 < z < 2: This is found through 2 1 2 1 I=//J-nl daé/ / J-ay, dmdz=f / -104cos(2z)'e_2da:clz 5' S o 0 Fl 0 0 - . 1 = 404(2); sin(2a:)loe'2 = —1.23 MA b) Find the total current leaving the region 0 < 1:, a: < 1, 2 < z < 3 by integrating J ~dS over the surface of the cube: Note first that current through the top and bottom surfaces will not exist, since J has no 2 component. Also note that there Will be no current through the m = 0 plane, since Jz = 0 there. Current will pass through the three remaining surfaces, and Will be found through I=/23/(;1J.(—ay)ly=oda;dz+A3A1J-(ay)ly=ldxdz+A3AIJ-(ae) 3 1 3 1 = 104/ / [cos(2a;)e"° — cos(2:r)e_2] da; dz — 104/ / sin(2)e"2y dy dz 2 o 2 o . = 104 sin(2:z:)i:(3 — 2) [1 — e-E] + 104 (l) amen-214 z=1 dy dz 1 0(3-2)=Q 2 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 ' _ 6.7., 8],, . __ ___.___ ‘4 -2'y__2’ —Zy = V J aw + By 10 [2cos(2a:)e cos(22)e ] _0_ asexpected ‘5 .Lqu _ . v I J=4ogsmearA/mg @ 7' +4 ' ' a) Find the total current flowing through that portion of the spherical surface 7' = 0.8, bounded by 0.17r < 9 < 0371', 0 < 43 < 2w: This will be 2" 3" 40051119 2 4OO(.8)227r f3" . 9 t = . = .s sm9d9d¢=—— sm‘dfi I //J H‘s da ./0 [in (-3)‘+4( 4-64 .l7r .31r - = 346.5 f g [1 — cos(26‘)] d9 = 77.4..x .lvr' - b) Find the average value of J over the defined area. The area-is Zn .311' 9 Area = f (.8)2 sin 9 d6 dqb = 1.46 m‘ 0 .l1r The average current density is thus J Mg = (774/146) 21., = 53.0 a, A/mz. (19' Let '2. 20 J=— — . pa” p2+0.01 a, A/mz a) Find the total current crossing the plane 2 = 0.2 in the az direction for p < 0.4: Use 21r .4 I —20 I: J. d = . d d . nz=.2 a A ./o-'P2+.01p p 45 1 _ _ (E) 201n(.01 + p2)l:(27r) = —207rln(17) = -l78.0A b) Calculate apu/Bt: This is found using the equation of continuity: 3”” = —V-J = 1.3.03.5) + 6‘72. = 13(25) ' a < ‘20 )=Q c) Find the outward current crossing the closed surface defined by p = 0.01, p = 0.4, 2 = 0, andz=0.2: Thiswillbe ~ - .2 271' '2 ’ 2'” I=/ / -—ap - (—a..><-01)d¢dz+ / —a,. - (am-em 0 0 .01 - o o .4 2" '4 -20 d d 27r .4 _20 +_/c; A p2+_01az‘(—az)p p ‘/O‘ maz'(az)Pdpd¢;-Q since the integrals will cancel each other. d) Show that the divergence theorem is satisfied for J ' In part c, the net outward flux was found to be zero was found to be zero (as will be its volume integral). is satisfied. and the surface specified in part b. , and in part b, the divergence of-J Therefore, the divergence theorem Then the electric field is found by dividing this result by 0: ‘ 3 9.55 — 27ra’pl a” — 7a” V/m The voltage between cylinders is now: 3' a -— — .- ' 9.3:) 9.5:; a 4.88 7ap-apdp—Tln(§>—TV Now, the resistance will be b) Show that integrating the power disSipeted unit volume over the volume gives the total dissipated power: We calculate 4. .4 P=VI=-§(3)=—146 W ‘ l l which is in agreement with the power' density integration. W ...
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This note was uploaded on 05/26/2008 for the course ECE 302 taught by Professor Ferguson during the Spring '08 term at Cal Poly Pomona.

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Hmwk_6_Solutions - (4»: J04 FHN “(a bw Let V = 10(p +...

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