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Unformatted text preview: cauEQt-uua 57%th hereafter-astute numeesw‘e ears/ices _ - flea-neat e. Cementee Engineering Sepertrnent ice, see “sea a . memes .I l... eat an I Let V = lO(p + 1):2 cos 45 V in free space. (a) Let the equipotentiul surface V = 20 V define a conductor surface. Find the equation of the conductor surface. (1)) Find p and E at that point on the conductor surface where qb : O.er an a z 1.5. (c) Find lpgl at that point. Given the potential field V = lOO.r;/(.r2 + 4) V in free space: ((1) Find D at . the surface = 0. (b) Show that the : = 0 surface is an equipotential surface. (c) Assume that the a = 0 surface is a conductor and find the total charge on that portion of the conductor defined by 0 < -r < 2, ——3 < y < 0. ® Atomic hydrogen contains 5.5 x if)25 atoms/ni3at 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 effective length of 7.1 x 10‘” m. ((1) Find P. (b) Find 6,. ' ® A coaxial conductor has radii a = 0.8 mm and «£7 = 3 mm and a polystyrene dielectric for which er 2 2.56. if P = (2/io)ap nC/m2 in the dielectric, find: (a) D and E as functions of p; ([2) Va], and Xe. (c) If there are 4 x 10'19 molecules per cubic meter in the dielectric. find p(p). The surface .r = O separates two perfect dielectrics. For .1: > 0 let e,. 5.1 = 3, while 5,; = 5 where x < 0. If E1 = 80213. — 60a. ~ 303: Wm. find: C l= rt ((0 EN}; ([1) ET}; (c) E;; (d) the angle 9, between E and a normal to the c; «,4 .9 surface; (e) DNZ; (f) 073; (g) 1);; (/1) P2; (1') the angle (92 between E; and a (9 $61 3 4’ h normal to the surface. 8 4, (e iw *5 t G A t *5 Two perfect dielectrics have relative permittivities 6,, = 2 and 6-3 = 8. The planar interface between them is the surface .r -— y + 22 = 5. The origin lies in region 1. le; : lOOax + 2003), — 503E V/m, find 5;. (given the potential field V : (Ap‘l + Bp—4)Sin 495; (a) Show that VZV _ O (7') Select A and B so that V = 100 V and E : __ _ ' (fir—22.59.32”. , l 500 V/m at P(p— i, Let V(-r. y) = 463" + ftx) -— 3323 in a regiOn of free space Where pp 2: 0. _It is known that both EA. and V are zero at the origin. Find f(.t') and V(.r, 32). Let V = (cos 2q§)/,0 in free space. ((1) Find the volume charge density at point AtOj. 60: . l). (b) Find the surface charge density on a conductor v surface passing through the point 30.. 30“. i). Coaxial conducting cylinders are located at ,0 = 0.5 cm and p = l .2 cm. The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at lOO V and the outer at O V. find: {(1) the location of the 20 V equipotential surface; (1)) 15,, "m; (c) 6,. if the charge per meter length on the inner cylinder is 20 nC/ni. Concentric conducting spheres arelocztted at r = 5 mm and'r = 20 mm. The region between the spheres is filled with a perfect dielectric. If the inner sphere is at lOO V and the outer sphere is at O V: (a) Find the location of the 20 V equipotential surface. ([2) Find Erlnm. (c) Find 6,. if the surface charge density on the inner sphere is 1.0 uC/mz. ' @ we went in design a smartest yarns-unthdatiacitdr with a siren fail-‘55.? “liaiiiifigfiuiar‘ sphere. whiten wit! he atria tn atnra the-greatest amount or electrical energy $135933 in the ennetratnt that the electrinfiald strength at the euriaca’ at the inner sphere may not exceed Ea. What radius h' ehnuid he. chosen. tar theinner spherical centimeter. and i hora much energy can be stirred? The potential fiifi'erence A's? between the A * L1i( V) l . plates of a spherical capacitor isltept' constant: MN“ ‘ A b Show that then the electric field at the surface of i the inner sphere will be a a = fir. Find this minimum value or” E. lg —— ——-—-——- 5 l3) A quindrteai. capaettnr has neuter art-dinner sundae-l} anyway 3/1 tiara whose are in the ratio or? sir/n =é/l. 