ph132s04_hw2

ph132s04_hw2 - PH 132 SPRING 2003 HOMEWORK # 2 Assigned:...

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Unformatted text preview: PH 132 SPRING 2003 HOMEWORK # 2 Assigned: 01/20/03 '22-]4? 50 SHEETS 22-142 IOO SHEET’S 22-1134 200 SHEETS 3 mm ' - I :5ch fields dd: 40 Pom C Chocae a. mahhak, mfskm. A mm: one MW? Ike «5:31 a! line (MW of .symre. 60-6, agave“, Wkfifizdmpmmls: _ a a gar: Lax-4% * 53x *% =Jg§ {-éli‘z. =4 ,2. (2%)" (“Ear g: a? 6214:5444hecw1’erof #12 Sfum. (WfiM‘ JINC'I‘IM Able: End: dug: £5 a dzsfmcc of EO- 5M 4"! (mkr o‘F Sffidm. huge:- is drawn on Me dagam, _, 2(7fi?x/0'%E‘J(1.0x/0‘8c.) ( 0. 05m): = Z X/Oq’V/Q, ———————..__ m _ __ k K lit-‘31 - 72 {Ow/c, ($47 2a" a,”- (zzxzowoa + (7.2270ch -—-——.—».—_..._.__ 22141 so SHEII'S 23-142 :00 sunrs 21.144 200 sugars ® , FIGURE 23-39 Exercise 28. [BE Figure 23-33 shows two parallel nonconducting u with their seam] axes along a comma line. Ring 1 l .1 charge qr1 and radius R; ring __2~h_§s__ A “H " Lmifonn charge fiend the same a radius H.111: rings are separated _ by a distance 3R. The net electric " I field at point P on the con-1mm line, at distance R from ring 1’ is _ zero. Wlm is the mio q];ng 5' 92 P R Ring2 Ring] R J E‘qu Elam: 5W m +2» :2 Wham mag dish‘bufim Charged N303 E/ecm'c fiua’ a? a. Cj'WjeJ n}? .- Z?_.__ 1332 E: warm" (2‘4R‘Y’z_ I. Z (“ml/ll" I:ka a’z‘sémac 0F Ipfivm 1% fin}:- QS'hj 141C- Wressm above in 19nd Mefib’drafpom-F pk ’5 a“ n ' law g kg 2 k4, ' 6%, 25% , P. 562) ' (Rwy/5 cat-49% 2%? £2: kiJZR) 92sz - 2%: ((2K)‘ 1- ’2‘sz ~ Rf)”: - 5%, R2 5:3 FVEZ =0 I E, and/6; mas; 1m: owe dreams 1b I 541534 7%: landfill“, so 6: =55. Kama??? F 5%.}? it: 24,, 23!; “5'5: Q! = fizz 4E; as?» (5% 22 h" 50 SHEETS 22442 IOU SHEETS 22 144 200 SHEETS . 19?. Auden-misconsme : .thccenu-alaxisofmelingof' _ chm-go of radius R discussed in Section 23-6. Show that trostatic force on the electron can cause it to oscillale n centeroffliefingwiflianangularfi'eqnency ll] '- sq V 4momR3’ amPIa‘udt osa‘lkfi'dns = 2«R. a‘m‘c R‘. _ 2‘0 i3 #36 Effill'bnhm Palm" 010 #1: 91mm (no m4 firm wing 0“ 4‘0- when Eat-0"I-Pulls +512 decked back-ha steamed) like a, "5Prlh5" Force... Rember Mflr a spnhfl, 004%: w =]/ 9‘52. Linc-rm? ‘Bth-likc” £265??? of F makes 41:6 ekcflm ,«we. we; a simptc I'M/mm}. Nwh‘orl. . Some will use 44:5 apfma‘mm: SO SHEETS 22442 100 SHEETS 22-144 200 SHEETS 22-‘Ifil a mm '4 _ H297” cams Ire-:1 755; E” F152; bu»? 7c: 7795' ' cousmEZ. THE FIELb I Ffi—om A. 5 LL PIECE :3}: C-HAFG-E NOTE TIME; CDM?DNENI$ NOTE W47" 7*»; X—cam?ov~&EQT$ 8F (95 CANCEL . 36)” TH: 2—wmwwzm-s Am: 60‘. . h I . ' '.—_ like «99 R suMMmc, 7-H; —CDr-§?S. B‘f 13:15an 14;: - 7f ww—h- 22 It“ 50 SHEETS 22-142 IOO SHEETS 22-144 200 SHEETS ® AMMO I “.4... m._...._._._..m ........._. .__ GIMME lenfl; o'F rod =L .3214): nffl .44: x2. Hmfi/Mafilufiefidw wasml/amdcfza ,df- 75c Mt The. dEx-componenis WW cam/de i0 fie symmafiy file prablm. 71nd 9155 mammals will add 1b 5.3m 5 egg =d€cos8’ {2?}? 6059 _ 5 dx wfl/ 4&0 he d'mtIt/f 75 {371671319 ,So {'1‘ (MW 5e: 7b " Chél? 1e van‘aé/e: 14m 7C 9'1: 19- - -+ 75%! am‘FDm? diary. on (ML-‘2 1 HM Make! 0/? 