homework 3 - " PH 132 SPRING 2005 HOMEWORK # 3...

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Unformatted text preview: " PH 132 SPRING 2005 HOMEWORK # 3 Assigned: 01/24/05 i J L v.70," F: a) (9.006 ~45; b) 72.60”? A c) 6.00% f r Wodéb find: @ +hro 10 file n’hf Face, 6r 6?), 15 dfldujc. j AQQML .u W‘s cube in Fig. 24-26 has icoulomb, is given by (a) 619% cdgelengths of1.40mandis ori- i (b) —2.003. and (c) -3.00i + : canted as shown inaregion 0f Lmi— 40612, (d) What is the total flux form electric field. Find the elec- _ through the cube for each of . trio flux through the right face if thcsc fields? _ the cleqfi'ic field, in nemons par man .—.-5, 0095 fir each E 55% Ear _ 5 Nate: @E‘jgdfi‘ Eris Cam-321%}?— In his Pwmgsv Hu‘s Q; $3135" . - _ =5? - (5‘) AMI-inn _ I N343: 593 ,3 gym by a BM" PflobucT,5c. a// I Hednc F/ux gm m; a was. 61) Q: E"? 3g defihfdhnfln area Vedar MW 5 pomfi owfwam' “For a flayed Jarfflcg Wed (35 arr/w .rperpemfl‘cu/arfv #:e. 5% .' 50, Pfir flail/1+ Face =AJ~ =(l4ygn)2j\ é C2=§573P= Ké».oo"’é)3 ' [Ma-Qt» “ember, 33%;);me = {MUNQM VOMYWZE ANS. :gmw” ; an - b) @EWE flab: (“-2.00“é>"‘ " (Mom)? :g'ximso' J I i, : (*2.oo~{,_._)(/.'/0M)Z : H.392“ Nmfc [CR/U5, a) fisféifis (-3.00%3 +450%i>- 0,40,“)? _ . ° : : [(—goo~é)(£¥om)z]% + [(¢Oo”’£)(fi40m)fl%§ Ants} So when 1414:? (1;. no Charge enc/asec/ d) @E-E : fiend ' («45 I37 #713 pmblemfiJW W/ Wax 07561 % Spaa'afly Um'firm fi’e/a/ #wuj/r a. c/asea’ ! Stuffing f5 zero; ' I; {2 (EM: 0 1979f 4) 9 b)? and a) . ANS. _ I I HRLO . ' .______T I l J A point charge q is placed at one corner of a cube q f hat is the flu): through each of the cube faces? (Him: U I law and symmeny arguments.) r'" I ‘— .“- ll i #297” L57 6 LABEL. 779E ,z—Au-Ls. i a a 5 NOTE : ‘ ass _ H ”" 7H5.” wa Tflmfi 2,525 g - RIGHT“ R3657; EDITION» + BALL gag 9:31:79 tat—7in BE 2510 mg ggmcE 7/95 RAD/‘AL— % F752,; away; ammmzifie FPZ—OM i DUN/L.- BE.— PAMLLEL 7‘0 THESE Fame. '70 Push THE FLUX T/JJZDvé’H Erma 2 0F 777?? ZEMNNG. x—ACES) com-mu Q0355 97A¢L8b so 7M7" $5 504” "9 4.004.723?) E t E g E s I l F l l l E E k .4 ,- aw 52am FME 0F “Du-3; “L’Lg rpm“ xfimi‘g 77+; 9% “‘) g v )> E T“ 9 % fl NW $25“ 1': Ma Q 1L, 50 SHEETS 22-142 IOU SHEETS 22—144 '200 SHEETS 22-141 .‘ AMPAI: a. @- A uniform? Charged c_‘3'I‘1du<:ting sphere of 1.2 m dial-[i3 ' a Surfs!“ Charge density of 8.1 #CfmZ. (3) Find the net the sphere. (b) What is the total electric flux leaving the s"? the sphere? 5:3Wc/M' =-3./K/J°C4nz a)? We 6‘ www‘ a) r: Charge _. 01161 30 5’9 = yam“):4ff(%")z(3.lflo“c/m?) Ms. b) Use/P éagSS‘ Law: pw‘ 6% 6620mm sarézc drowa/ sphcre. /- - x @=§mr=@fl ( 1 E - 6'? K \ I/j @ = it”, ff. :2 i x/O-Sc‘ e '__ ,x E 6. 