'l"rt'e'=i inner ennduetnr ' is to he replaced by ass-ire entree; . I radius is one—hafi the radius of the orig-inalinner ennduetnr. E? tel-tat factor should the length he in—_ creased. in. order tn obtain snap-animus to that . n? the original capacitor? ‘ ' ' Current-carrying components in high-voltage power equipment must be cooled to carry away the heat caused by ohmic losses. A meansof pumping is based on the force transmit— ted to the cooling fluid by charges in an electric field. The electrohydrodynamic (EHD) pumping is modeled in Figure 6.1. The region between the electrodes contains a uniform charge p0, which is generated at the left electrode and collected at the right electrode. Ca1- culate the pressure of the pump if p0 = 25 mC/m3 and V0 = 22 kV.' rigtre 6.1 An electrohydrodynamic pump Example 6.1. Argo-2m: E = 550 'M/ml l Area S / ' A point charge Q is. located a distancez‘r from 3. Conducting plane. Deter- mine the locuscf paints in the plane at" which the magnitude of the surface mmgdmmyaogmt AMWflJ ;{-cigggk ED} 4 A. posztive poth charge {2 is fixed '10 cm above a horizontal CUHdUCHng plane. An equal negative charge -Q is to be located some- Awsz: 1when: along the perpendicular dro ’ t . V pped from Q4 to the lane. Where a. . am can —Q be placed sot-hat the total force on it will he 23m? 1 ‘3 ‘ 3.0 g C l6 a.) In the field of the point charge over the plane (Fig. H i, if you genesis field line that starts out from the point charge in a horizontal .1 ' - _ FT" S Irectlon, that is, parallel. to the plane, where does it meet the surface AM i R a l 3 n of the conductor? (You’ll. need Gauss’s law. and a simple integration.) HGURE 3 Some .field lines for the charge above _ 7 V the lane. field strength at the surface, p The dete ‘ the surface charge density Pf . rmmes b) A charge Q is located 12 cm above a conducting plane, (1;; £310) Asked to predict the amount of work that would haveto be done to move this charge out to infinite distance from the plane, one student says that it is the same as the work-required to separate to infinite distance two charges Q and - Q which are initially 211 cm apart, hence W =KQ2/2h. Another student calculates the force that acts on 'the'charge as it is being moved and integrates F dx, but gets a different answer. What did the second strident get, and who is right? "AN: W 2 lV’tra-gthZ/‘thl ® Given the current density .l = - lO4[sir1(2:t)e‘_3—"aI —t— cos(2x)e‘3-"a\.l l(.A../‘11‘lzi ((1) Find the total current crossing the plane y = l in the a}. direction in the region 0 < x < l, 0 < z, < 2-. (1’)) Find the total current leaving the region 0 < x, y < l, 2 < z < '3 by integrating E - 515 over the surface of the cube. (0) Repeat part (b), but use the divergence theorem. Let 3 = 400 sin 19 /(r2 + 4) ar A/mZ. (a) Find the total current flowing through that portion of the spherical surface r = 0.8, bounded by 0.17: < 9 < 0.3%, O < (15 < 272'. (In) Find the average value of} over the defined area. ‘ Let 3 = 25/pa,, —— Ell/(,02 + 0.0!) a: Almz. (c1) Find the total current crossing the plane .: = 0.2 in the a: direction for p < 0.4. ([1) Calculate apt/lit. (c) Find the-outward current crossing the closed surface defined-by p = 0.01, ,0 = 0.4, 2 = 0, and z :4. 