1‘5 0’6: Once we hm an ago/$513M fir déj we will «Se. an r‘ m/ib jehfic WM #36 (dda‘kjtfl'l/IESM/ places bad m ' k0, dz fs‘dtg‘fifidf/fl,» i _‘fik1£’clddfPa’4cfi@/IS= dg: T; ' . dé‘, ’2th 6,6334% r‘ i ) 22 Ii" 50 SHEETS 22442 100 SHEETS 22-!44 200 SHEETS 0 mm a) dun? ' r __ fine = a): éfi‘om +17! d'ajmm) Change 0F mn'abbs: x= g'fana dx =d(g+ane) = dame) a gseczeda @fle:§ firm d9 = - .. .hhcosedx; IR?! 0059 . “’Efl ‘ fie -‘— QstdQ fl . - ._ 6:9‘ Because. O‘FHIC um? Wt. can In}! Mk over-‘1 54 1“: £3 ' 2 gdgg mo! Md mum}??? at! f ‘1 =0 " [(EQEUZSGOIGJ = 2'” 6 __-——-5"9 ':.- o 3 "Jo «mike dfdymm, 5:3190“ 5%?- and 5:73: (3)“ Efl’ECPF M. 42 ' ' ' :1 Now some reamm +9 mice #9:: re:qu look m M: an: m fie yam”: =Q.L'=!2_a,1— - 9 3 "EEG-2+{Zgrj fl = £— . 2_._ h: "J— " .. (1)2) flrrqu 6{ 1/176, ad Lh'i =.._2__.. ‘5 #776. gm 54)): € In_— 2176, 2.2+ 4g 1)” 22 14] 50 SHEETS 22-142 100 SHEETS 22444 '200 SHEETS ® 'c disk of radius R is the nugnimde of the electric field equal a __I One-half the magnitude of the field at {he center of the surface Ext: what distance along the oenual axis of a uniformly charged j Chdfgt’d Elecfn'c. field due. 1b a Chfirggdd/jsk: 5Q) zgop .. LJQQ 25-24) I Ear :20 e»[*-%l=é(% e 3 =1 ’ z J:——L. _lfi ' R: widths ofdl‘sk ' IF e_(z)=iez¢),+hm z =€’ _- ewe aw due '4» a. mam dash-{bum of W55: 65% "58?ko of dfsk is at £—-o . 50 SHEETS 22442 ‘IOO SHEETS 22' 144 200 SHEETS 22-h“ 5 mm 4UP. At some instant the velocity components of an electron mov- ing between two charged parallel plates are v: = 1.5 X 105 mils and v, = 3.0 X 103 rats. Suppose @211 the electric field between the plates is given by E = (120 NfC)j. (a) What is the acceleration a @ t:o, _ of the electron? (b) What will be the velocity of the electron am: 1. 5 its x coordinate has changed by 2.0 cm? V310) ’- 5 ’00 "'2 v.1 (a)? 3.0 alt/03 “£5 ' a4 a batter Hm tflhc {100%ka ox‘ {lea‘rm change; by 0.02% Force Jar +0 an eledrfc. field , Conskzm‘ mania}: mafia: a =.E : -_;e;§".be/o""6)(120”/ ~) )3? m m” alive-3mg: ANS . Nate: 25‘? 033 , ago SMZé’WBELS/uo ex Combhfnj (a) and (3): __ -.- IE, I v3 1+) van) + a, ( Vac/0) ' _ 0.02m ' ~ (aux/03mg) {- [f—ZJI 75/0. 3'16: rug/03w = '2, 7270" M3 2-0—52.- NI‘J‘ - II 5 I20 cJ ! I :0 {1 1‘2 141 50 SHEETS I42 100 SHEETS 2 22 I46 200 SHEETS 42F. A 10.0 g block with a charge of +3.00 X 10‘: C is placed 11: electric field E = {3.00 X 103): — 600;, where E is in newt-ms per coulomb. (a) What are the magnitude and direction of the force: on the block? (b) If the block is released from rest at the origin at I: O.whaxwillbeitscoordin.a1es an: 3.00:? A = 0.2'1’0'N3 - 0.04.90/u3 C : ‘ m = )c/o'z - (a =~ + Eva'sc - Ex: 3.01) 2:103 We is = -600 “(a '52 ~+ O fwd; a) maflnflude and dlir'eoflan of 5) wordsmk of oégjed' 3-005 antler rabies fun res-f eaf (o, 0) 6L) Fx = €613 (3.00n/0—SC)(5.00mom/c.)3 0-240 N F3 % (KG-.5 = (8’...00x10‘5a)(-(ooo Me) = I -— o. 0WD N 1: WWWM. e=+aw(%)=+an*'<w) b) x=2<3+ ,2 game“: 1 fr e2=1-(—°—'?-i’i?i“—Z(zms>“=lo<2m “N5 2m t :10, 221 F 1:16.001430) ML “W, 473% MW"; ANS. (if-“35):: “21.0;14 Jim‘ffiii; 42. = +2 '€2 = “if 1‘ m= 6363,68 e; 56-81%; SW 601? is 114061.! i» do pad“ 6) firs? so we. know whrch diedng 'H'le. elean field fine; Pom! :‘rL'TWs will an»: as “It proper (1‘) and (-) Sa’ 115 2k #2 qua/53m ‘fir #1: "e!" WI: fickle. - . Ne. wit! «(50 Med #2 expressibm 5r Hie eka‘n'c. flea/d due 4:: a ppfn+ charge: 635;; . at will “‘14 =“RQ Zia EA o"- ??? #27? 200 SHEETS I 22 l42 IOO SHIITS @ 2:31. 