6., SSSWWMZM: l: fiéix/owg-g IAN-3 M: d= 2r: [ZZW—w I ‘ I I. surface Charge dwiflj (chdfi%m> ! g f ! 6) AMDIID 50 SHEETS IOU SHEETS 22-1 44 200 SHEETS 22 l4] 22-142 t? 'Spaoe vehicles traveling through Earth’s radiation ' ercept a significant number of electrons. The resultin buildup can damage electronic components and disrupt operg Suppose a spherical metallic satellite 1.3 In in diameter ac lattes 2.4 ILLC of charge in one orbital revolution. (a) Find suiting surface charge density. (b) Calculate the magnitude electric field just outside the surface of the satellite, due: surface charge. E73 507‘F. ARE/3r D?- A 9?H£}:g: A = 17min: Z. 2-4/eron :: flkr‘b-L 90 1r: MEN-’5 I; E “Dime-73% l_._V,——J 777;; 991,75. a.th bi—SNSW? 361 L05 mow Wm" WEE 7H5- g-F/£L.b A) - - I __-___m-h—J‘ IWIH Fig-fl. 24—3—3: a:mall, nonconducting ball of mass m = m ; m: 0 75/0‘3 _ . mg and charge q = 2.0 x 10-3 c (distributed uniformly m through its volume) hangs from . 2A0 X/O—c’ an insulating thread that makes; an angle 6 = 30“ with a vertical, _ uniformly charged nonconduct- ing sheet (shown in cross see- :7 tion). Considering the gravita— . tional force on the ball. and assuming that the sheet extends far vertically and into and out of the page‘ calculate the sur face charge density 0' of the sheet 50 SHEETS 22 I42 100 SHEETS 22-144 200 SHEETS 22-14] Id due 40 a chafiedshm‘ and/UM? ZMLa-w Draw a “14-63 bodfl dafim: I I M I 1g is the 5m due #0 the death; I 196161 of shfetf Iii—"ZE- IE: is "’[D file nah? tame the shes? )5 repel-15:? file, 3- V3 c_.£ °§ TE 3‘ “g a f .asmj Newt'm's 2"‘Law: ZFam4=o ! {de'ffC‘IzJ'M2 €E~Tsfn9=o gdr'recfim: 7—6036 '"Mj =0 I E:- T—LW (I) T: -.m—?L (2) "’2 he ‘ oose :- Plaggfnfi Mfg (I): f: %—2 ‘mi’la i _ m E: 2%, (63. 241- (5 9.5%) i“? 2%; -I game- 6‘ = ZEWQWLagéz Z 355 ’Vo-nCy/U‘mt>0«0x/0h¢ )fi?“3‘)mfi50° I 7 Z-Ox 10% as. 50 SHEETS 22-142 ‘IOD SHEETS 22-144 200 SHEETS 22-14] ‘ ' __l Mia?» h - d (:0 —' h r " "'“" ' ' H m H - m" 1 o c arge ncentnc sp eras haw, radii of 10 O C ‘ .. 15.0 cm. The charge on the inner sphere is 4.00 X 1 ‘ r; h that on the outer sphere is 2.00 X 10—3 C. Find the e {a} air: 12.0 cm and (b) atr= 20.0 cm. 6 r _.. = 1(00 V046 EM: mar) 39» r= £950“) for r=20-06M— we curl! pm a (woman sphem +hene. wa= $3 Al's Jheareaofd’fie QQLLKSJZM Sphtrc . E. _ ‘2; “germ/Ema ()1 Me 5am: damsmf’rs A H H (M545 ,r 50 '_ ’1" m7"! firm "(I ‘ ea ‘ a g a .. E: 4min = Egg ’ {—x—L—M #0 )(wow ‘2):: Z- QX/oyN/C ; f" (0‘ ' E Now we an: fm‘mskd :‘n r‘=20-0¢m (r >13), We CM? MBNSW I"! 7%!— rflfl’b?“ F=/Z.0an(n<r<r;)so I \ 777+: Gama/3 Sphere surraumlr boHa SpluretsKHZXL l ' "~53 {and __ . J @g-dn “a m— (2,sz EA= (a? - 5(47Tf‘2’): (a? __L. Qfigz = thfr-qz) I - €= Wet, “F? r-L = (Km/0’” ‘”%=) (44 00 9.00 x/o‘ec /0. 200m)2 =A5. 