0.2. ((1) Show that the divergence theorem is satisified for .i and the surface specified in part (c). 69:) Two perfectly conducting cylindrical surfaces of length E are located at, p = 3 and p :: 5 cm. The total current passing radially outward through the medium between the cylinders is 3 A dc. ((1) Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having or 4-— 0.05 S/m is present for 3 < p < 5 cm. (17) Show that integrating the power dissipated per unit volume over the a. volume gives thetotal dissipated power. fins 3 (a) E = [(9'55ypmap Wm, V = _ R = (1.63)/ Z S2, wherel is the cyfinder_lengt A“ wag ; (not given) (19) 14.64/l W '. 1. Mfg...“ (c210: “1221:0311: = 2 .. _, . (12);: = 0.10,E(.10,.27r, 1.5) = —1s.2 145%. — 26.7az V/m (c) 1.32 ire/m? (a) D01 = 0) = —(100«sox)/<x2 +49? (a) 12.5 mm (b) 26.7 kV/m (c) 4.23 (with given p5 = 1_0,,LC/m2) ’ (LO-0.92119 me) long: Gaga (a) 6.26'pC/m2 (12) 1.000176- 0 ' " " " ' ' ’ L1 ((2)133 = [(l44.9)/,0]ap V/m, D : ' _- 21 i h "-t 3 (3.283,:VP nC/m2 (b) Vab =192 V, Xe 21,56 ‘3) Wmm‘i’i_ 511(H‘TC’9 be a" (C) x 10—29)/p]ap C h m 7 . ., i ,. , ',,.._,__-_ ® (a) 80 Win (/7) ~60a 1 — 303s V/m (c) 67 l V/rn ‘ t r I .1 l. - a »—1.23 (d) l04l.4 V/m (a) 40.0“ (f) 2.12 nC/m2 (g) 2.97 @( ) M (b) 0 (C) 0’ as expecmd‘ 7 . 2 , nC/m; __ 2.663)! _ nC/mz (a) 53.032' A1111 (‘1') 1.70211~ — 2.132}. — 1.063: nC/m2 (j) 545° ! (a) "—178'OA (b) O (C) O @ 125.2. + l7Sa_\, V/m @ ;f(.1‘.‘.\=) 2 ~49?‘ + 3x2. V(_x, y) = 3(x2 — y?) 6‘) (b) A = 112.5, B = —12.5 or A = —12.5. s = 112.5 (a) — 106 pC/m3 (b) :l:0.399 pC/m2 (depending on which side of the surface is considered) ® (a) 1.01 cm ()2) 22.8 kV/rn (c) 3.15 302 HM e 7 558% Let V = 10(p'+ 1)z2 cos 45V in free space. I; 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: (,0 + l)z2 00345 = 2 b); Find p and E at that point on the conductor surface Where gt = 0.271" and z = 1.5: At the given values of cf) and z, we solve the equation of‘the surface found in part a. for p, obtaining p = 4Q. Then ' ‘ V E: 471/: $273.9- lfiagb- alaz —1022eos gs ap + 10 Then E(.10, .27r, 1.5) = ~18.2 ap + 145 at — 26.7 az'V/m c) Find {pal at that point: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: p5 = EOE-n M = eofiffif) i. ecu/E - Egy’ (18.2)2 + (145)2 + (26.7) = 1.32 nC/xn2 S ace - I m "Z . Given“ the footential field V = 10032 / (a:2 4? 4) V. in free space: a) Find’D at thesurface z 4—- 0: Use * - . a a; ' 100:5 Ef‘VV- ‘100245;($2+4) are—“armzméévm At A: 0‘, we use this to find D(z = O) = 60E(z = 0) = -—1005;J:z:/(x2 +4) a2 C/Inz. b) Show that the'z L—— 0 surface is an equipotential surface: There are two reasons for this: I 1) E at z = O is everywhere z—directed, and so moving a charge around on the surface involves doing no "work; 2) When eValuéfting the given potential function at z = O, the result is O for all :2: and y. ' c) Assume that the z =1 Oilsurface is a conductor and find the total charge on that portion of the conductor defined by 0 < a: < 2, —3-<: y < 0: We have ' £00603.