ece302hw6 - ‘ CALIFORNIA STATE FOLYTECHNIC Umveaezw...

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Unformatted text preview: ‘ CALIFORNIA STATE FOLYTECHNIC Umveaezw; POMONA Electrical & Computer Engineering Department L. Ferguseh I ' - 7 ECE $32 . ' ‘ HOMEWORK e 5'" ‘ , y ( Q . Determine the work done in carrying a 2-uC charge from (2, 1, ~41) to (8, 2, —1) in the field E = yax + xay along (a) the parabola x = 2312, (b) the , hyperbolax = 8/(7 — 3y); (c) the straight line x = 6y —— 4. , charge on a circular arc, the circle centered at the origin, from x = a to xi: y = (IA/2.. A potential field in free space is expressed as V = 20/ (xyi) V. (a) Find the total ehergy stored {within the cube 1, < x, y, z < 2. (b) What value would be obtained by assuming a unifoxm energy density equal to the value at the Center of the cube? (9 (2) Given E = —-xax —l_— yay, find the work involved in moving a unit positive (3 l @ F0”? 99in: Charges: "53511 05 Value lie-ea; C arekeituafted' at the vertices 14,553,, C B (loeatedfin that sequence) of a square ef side's 1m. Find the foilowing: {ai'the : I..'ih’t3‘t‘k required ta assemble the chatgecdistt‘ibutioti; (53} the additiona'i 'wetit re- quired to move‘the charge at i} to the ceeter of. the Square; and"{e} the additienal work. required to then metre the charge at C to. the center of. the side 5354 4 Am. “(3);” 63.93739 3; (313)192961533; ta) 36.1746555- . " * 5 Three peintcitatges ef values 1; 2, are situated at the eernete ofan equilat- eral triangle of sides 1 m. it is de'sited? te find the work required to move these, ' charges to the corners et‘ ea equilateral triaegle ef shorter sidee was shown in Fig. 3?“ -" .. ' ._...__..—-—_._ Ahshw‘i we: n '59 I . . “itfi‘eo ": fig?! Etingingithree point? 4 chatgesimmjhe WES e: a target .. ., equilaterial ttfieegie ta the eeteets of; 3‘ stnafleijmufla‘ieml tsiengie. ' ' j :Two I'Sphericaiflchargem each of the ariiriémrgciius a and this same; uniform charge ~ density p0 C/mJ, are situated infiriiteiy' Egan, The twn Spherical charges arr} ‘ brought together and made inm afisirigle spheric'él ghargeof uniform density. Find . the following: (a) his mark required if~the radiuspf the new spherical charge is'a: f (h) the work required if- i-he charge: density cf the, new spherical charge is pa; and '.= ‘(ciihe radius and the charge density. of the new Spherical charge {Dr whichihe: WOri'i required is zero. ' K ' ‘ ' A215.- .(a)‘-.1.57sspsa5/sdi; (h), Maggie's/.503; ($251,025,063 ® A Viiirnnc charge'Llistrihutimi Ls given in Sphuricni cnmmiinzncs by, .='__ huh/u) {hr 1' <1: . 1r, ii)" ‘ [hr 1‘ >u in) Find the wmi». required in rearrange the charge with uniiiirnr dunsny within ' ihu rcgiun r- < in ‘ I . (hi Find the mum's (if the- Sphcricui region and the charge density liarwhich nci‘ thcr additional 'wnri-L is rcquirqd hur wnrir. is mth zwziiiubie in redistributing (in: charge uniiiirml‘y within tin: region. ' .... _...—.. .._— .... q-u..,..,-.,._.-. , ... . .. ...-.._....K.. V... . .__.._.._ -._.._. .__...... ..... ... .. ., iChnrgiéis-T and -Q are distributed Uniiiirmly 0n. the surfaces of Lwh I Tsiihcr23"n't“rzidii‘Z! Enid“b"(>71)1'rEEfiEEiWééf."'TiiE'Yifitéi‘"§fifiéf€iii-ifizidé‘up at two, scparuhiu hemispheres. Find the: work rcquircd in carry me two hemisphere-$- in an 1 infinite distance i'rmn tin: inncr sphtrc anti juin them to harm it sphere ui umiuriniy ilimri‘hutud charge -{_). i..———-" fl“ ._ a .. 1’.“ —" @ A sphera of "radius Rama; 5. charge: Q‘dis'tributed unjfbrmly Omar its sill-fact How large a spheretcntaing 90 percenth thegnergy sibred'inihe alectrosiatic figirilaf ihi; chars; histnb‘utmn? Z Q w: ’28“? @ w: ~a2/z saw—:38??? is) WZZOiPT '” 7a; ,W-z; 7T3”; 0.. j—_ Yr: a ck, _. 1°00 2 *7“ Q ‘. ’5 8 W 2 «T U erred?