50 S H EETS 22-1 44 200 SHEETS 22-14] 0 22-142 IOO SHEETS L 1 0 mm given: share: H67, *3 333$; I she,” r:- b far z'nnerSurém-f I {G c For aufirsur‘éa: I N I u 24—34 3 sphere, of radius a and charge +q uniformly "ni‘ ; througfiflm its "01' the sphere {r < a]. % Simncfinmc “1h 3 Sphefi‘ ' ' Tit-tween the sphere and the 5113 She“ of inner 1'3" $1 E; < r < b). (c) inside the ouEeI radius c. This b < r < a)? and (d) omside _s a net charge of -9. Find _ _‘___ions for the electrie field, flfifunction of the radius _r, _1l(r > c). (e) What as me S on the inner and outer ' “s ofthc shell? I “(a (ma’acfihj) 35%" @5040 - b) E(&<r4b) -* c)ECb<r<C) ‘ . ' 4)£(r “>c) ' 6/1626 an Miler and safer SW (5.: 01‘3th a) Place a fiawan 3pm m ‘HM r {on r<a,.77ie charge of m: sphere is 64191160149 dfkfr’fbH/Q/ WWW A I56 volume. 75 90d and 1%,» 447:3 re 1297, we. wffl have 1b look «4‘ Me 5pm}: Charge “Mk—5'7 9‘9" {9 I3 am‘flrm 6r- Me :phmz: - " fl @r—ra. =pér‘fi6mm Sphere. 7‘? en ffizjif; :>€€nd:€(a£) How we are few/g 15 we 6,4663% Law: I §§iflf= 163d Neale: fir a more defiF/m’ dwwmq 0.34/5, we 01: 6° 64”“? MW; see Hze So)th {a CHM-we H g _r: 5 EA ' €061) EWW‘VEQY egg; PL; = 753: Ame. Q a<r<b Plate :9. 6mm sphere in {he rejan awed): kw“ €254 ' 2, = ‘20 7:.) E:- bL, 3" '1 if: g: r2 ich b<r<c The she” (’5 acma’amq 50 add-=0 (h5r'déW6/76/Z I Am. . : I' 50 SHEETS 22-141 22-142 100 SHEETS 22-144 100 SHEETS M 6i -P.|'\ /_ n i 1 _/ New] Charge is distributed uniformly throughout the volume of an %itely long cylinder of radius R. (a) Show that, at a distance r from the cylinder axis (for r < R), pr E = —— 2.90, where p is The volume charge density. 0:) Write an Expression far Ewhen r 2: ECO r-‘é-l'b 52.55:)"th FIE-Lia 42,455 (nor 5 H mom) ' ourme I APE RADth IHF‘IMYFEM LONG 57:. MVQOpr QVHNDEZ— w A VOL. CH2KRQEW, "19% poogyg A GaAO‘EEJm Chg”! "34,35 M51) APPLY. 6325055. [-072.57>El LDC/ATE @075 the: MD A emu-g9 { AM 06!... AWL? ekuss/ 2' z E gem) ; 22 I41 50 SHEETS 22-142 ‘1 00 SHEETS 22-144 200 SH£ET§ (:9 AMPAD is Lhe net electric flux through the cube? r’? . . . . . lii/ \n electnc fieid given by E = 41 - 3(3:2 + 2); pierces the aussian cube of Fig- 24-5. (E is in newtons per coulomb and: is in meters.) What is xhe electric flux through (a) the top face, (b) the bottom face, (c) the left face. and (d) the back face? (e) What _ -X, =10m M/ )9} 5.0m,so +519 ZenflM we each $1316 01‘ fire Cube) car's 2.0m_ Emi- @E +hf0th/N 6044): hop as to) +41; boHam 191cc c)+hc {eFa‘ 1%: d) 14K back «Face. ' a) finer ,xz E/ecfrs‘c F/ax - 52,) #3,: Page: _@E=5§'“’-d‘h" We My fang/amber hbkfice: dz} = dAj 7 3:20,.“ on #1: +op Ewe-So 272% 2/3 #33 cause @E=j?