- ' z=O " £62 + 4 p; = D - az C/In2 160$ " . " 1 ' *2’ ' 2' . -- .j 932 +4 dz: dy. (3)(l0f))eo <2) +4)|0 15060111 _ n. I ~ g. . Atomic hydrogen contains 5.5 x 1025 atoms /m3: at a certain temperature and pressure. When an electric field of 4. kV / In is applied7 each dipole formed by the electron and positive nucleus has an effective length of 7.1 x 10—19 m. a) Find P: With all identical dipoles, we have b) Find er: We use P = eoXeE, and so P 6.26 x 10-12 e=__.=____.____——_=.6 *4 X 50E (8.85x10—12)(4x103) 17 X10 Then er = 1 + X5 = 1.000176. A coaxial conductor has radii a = 0.8 mm and b = 3 111m and a polystyrene dielectric for V which 6,. = 2.56. .If P = (2/p)ap nC/rn2 in the dielectric, find: - a) D and E as functions of ,0: Use P (2/p) X 10‘9a 144.9 E : ———-———— = —-——-————————p—-— : 60(5. -— 1) (8.85 x 10-12)(1.56) p 3” Wm Then 2 x 10-9a 1 3.28 x 10-9a 3 28 D: E+P=—-————!’ ——— =____._P 2: - 3p 2 60 p [1.56 + l] p C/rn p nC/m W b) Find Val, and X5: Use 0.8 144.9 - Vab = _/ T dp=144.9111(—3-) = 192V 3 X5 = er — 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 P (2 >'< 10-9/p) _ 5.0 x 10-29 P=N=Wap p apc‘m P = qu = (5.5 x 1025)(1.602 >< 10‘19)(7.1x10_19) = 6.26 x 10~12 C/m2 = 6.26 pC/m2 fr“ : «t / \wwfl The surface a: = 0 separates two perfect dielectrics. For a: > 0, let er = 67-1 = 3, while 6,-2 = 5 where :c < 0. If E1 = 80395 -— €3an — 30a; V/m, find: a) Ele This will be E1 -a1 = SUV/m. b) Em. This has components of E1 not normal to the surface, or E111 = ~60ay —— 30a; V / m. c) ET1 = «(60)? + (30)2 = 67.1 V/m. d) E1 = «(80? + (60)2 + (30)2 = 104.4V/m. e) The angle 01 between E1 and a normal to the surface: Use f) Dm = 13m = ErleoENl = 3(8.85 >< 10-12)(80) = 2.12 no/mz.‘ g) 1312 = meoETl = 5(8.85 >< 10—12)(67.1) = 2.97nC/m2. V h) D2 = ErleoENlaa, + ErzéoETl = 2.12am -— 2.66% — 1.33512 nC/mZ. i) P2 = D2 — 60132 = D2 [1 — (1/52)] = (4/5)D2 = 1.70am — 2.13ay — 1.06az nC/mz. j) the angle 92 between 1E2 and a normal to the surface: Use ' '9 =—-’—-=-———l—=———————'—-————=.581 'COS 2' E2 D2 (2.12)2 = (2.66)2 +0.33)2 Thus 92 = cos“1(.581) =_ 54.5°. ' Two perfect dielectrics have relative permittivities 67.1 = 2 and 5T2 = 8. The planar interface MM" between them is the surface a: —y+22 = 5.7 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 cenditions. The normal component will be EN1 2 E1 - 11. Taking f = a: — y + 22, the unit vector that is normal to the surface is ‘ _E__1_ _ “WWI—«Elam “2&4 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 z = 5, y = —5, and. z = 2.5). So Em = (l/x/EMlOO — 200 — 100] = —81.7 V/ m. Since the magnitude is negative, the normal component points into region 1 from the surface. Then Em = ——81.65 — a1, + 2az] = —33.33a,, + 33.33ay — 66.67az V/In Now, the tangential component will be ET1 = E1 —— ENI = 133.3aI + 166.7113, + 16.67az. Our boundary conditions state that Em = Em and ENz = (EH/6,2)EN1 = (1/4)EN1. Thus 1 E2 ; ETZ + ENg = ETl + zENl = 1333-33 + 156.731, + 16673.2 -- 8.38% 8.38.24 -— 16673.2 = 125ac + 175.3y V/m Let V(:r, y) = 462$1+ f '~— 3y2 in a regionoffiee space where p1, = 0. It is known that both} Em and V are zero at the origin. Find f and V(a:, y): Since p1, = O, we know that VZV = 0, and so ' ‘ 62V - 82V d2 f 2 = __ V____ = 23 __ __ = V V. 81:2 + 83/2 165 + (£032 6 Therefore d2 d I _];=-1652”+6 => i=~862m+6z+01 da; d1; ‘ NOW av ‘ df _ __ __ 2m __ Ea: _ 8:5 86 + d2; and at the origin, this becomes df 1 . —— 8 + a; z=o — 0(as given) Thus df/da: [3:43 = —8, and so it follows that 01 = 0. Integrating again, we f(a:, y) =' —482‘” + 39:2 + 02 which at the origin becomes f(0,0) = —4+ 02. However, V(0,0) = 0 = 4 + f(0,0). So f(0, 0) = -—4 and 02 = 0. Finally, f(:z;,y) = —-4eg‘” + 33:2, and V(a;, y) = 4&2” —— 4e233 + 332 — 3312 = 3(ch -— yz). _ ' Given'the potential field V = (Ap4 + B p4) sin 4¢i a) Show that V2V = O: In cylindrical coordinates, 15 8V5 182V 2 __,____ ____I __ vv‘paplaplfifé‘aw 13 _ . 