22 I“ (To field?) Maw f“) Slat Able.‘ Ue used-4v. Mlam'hdu J hm ’1 (ad MM Way/M qppmprkvfc 5‘ As “’0 He 46(025 9.91:3 AWWM‘ 14k: elem: 45%: mm. - .E = E. téfii = 4(3kg)= 12k AN; B (a: z (“I/232' '5 3—52 (73 file n'jhf') QM= {4&2}? _' i ' odx‘amokrafdrop, b = {.ZONO M ' ' " ‘i 1%. ‘0)wg'fih1‘ 0?an ,W b)“ cn'ra elem, VI, or: drop Nola: Since «Z is neflafme, 1%de - Caven'tmes a Coulomb fine. in HR. direc‘b'm b E". Ouanh‘ea-h'on 010 charge ,Meuflan‘s 2"" Law ' 67) W=mj ,so we, need #12 mass 1% HM drags. We are nof awed Mmess, bu-f we do kmm} {he dgmfij of wag/j '5 [Osha/ma. [93% So M3/0V. 3 3 W= (ova = (Ugh-"b = L’gflE: W0 afi’X‘ZZE‘XLZws/J‘uf' ' b) Llsfnj NEWS 2"! Law: W'- ‘ZE =0 (“HR drop is .suspmded) 3 W=g€ g W=nle|€ ' E ' n =._"_\./_= :- ANS. lelE O.bxzo"‘¢)(4(oz“/c,) in calm air owingw adownward—direclcd atmospheric electric ficldE=4fi2N!C.(a)Whalislheweightuflhedmp?(b)l-lnw mycxoessclapmsdocsithave? 22 II" 50 SHII‘I'S 22-142 100 SHEEYS 22-144 200 SHEETS fl _..._._...-.-—_....._._..u mu._.wmm_—_W. .. PH 132 Suggested Problems llll’. (a) In Flg. 23-27, two fixed pointchmeesm = “Sandro = +2; are separated by distance d. Locate the point (or points) at Which the net electric field due to the two charges is zero. (13} Sketch the net electric field lines quali- tafively. At points left of q; (on the —z axis) the fields point in opposite directions, but there is no possibility of cancellation (zero net field) since is everywhere bigger than in this region. In the region between the charges (0 < :c < (1) both fields point leftward and there is no possibility of cancellation. At points to the right of q2 (where z > d), E} points leftward and E; points rightward so the net field in this range is - 133;.“ = — in that direotion. Although 1o}: > q, thereisthe possibuityoffim=0sinoe these pointsareclosertoqz thantoq]. Thus, we look for the zero net field point in the a: > (1 region: on IE! = Isl __ ; 1 in! z 1 s - 43'on 1'2 41.50 (I - :02 which leads to “It ffinfl 5‘ [91' 5 . Thus, we obtain a: = ti? o- 2.7a. A sketch of the field lines is shown below. _....~.—.......". .. .__.. .. . . 22 141 SD SHEETS 22-142 100 SHEETS 22 14‘ 200 SHEETS 11?. In Fig. 23-28. whm is the ' -uP +5“ I of the fielde 1 fly— ' charges shown? pom ®’ e "' 4' 5‘% ®°<i°q +' (.23 +302 3 g6 E3 9 t: x +5.09 q . new]: 23-32 Problem 1- - e’®._~a fin—d an. MC d4: ' PM P : EP Eledn'c field: due 19; $5.344 charges 26E. A disk of radius 2.5 cm has a surface charge density of 5.3 pCIl'nz on its upper face. What is the magnitude of the electn'c field produced by the disk at a point on its central axis at distance 2: 12 cm from thedisk? '- 50 SHEETS 22-142 100 SHEETS 22 144 200 SHEETS 22-141 I E... E 5:»:5? 156°“57:) “ fr; mus-77 So EX 59?. Aunflmmehcnicfiddmmamgimbemmmop. posiuelydmgedplates.Aneiewonisrclcasedfi-omrcsnrme surface of the negatively charged plate and strikes the surface of flnoppositcplaze.2.flcmaway,inatimel.5x10‘35.(a)Whal isflrespéedofthedocfiopasitsflikcjflleswondplateflmwmt istiwmagnimdeofgheflecfiieficldflfi ‘ aha-0.02M THE AacLEzé-R/mm 1‘5 77ft)": me: cm as: apnea". Add—5L. Em; _ - m9 = O M ' x '2': = ...
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ph132s04_hw2 - PH 132 SPRING 2003 HOMEWORK # 2 Assigned:...

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