- L. Ferguson CALKFOHMA STATE PoLweeHme 'Umveeszw', PomoNA Electrical s Computer Engineering Department ' eoe sea" , p , HQMEWORK #g' stem“: ' ( . , . , l _ , ' V ' ' © Determine the work clone in carrying a 2-pC charge from (2,1,4) to, (8,2,4). in the field E a) the parabola a; = 2y2: = yam + may along - As a look ahead, we can show (by taking its curl) that E is conservative. We therefore expect the same answer for all three paths. The general expression for the work is ‘- - ' ' _ , ' B I 8 2 =“Q/ E'dL=-q[/ ydm-t-f catty] A 2 1 In the present case, a: =, 2y2, and so y = «Av/2. Substituting these and the charge, we get . 8 2 _ A a ’ . W,=_2x10‘5 x/x/2dz—l-f 2y2dy] =—2x10—6 [3W2] +3y3|2 =_2s in 2 1 3 2' 3 1 b) the hyperbole a: = 8/(7 — 33;): We. find y = 7/3 —— 8/353, and the worlr is 4 @ 8 . 2 __ 7 8 8 2 i 2 3 3:1: » w+ 1 7~3ydy _ 7 8 ' 8 .8 ~ '2 =—2 106— — ._~ _ __ _ I =_ X‘ [3(8 2) 3 ln (2) 3 ln(7 3301] 28 it} c) the straight line a: = 63,: ——' 4: Here, 3; = :E/fi + 2/3, and the work is g 8 , 2 W3 = —2 x 10—6 + dart—f (fig—4) dy] = —28 [,LJ ’ 2 1 ’ Given E = —:ra,,, + gay, find the worlr involved in moving a unit positive charge on a circular arc, the circle centered at the origin, from at = a to :1: = y = (1/ In moving along the arc, we start at qi = O and move to 45 = 77/4. The setup is W=—q/E'-dL=—— .E-ad45a¢=—/V (—zazva¢+‘yay-a¢)adq5 0 0 .W-Jv W4 —-sinq5 cosqb -. 77/4 1r/4 » ' _ ' '7; — f . 253 Sin qbcos qb dd) =n—— / a2 sin(2¢) d9!) = -—a,2/2 a D 0 i5 where q = 1, a: = acos <25, and y = asinqS. Note that the field is conservative, so .We would get the same result by integrating along . a two-segment path over a; and y as shown: 'f W=_/E.dL=_ [f/flrmn/Wfldy] Lag/2 0 A potential field in free space is expressed as V = 20/ 3.) Find the total energy stored Within'the cube-1 < 1:,y, z‘< 2. We integrate the energy density over the cube volume, Where My = (1/2) 50E - E, and Where ’E_=;VV=20[ 1 am+ mZyz The energy is now b) What value would be obtained by assuming a. uniform energy density equal to the value at the center of the cube? At C(1.5,1.5,1.5) the energy density is we = 20060‘(3) = 2.07. x 10-10 J/m3 ‘ This, multiplied by a. cube volume of 1, produces an energy value of 207 pl # ) I Four point charges, each of .value 4414-55 Crete situated at the verti‘ces A, 'B, C, and; D (located in_that sequence) of a square of sides l-m. Find the following: (a) thei ._'W0.rk required to assemble the charge distribution; (b) the additional work 1":-j ‘qured t0 move the charge at D torthe cenier of. the Square; and (c) the additional? waif rggpiged g0 {hgggzgygthe Charge- at'C to the center. of the side AB. 3 A: 68.03780 1 @— (b) Energy stored in the new charge distribution =%[4zso(1+7%+«f§)+4zso(1+l+fi) ' +4fl$o(-\715+1+J§)+4n£0(fi+fi+~/§)] I = 37.333130 I . Additidnal work reqmred A » i = 37.333150 — 63.03720 = 19.296150 J .(c) Energy stored in the new charge distribution = «[2 X 87.333180 =.123.5077£g J, since the geometry is proportional to that in (b) by the factor 1/45. Additional wofk requii'ed _ =123.5077ao— 37.3331ao= 364796503 ' g i /) Three point charges of values l, and 3 C are situated at the corners of an equilat- eral triangle of sides 1 in. it is desired to find the work required to move these. charges to the corners of an equilateral‘ triangle of shorter sides i an as shown in Fig. l ‘ «,1 ll ‘ l Figure l Bringing three point charges from the corners of a larger :3 equilaterial triangle to the corners as smaller} Waters! triangle. , l __ "the notential energy stored'in the system the larger equilateral triangle is given by -1-‘ 1(‘2-43) (r 3) (1' — iv =— l + + + + + 2 Q J 2i: 41750”. 47780 2 471'50 47780 3 41750 41750 _l[5+8.+9]_ 11 Nm 1 _ v, 2 477750 ' a i L I 47T 50 The potential energy stored in the system of three charges at the corners Of the smaller equilateral triangle is equal to twice this value since all distances are halved. The increase in potential energy of the system in going from the larger to the smaller equilateral triangle is equal to l 1/411'30 N-m. Obviously, this increase in en- i ergy must be supplied by an external agent, and hence the work required to move a the charges to the corners of the equilateral triangle of sides i m from the corners of g the equilateraltriangle of sides 1 m is equal to ll/47rao N-m. % Two spherical charges, each of the same radius a and the same uniform charge-i density p0 C/m“, are situated infinitely apart. The two spherical charges are "' brought together and made into a single spherical charge, of uniform density. Find the following: (a) the work required if the radiusof the new spherical charge is a: (b) the work required if the charge density of the new spherical charge is p0; and (c)r'the radius and the charge density of the new 'spherical charge for which the work required is‘zero. ‘ v— ..