1 . l . ‘ _ pap (in/)3 -4Bp 5))51n4qb— Fla/1,24 +Bp 4)sm4¢ 16 . - 16 _ . = 7(Ap3 +Bp—5) sméqfi —.;2-(Ap4 +Bp 4) 5111445 = O b) Select, A and B so that V = 100 V and = 500 V/m at P(p = 1,45 = 22.5°;z = 2): . . First, V 8V » 5V 1 E—~VV————B—p—ap—;E-¢—a¢ I I = ——4 [(Ap3 -— Bp-E’) sin4q5 ap —{—'(Ap3 + Bp's) cosélqbafl and at P, Ep = —4(A — B) ap. Thus 1131:} = i4(A — B). Also, Vp = A+ B. Our two equations are: a . * 4(A f 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 fl @ 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 Let V = (cos 2¢) / p iii-free space. : 3.) Find the volume charge density at‘point ‘A(O.5, 655, 1): Use Poisson’s equation: . 2 p” = ;Eov2V =._ED 1 a pap 5—p— 35—332- ‘(1 B (—coqub) 4 cos2¢> 360c052q§ :: ~50 —— —- ——2- = pap p p p 9 So at A we find: 3 1200 pw; = W = —1250 = —106pC/m3 0.53 b) Find the surface charge density on a conductor surface passing through (2, 30°, 1): First, We find E: ‘ 7 8V 1 6V L E=~VV=———a ————a 6p P p aqb t cos 2d 25111245 = a + a p2 p #2 ¢ At point B the field becomes EB = 905460 ap + 25mm an = 0.125 a,;+ 0.433 as, ' 4 I The surface charge density will now be pa); = 11133; = ieglngl = 4—0.45150 = £0.399pC/m2 The charge is positive or negative depending on which side of the surface we are consid— ering. Theproblem did not provide information necessary to determine this. and the outer at 0V, find: V a) the location of the 20V equipotential surface: From Eq. (16) we have __ ln(.012/p) W”) T" 100111(.012/.005) V We seek p at which V_= 20 V, and thus we need to solve: ln(.012/p) .012 I = ———————-———— = = 1.01 20‘ 100 1n(2.4) é p (2-4)” cm 1)) Ep mm: We have E _ £2 _ AZ _ 100 P.“ 6p " dp ‘ was whose maximum value will occur at the inner cylinder, or at p = .5 cm: s E _ 100_ Pm _ .0051n(2.4) c) 5,. if the charge per meter length on the inner cylinder is 20 110/111: The‘capacitance per meter length is C _ 27reoer Q — ln(2.4) V0 ‘ We solve for 51.: - h a (20 x 10-9)1n(2.4) = -—-————_————— = 3.15 2t§o(100) 57' ‘ = 2.28 x 104 V/m = 22.8 kV/m @ .ic cenductmg spheres are loeated at 1' == 5 mm and 7' = 90 mm. The regien between - ietre: ‘1; filled With a. perfect dielectric. If the inner sphere is at 100 V and the outer a a. : - Find the location of the 20 V equipetential surface: Solving Laplace’s equation gives us V(7') = V0 Where V0 = 100, a = 5 and b = 90. Setting V(7‘) = 20, and solving for 1' produees == 12.5mm. b) Find Em”: Use dV V034— E— vv_—H7ai,_rg(%_% V 100 Er,ma.m = E0" = a) = "fl _ E’a/m = 5(1 — (5/20)) = 26.7 V/mm = 96.7 kV/m c) Find eq- if the surface charge density on the inner sphere is 1.0 [LC/13121 p5 will be equal in magnitude to the electric flux density at 7' = Q. So p5 = (2.67 X 104 V/m)ereo = 10‘6 0/132. Thus 6,- = $2.; ' We wem is design a sphericai eewum'eaeecéteé' with a given fadiue e ier'ihe cuter. eehere. which win he able to share the greatest amount ei eiectricei energy subject in the cansiraini that the eieeiqiafieie avenge-a at the euréace of the inner genera may not exceed E9. What radius h' etmuid he eheeen ier iheinner eeheficai centimeter. and haw much energy can he steree? V, I ——~~~~~~~MW"""‘*‘“"" H N ,. 