-4’A”=j(4z‘-/gj)- Mj fiflwfl- Iggy/35% = Wflé- JA =-22W¢A=-;g%(2.om)2= - 2% Am. 8 hm "Face" Cm; J. €(g=o)= 4:? " 62d“ Nde" E13 eonsf- 6356M @E=jf-W=J(4f—oj)‘ (gm {[46 ° -: .Jr) -(D(<‘--j‘)]dA = (We, 0M =<o~£A= e"é(2.omY=H/V5- 0) {9919’65‘ “V: N?) am @gjidfifldi‘ —3(g"+2)f] 0 (-MA =j?! ~3z¢+zx§4€fldn - ‘ 1/ ‘ r_ ‘ . = “Me = W Wm ms, $me 19165: J3: dM' l2), (4 wide; over #7: vfiace —.5. A A ? JlO @E-jeaa =j[4f—sc3=+z>3}c—1)JA {[4 "-Io)-3(H‘+ZVL)]d/4 f); éhroujh 1%: fiénf {hie 3: We 0. b\ 11...; T I' 7' g .f D /.. .. 4. \ I -l .t.l'l.1|. givenué’: gig—(9423f _ | l ! ! i i ! i ‘[ _E {3.25 wo‘jfiJé/gflf) PH 132 Suggested Problems ,‘r'hen a ShOWer is turned or: in a closed bathroom1 the splashi ' fthe water on the bare tub can fill the room's air with negative f charged ions and produce an electric field in the air as great 1000 NFC. Consider a bathroom with dimensions of 2.5 m 3.0 m X 2.0 m. Along the ceiling, floor, and four walls. approxi- mate the electric field in the air as being directed perpendicular try-E the surface and as having a uniform magnitude of 600 NFC. Also, treat those surfaces as forming a closed Gaussian surface around the room's air. What are (a) the volume charge density p and- (b) the number of excess elementary-"Charges e per cubic meter in the room’s air? __,_.._..._._._.._..._ .__,. 3’1 4. ll U\ at :3 3. 1‘ “ ('3 f‘flw'flfl "" \ I03; (3 >5 V 1;, ____ I ’2. )2 IQ ..-u-‘”" l! 5 Cleaner 7- " a I Vie/5 ’ i l V _ J Were-PH"; '— L____g# _______ __ -_ M f - ‘8’ c M to a r” ‘ f f .r a i a few 1" ,r W {a l f ri‘fi 2“ F H__u_____,_,.. m_ I; 5 1-6 *4 W L’A‘aaxmfit‘ LII/fl theme“ \ “ Mt? t we": -4: “Ml-Z : IA 6 l5 f; f ‘3’... f? \ ,0 Mfu- 5— Mme l r F”; i #5.. :1 fly.“ m.»- 1 f i you A Itmg, straight Wll‘e has C; 1V5 N gative charge with a 1111— C ge density of magnitude 3.6 anm. The wire is to be en- > -;, 5: é X 1/ D "" y a thin, noncomiuctmg cyhnder of outSIde radius 1.5 cm, with the wire. The cylinder is to have positive charge on E. z 0 J of 5 M e surface with a surface charge density 0' such that the net i ._al electric field is zero. Caiculate the required cr. ' E’mrfi‘ ZALTHOW tug: maze: no“? Agni-Q} . I “79 CALC. THE E—F/Ez—b 2:02: mm FAR BUT £X7£72MHL 71;: 73.27 i-I-I- I -! NOTE THAT THE: ELECJ‘JEJJ, “23 F/ELb Hm£$ Con bleach)“; grw { H525) POINT TowAZb § 712.3: 3 Mug 9?; CIMRQE ass + 6. THE: MEG:- ..m 2 I..€.';§) w i 3! vaééxm a L. TMC’K- 1: WE: (Ir/{LL (4AUE. THE M95 AQK—Eb Fwe large, thin metal plates J parallel and close to each other, as in Fig. 24—16:. On their inner faces, the plates have exces surface charge densities of Opposite signs and with a magnitude 7.0 X 10’22 Cfmz, with the negatively charged plate on the-left. What are the magnitude and direction of the electric field E (a) to the left of Lhe plates, (b) to the right of the plates, and (c) between the plates? 