-___...,__......._._........._.._ _....».—n‘.. “cw—"us ,.__..,, _. _. . _ V 5 Energy stored in the initial charge distribution = 2 x 4‘93“ = ____3"p%“ (a) 2 x % m3po = m3p’, where p’ is the new charge density. pf: 2pc . 2 5 4 -‘ 162: a Energy stored = Ts—g-(ZpQFaS = 75%)— lt’mpgas Expects .. pfia5 Work reqmred =‘ 1580 —- 1580 = 1.6753 50 (h) 2 x g- 7ra3po =§ 75(b)3po, where b is the new radius. b = 213a f ' I 2 5 _ 42: 2 1/3 5 _ 12.69927L'p0a Energy stored —- --——1580 po (2 g a) — 1580 . 12.69927: 2a5 87: 2a5 Work required = ——1-5—8(%)—— — lggo 2 5 =o.9s42 319— so 27*” (c) If. the required radius and charge density are he: and kzpo, respectively, then; 4 3 4 3 3 2><§zra p0 = Emma) kzpo or, k1k2=2 2 x 4;:p3a5 = 47:(k2p0)2(k1a)5 1560 1550 ' or, = 2 Solving, we get k1 = 2 and k2 = 114. Thus, radius of new charge = 2a Charge density of new charge = 0.25 po nhnne charge distribution is given in spherical Ctml'kiillillCS by r <u 1‘ >11 Muir/u) tiir ) = - ’v in ifm (at) Find the work required in rearrange the charge with unitiirm density within the reginn r < u. (it) Find the radius (il' the spherical region and the charge density for which net i ther additional wuri; is required nor work is made available in redistributing ' the charge uniliirmly within the region. ‘ 2 Wommuimdgfifljfmigefi ‘ 2080 I - 780 14060 ._,_~...._____ _ . / __ _ U ‘ 'r 2: 2:: . r _ e I J. I (pg EJrz sin 6dr d6d¢ forr<a i i eoLiTW'fi'Ef ferr<a ' a 1: 22:” r : 4804 i I j I (mzjrzsmedrdedtp forr>a . r E, = . E r=o 9:0 ¢=o » , , 3 K “‘7 for r > a i 80" mgr-“la for r < a m W w N I > ‘ > WE: ‘ forr>al i (“3) Energy stored in the electric field of the charge distribution a n: 2:: ' ‘_ ' ‘ i =j j j §eo BEETrZsin adrdedq) ‘ V r=o a=o ¢=o .4eoa- e _ 4 6° 1: 21! 3 2 '_ I +j j j %£o Elia—J r2sinedrded¢ r=a a=o ¢=o 450;? I -xégs . When the charge is redistributed with uniform densrty, sayi p , then :3- :ta3p = i 3 P’ =zPo >. From find: 37 , energy stored in the new Charge distribution 3“ term 3 6’} yo E -2223 _ )2_3"_P%a_5 . i 7 - —' 1550 p0 2050 A i i kid {he radius and the Charge density of the new spherical charge. Then gemklafikzpo = zp0a3 5 ji- (k10)5(kzpo)2 = FEE 15:0 720 01', 3 W52 =3; 15 Solving, we obtain 9 28 21 k1=fixi§=§6 3 ‘ o 2000‘ kz-zxé—x =fifé‘7 _ ‘ . . ' 0 .- ~21 ' 2000 A Thus the radlus and the charge denmty of the new charge are—20a and 3087 130, respectively. ChurgesQ and -Q are distributed uniformly on the surfaces of two concentric spheres of rzidii a and b (> a). respectiveiy. The outer sphere is made up of two separable hemispheres. Find the work required to carry the two hemispheres to an " intinite distance from the inner sphere and join them to form a sphere of uniformly distributed charge -(_). For the two concentric spheres, as shown in the figure, 0 Vforr<a 4,325,: Q/éafora<r<b O forr>b 0 forr<a 0 ‘ Er- —=—4W2 fora<r<b O forr>ib b 7: 2111 Q i W = - r23ix16drd9d¢ ‘ L=aL=OL=02£O(4W2)Z _ 2 (1.1) 31.580 a b With the two spheres infinitely apart, v 'oo 7: 2:: . ' I 1 Q . 4 We 5}; L=of¢=oiao(4 zjrzsmedrdedqb +r fit I” éeo [—LT r2.sin 6dr d9d¢ 17 9:0 ¢=o 4W2 =§ie+s mo 2 1 1 2 1 1 Work reqmmd ~8%E—O(E+E)—%(Z—E) (a I A sphere of radius R has a charge Q distributed unifcrmly i1 oyer its surface. How large a sphere contains 90 percent of the energy stored in the electrostatic field of this charge distribution? ...
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ece302hw6 - ‘ CALIFORNIA STATE FOLYTECHNIC Umveaezw...

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