5 z y I ‘ . t) The Potentialzdifi‘erence AV betweenw‘themw w VVVV ' plates of a spherical capaeitor is-kepr constant. ‘ Show that then the electrid field'at’the surface of ‘L the inner sphere will be a minimum if a = 117. I Find this minimum value of E. ' 2 Q 1-; a twaQ‘E r ‘3 . dE. ~ 19‘ x H ’ 5ma<o for a<§. andgoo fora>§ . a2 2 [ b1 _ ‘ ¥ ‘ ' 4% I “ a ‘ E— . item is ndnimum E . = _b_ ‘ ‘ A 6ffi;dfigai capafiit‘o’r her enter and inner canduég tors whose‘radii are in the ratio of b/a = 4/1; T’nei : ' V inner conductor is to he replaced by a wire whosei‘i radius is one-half the radius of the originai irmeriE canductor. By what factor should the length be 3111—}; creased in order t0 obtain a capacitance equal to that ’ of the original capacitor?- ' ' 'wherea2=g21 '5; 1:14:— at Current—carrying components in high-voltage power equipment must be cooled to carry ‘y away the heat caused by ohmic losses. A means of pumping is based on the force transmit— v ted to the cooling fluid by charges in an electric field. The electrohydrodynarnic (El-ID) . pumping is modeled in Figure 6. The region between the electrodes contains a uniform charge pm which is generated at the left electrode and collected at the right electrode. Cal— culate the pressure of the pump if ,00 i 25 rnC/rn3 and V0 = 22 kV. Figure 6: An electrohydrodynaniic pump; Area 5' Since pt, 7‘5 0, We apply Poisson’s equation _ VZV = ~& 8 The boundary conditions VL = O) = V0 and V(z = d) = 0 show that V depends only On 2 (there is no ,0 or qb dependence), Hence dZV : ‘po dz2 8 Integrating once gives dV — z -— = p“ + A dz 8 Integrating again yields ’ _2 V = 1”” + A: + B tions.Whenz = O, V: V0, Vo=—O«l-O+B——>B=V0 Whenz=d,V=O, podz o = — + Ad + V0 28 01‘ _ pod _ Y3 A 28 d where A and B are integration constants to be determined by applying the boundary condi- f The electric field is given by The net force is z=0 .‘ Voz _,_ po 2 ] d A‘ = — I _ d .. poS[ d .2 (z Z) ’03; F=poSVoaz The force per unit area or pressure is p = E = poVo = 25 x 10-3 x 22 x 103 =, 550 mm:2 2-9 A» § ff._;_'..‘>i L" 3%“2‘”? " 16'?“ ‘ “i . .4...— 171 J u _/ .4. _.__ , ._.-H_. E " ' Force req’uired +0 move +his‘ ‘ c @- Charge ‘upword =E€Q2/(27L)Z g T . The second Student W 777W calCUiG—i-es as Fol iows : l . i I °° 2 I ' KG Q a! a, 1‘ Work ~de'x. ifqu x = as initial state C." . h ' ’ This is the correct answer. Norei-hcrt H3 two real charges Q and -Q were being pulled apart Symmetricoliy, the +0173! work done " ‘Wouid beKQZ/Zh, but the agency moving Q would} Supply only he}? OF H‘; _ 4 Given the current density J = «104[sin(2:z:)e—:2'yam + cos(2$)e‘gyay] kA/mzz a) 'Find the total current crossing the plane y = l in the as, direction in the region 0 < a: < 1, 0 < z < 2: This is found through ' 2 1 2 1 ; I=//J-n] 'dat/ / J-ayl drdz=/ / —1O4cos(2a:)_e_2dxdz 5' 3 o o y=1 o o . v l = ~104(2)%sin(2$)‘oe_2 e —1.23MA b) Find the total current leaving the region 0 < :17, a: < 1, 2 < z < 3 by integrating J ~dS over the surface of the cube: Note first that current through the top and bottom surfaces will not exist, since J has no 2 component. Also note that there Will be no current through the :1: = 0 plane, since Jz = 0 there. Current Will pass through the three remaining surfaces, and will be found through I=/:/01J.