50 SHEETS ~07? £125»: TH‘M'E “fig; A15 @wbfi" swat-:75 SIM—E “’4‘ - - THE: oPMSIJE “Mia— FRDM P05. F39” 1705' AWQ" To ONE 5’ «——~ I 4—— :- m wee: a? .h-a". “'"' FP—oM MSG. mfigofm co m1) ud‘t- G .4; CHAR—(s "‘ H5575 ./ .. 51:0 4"- "0 iENSITIE$ +5“ wag A MDMCDM . . . 2%] TH; xii/£42); .oTész-C’sz' AN]; 7136:)»;— Abb- - 22-142 100 SHEETS 22444 200 SHEETS 22~141 E a” O_ .A. man; _ ago 1+ (Zia I P09: 'PLNT -!-3£-'é- RAMS 50 SHEETS 22-!42 100 SHEETS 22-»144 200 SHEETS 22-]41 (a) RMDAD 4 .Aé "51215;" Fe?- 7742;: éAuég-Im charge density ,0. Plot E due to the Shell for distancss r . center of the shell ranging from Him to 30 cm. Assume g 5—2:: WE M42513 NO EXP/ZESBIDM Fog 5 —7’—-—~—*#—“———'—“—"—, rfl A“ 774/15; flame“), came/2557i ,_.L_ HE/‘Q—KAL 6A0 (M SPHEKE 07: mbyus {*2 GMf’;Aw: 32/3 ac O AWL—WING @fiuss’ LAW “HEADS: Eéfi’h‘r‘z) ~ ' z W D . F071 * S, PPLLKHQC. érlwés" Mu.) w: 3 3 E (’7‘ ’77’r}\ : 97H5-FE JADCJU‘ELD pm— WW “3 BET a Mb [:1 j 2 25% i g LWCF-fi—ai 1 3 i I ::ES i J i r— ? LVKNG-r GAUSQX AALQ ‘7’254335. N075 ' '0?- "Ha éAvéerxw 3— 3 5 Gym} Ema-jazz moan-21> Age/77(3) F pig—70;. a i AT F>L=. - Eb _ r : H—“' E . $470336?) ‘ yg| g E 5:: ‘ g i am . EETS EETS 50 5 HEETS Ii 22-14] 22-]42 100 5 @ 22-144 100 5 above the center of a square the electric flux through the square? (Him: Think of the square as one-face of a cube with edge at.) 9E. In Fig. 24-30, a point charge +9 is a distance £1.12 directly HEW of side 5!. Whal is the magnitude of gap} (£6? C64 ba- I fffié? @E‘l'hreujh ‘HK 5?“?2” | @E=j-E*'d? jbw E’z‘s nficma‘afl‘? ciier’ W wafd‘ 2% 7711's Ihkjra/ :3 aé'éfi‘w/f 7'2: _ Oz/m/aé. “SJ/Pg GdLCCS‘" (an): Q =§§1djfa éfluss's Law {3 6f dosed Surfiggs) 50 we cmpwew’fie Cubeth *3 #35 cemler 7772 ekcfnt Wax (‘5 um‘érm Magi/l €620}: £25: 01” fire Cubersa 37L 1'5 I AX? é? flame” surface. .: gg=§ “111:5 £5 Hue “Iain/4%)! 6" o Hzmujh 1%! Gabe. 73 find fig +firwjh #1.: Sfaara,w€ fitfi’c/ flux mejh Me be : y@ 01’ fire W. 1-ch m" #1? ca ' ‘2 Q fir #1! Sfer "" E Our {ask {Is Mata/4... E’dflér by given: flea! W | 50 slum .22-‘42 HID SHEETS 22-1'44 3200 SHEETS 22-141 GB Wherea<r<b. f/Zm-L ) = Bw=§f( - FIGURERA- Problem 25. 25F. Figure 24—33 shows a seation through two long thin con- centric cylinders of radii a and b with a < b. The cylinders have aqua! and opposite charges per unit iengih A‘ Using Gauss‘ law, prev: (21} that E = 0 for r < a and (b) that between the cylinders. ins! 6:: ! ZWrL. ,1 Use a 651662922” cflfi'nder‘ 01c Gaussian mitts P J WWW? W374" bat“ “J 55,;er cgfmdea Jan-here r442. This will "’ a/aw as 1‘0 rhuas'fi'gm’e #115 fé'jfon P46?” -"'-"5 (fig—4" E :‘s Const and/1h {fie radiz/ dm/rfm, ‘14 4593;: A (‘5 file area. (:ow 6. We NV! 