(—ay)lfi=od:z:dz+1231):!-(ay)|y=1dmdz+L3AIJ-(am) 3 1 3 1 = 104 f / [cos(2m)e"0 -— cos(2m)e“21 dz: dz -— 104 f / sin(2)e'2y dy dz 2 o 2 o , I I I = 104 sin(2w)‘:(3 — 2) [1 — e-Z] + 104 Im(2)e—2yl:(3 — 2) =_ g dy dz z=l c) Repeat part b, but use the divergence theorem: We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume. We have 65:: + gag—y = —10_4 [2 cos(2:r)e“2y —- 2Acos(2$)e_2y] = Q as expected I V-J: ’ Lét' ~__.4DOsin0 J- 72-1—4 a7. A/rn2 ' a) Find the total current flowing through that portion of the spherical surface 7' = 0.8, bounded by 0.11r < 6 < 0.37r, O < 4) < 271': This will be 27r .31r éoosina 2 400(.8)227r_/.31r I 9 ‘ I“//J “lad” /O [1,. (.8)2+4( Hm ‘15 4.64 1,, g 371' 1 I = 34.6.5 ' 5n — cos(29)] d9 = 77.4A .176 b) Find the average value of J over the defined area. The areais 211' 371' 9 Area = f f (.8)2 sinfi d9 dd) = 1.46 III." 0 .17r . The average current densityis thus J Mg ; 4/146) 3., = 53.0 ar A/in2. @g Lat V 25' ' 2:01,. a . J=——-'a ——-——-—-az A/In2 p p— 102440.01 a) Find the total current crossing the plane z --.= 0.2 in the a; direction for p < 0.4: Use ‘ 271' .4 _20 I: J-n d = . d d . z=.2 a ‘/D j; ‘p2+.01p p gs I _ — 20m(.01 + ,0?) .4 0 (271') = -207rln(17) = —178.0A b) Calculate BpU/at: This is found using the equation of continifity: 8p”. 1 8 8.1,, 1 8 8 —20 6t p8p('0 ’0)+ 62 .pap(25)+az (p2+.01)‘ “Q c) Find the outward current crossing the closed surface defined by p = 0.01, p = 0.4, z = O, and z = 0.2: This will be ' ' .2 271' 25 -2 ‘37? 25 . V I=/0I/O TO—l-ap-(——a,,)(.01‘)d915clz+fO f0. Zap-(ap)(.4)d¢dz 2" ‘4. —20 2w .4 _20 z‘ — Cl d " _ fl = +./o f0 p2+.01a < 8‘2” p ¢+fo f0 pumaz (a.)pdpd¢ Q since the integrals will cancel each other. d) Show that the divergence theorem is satisfied for J and the surface Specified in part b. In part c, the net outward flux was found to be zero, and in part b, the divergence of J l was found to be zero (as will be its Volume integral). Therefore, the divergence theorem is satisfied. ' / Two perfectly—conducting cylindrical surfaces of length l are located at p = 3 and p = 5 cm. The total current passing radially outward through the medium between the cylinders is 3 A dc. a) Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having or = 0.05 S /1n is present for 3 < p < 5 cm: Given the current, and knowing that it is radially—directed, we find the current density by dividing it by the area of a cylinder of radius p and length l: 3 ___ __ 2 Jr 27ml ap A/rn :‘ rill-hen field is found by dividing this result by a": V 3 9.55 an = 721p E = 27ro'pl The voltage between cylinders is now: 3‘ 5 ' 9.55 9.55 5 4.88 Now, the resistance will be , V 4 88 1 63 RT; 7 .= w“ = T 9 ' b) Shaw that integrating the power per unit volume over the volume gives the total dissipated power: We calculate ’ ~ z 21 .05 I 2 ' . 2 ‘ 3 3 5 14.64. P: E-Jd = ———-———-—- d d d = 1 _ :— fv ” f0 jog (2w>2p2<.05)zzp p i z 27r<.05>zfl(3) z W We also find the power by taking the product of voltage and current A. which is in agreement with the power; density integration. ...
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This note was uploaded on 08/12/2010 for the course ECE ECE 302 taught by Professor Ferguson during the Spring '10 term at Cal Poly Pomona.

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