'b msflaar Hue. aware-mi prams of am- 4mm Imdtr, shag 44m}- mm radars pom“ 145:2an JL=kfl3M 0‘5" 65/:de Same file W78 0n Hie 1%na'cy/zkoéruk 1%an ar‘ Fm) 423M =0 149w r461. i... E(r‘)=0 ) ' in dféremt dream: =.—.' S 1657 1 .— o‘ 9 . '9 .5, . t4 5‘71?“an s‘?0‘+€/43mso‘= @331 Al m 2%. D . -_. _ —__._.._._._.—_._.._.._.-.—.__... WWW-‘— 22 Ml so suszrs " 22-142 100 5:42:75 22 I44 200 SHEE‘I’S ~39 Q6? Now we are :hkfflsied in Hie realm a<r<b,$o we wear éawwm Cam'de Mere. We already Worked am“ #16. in pant a) = -' @591 ' Ecr — 6, (27TH. The dreary, andcsed 63' #2: 64mm eff/MM :‘5 +32 Charjaaf’ +he inner cjlmdcr. we are (16% given f )bad- we. me 3km 7t. _ aha r n ?\- 7253i ,50 f- “\L - E0“): )a4r4lo _ '2‘. J... e/ea‘nt field fir Cgé'ndn’ad ' 50-SHEETS ' 22-142 loo 51-12515 22-144 200 snssrs 12-14! 47E. A thin, metallic, spherical shell of radius a has a charge 90. . Concentric with it is another thin. metallic. spherical shell of IE " a ) q - radius b {where b > a) and charge qt. Find the electric field at 9' Hra'dial points rwhere (a) r < a, {b)a < r < b, and (c) r 3' b. (d) r- : b Discuss the criterion one would use to determine how the charges ' b v {a are distributed on the inner and outer surfaces of the shells. F! n d a.) ECna.) b) 60% I’d-b) __...._.___ I (DEG- >19) d) charge on timer and ochr 5 «Ana; If slung éoluh‘on anuss's Law _ 4’) r<0~ Place a (mussc'an Surface (Sphere) :h H»: fegfm (“462. I §?-_dfi*= (“if 62W =0 ,50 Mi tbaéréb Plaza a Mm Sphcfe m Hes {6’th a<r<h 9930123“ ‘i-gg- More: +3.» a m mas/ed mm of m be 0F 4“ Game Law’see me stem 15; came/as: EH‘ 3; ' . I .._L.. .. k l EMM‘)‘ 3—; => E“ 4776. £12” é, ANS a) r>b Place a. éaussfan Sphere 50 hi .Surfvunds b0”) sheik. “id? = 22% - _L. ‘34*' + 5mm:- ‘agh => aware-E2: flats) HA5. d) use cm; Law: age-m {W 15W; .2“ =0 451— rm. +543 charge. on #2: innerSarézce afa‘ée mister she/H5 dwarf zero. _ E7716 charge on He Oukr .Sur'r‘bce. J HM rhnershcll Mimi be fa. I. I“: Sm“ 5’0 my! 1‘4"? “A” 5591/ 0+8 main/Atfl Cmt'fldm—l);+}}e— net‘ charja fine/W xh IL 667%???” .Suréte {4747‘ is pitted {Dew #26 finer and Jukerfivw of #2“! other shell is zero. g” f {gum/98 am fine rhflJf' Surfite of HR curler SML-mea'fibre, Hie clergy, an #26:, inner Surfing. at" five fl’wtw‘er’ .sheH mus+ be —{a. 17?. In order 1hr 44M WSW] a)» have a fl‘ddmge. 0+”- {6 ) Hue antler surfime ...
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This note was uploaded on 11/04/2009 for the course PHYSICS PH13100 taught by Professor Dr.wick during the Spring '09 term at Clarkson University .

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homework 3 - &amp;quot; PH 132 SPRING 2005 HOMEWORK # 3...

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