homework 2 - PH 132 SPRING 2005 HOMEWORK # 2 Assigned:...

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Unformatted text preview: PH 132 SPRING 2005 HOMEWORK # 2 Assigned: 01/17/05 cuzz,’ H) 12,24,20lffiij LflbL/QIJ‘KQ £435 64022138 4?? gm? Chm —- 2:2;(37/0jéo} 71/3' 22 141 50 SHEETS 224612 100 SHIITS 22 I44 200 SHEETS @9 Given: 'f—mx/O'Q', ' - . .a’: Eihd.‘ gar! Jhe CW 01‘ ’ #12 Sfm(m W4 I Iflc'h'on ' Nola: Edd! duty is a drshncc 43 $0. Han. 4‘" (cider m" 1%.: . 39mm. ‘ ' ' 4?ch fields Jae Jo POM cmrjq a. Cardth 5!. A Mlle/{bl} one ,3 flown on Me dawn, MW? 448 Oral?! a-f Me (en/er o’fk sytm . I§I=1f.='z:r+ rs; . 9=4arf Ear 45° ANS. 22 14! 50 SHEETS 22-!42 IOO SHEETS 12-1 44 200 SflEETS ® v _ ’ ' “igure 23-33 shows twoparalle] nonconducting with their central axes along a common line. Ring I I by a distance 31?. The net electric “131869! and M993: @238...- unjform change q: and the same radius R. The rings are separatgd _' ' field at point P on an common line, at distanoeR from ring 1, is _ zero. What is the ratio qlfqz? dim-1’23 a Cmfnnm charge afi‘fi'fbafim: Charges-l N313 ' E/eahz field an. ISM/fed! r637 : (%,' . 25% ., p.562) ' as"? ’1!” ekP’e-{SW about 15> 19nd {Ae‘fih/dC'fiM-r “P m 15 ad: - . n _ _ PETE-.- k‘ZJR) MK 3 MIR _ fl E} ?+E; =0 - E, made; mas; hm opplk dream 7b 1 2; 247. 2 3}: -— ——------ nu: __L_%+.___. .. _-.H_..,iu.fi_...?:9\m___ ‘ . F J . i ‘ i _ ff'ys- I 2,5541%.- andfim’so {=5} . - 50 SHEETS 22-142 100 SHEETS 22444 200 SHEETS 22-]41 -6 mm 0 In Fig. 23-3442, two curved plastic one of charge -q. form a circle of radius R in an 311975 cams/m 755 2:"- 724512) my; 71;) 7—»;— iL'U KAR/Afi' CDHSlbfl THE mm; ' ' ' FROM A. 5 LL P1545 con?o:~3£.y~a:$ I (9!: ~ 695‘ T- K - .dE NEXT cam. THE Memnubg DF ACE: I :3 TE? of?» Aid. - {Q 0 £55547» 095 = L if I R -———50' Q5 :Roge) :99 3.5 = 141 E? . team??? “391‘s. 8?: c953 a)»:ch New: 774v- THE x- BUT— meTDME-KLTS Abbi 9‘65- ‘ 2, = 3:1? @9169 7&0 ':._' ’41 We :96 §UMMIMG .‘THE’ flew-0’5. BY 135561.161 HQ: - _ “77' ' :1 HA 5' = .-.. " age =‘- 2‘51 :- g-JI: --—- 9? 777’; E WELD F072. 7 ° Lori-7E HALF 22-341 50 SHEETS a 22-142 100 SHEETS 6) AMM 22-144 200 SHEETS .'_. xi-o x 2‘ . I finsfi IooKa-f {ht fieldan 7‘0 51 swig/71am" oFdM dg I15 d6 Once we. hm an firme 6r d5 we m/foje-Hée. WW fieé/ (ea/Mn «if/Esau? Sew-crmied will “5% m: r’ ,df- 76¢ e/edh’c piece: 4D 39% W I led 6Q fidf/fiflo _Thc1 «Warm: 7‘» cw: dE= 4: 533.3%” “‘5 ’(mfié ‘ '” - ’9‘ “’9 W ' ,., 12W dgrza’x The. dgx-compm w!!! dance/(WW: f0 fie symma‘g affire I 5/63 =d€c096= kzdxfiow r2 *GOMpM MW add in 5232 E: ._ dx 0017/ M30 he dfifta/é‘ 7b {Myrafifio if 6017/ be Mpfit/ 7b _ 19 ‘ i Gila/fife Vdnké/flr 140m X 0‘0 . .5 _ mm ‘ -- fine = 55 «Fem #16 diagram) ‘ 0 Change. 0F variables: x= gw‘ane dx =d@7‘an9) = _ dame) == gsec?6d9 h dart 2 3525601,“. ragga?” = Qwsede 'J . ., eta" Bamaofl'heswndr mecca He kawhflk Elg- Zi; dEg‘ rode Muffiij (9.162). n n in: a 22 14! 50 SHEETS 22442 loo SHEETS 22-144 200 SHEETS =0 . ? -‘-' Z [(%)j259d9] = gflfsmeje' = 25—23th I ’l l i I f I 6 AMMD o O L/ romJ‘hc dfagram) sfan ~55 and a: 3: *C- a E = ecp): -&A. #2 ‘3 a Now some «mania-7 1% make flux :95qu Zack 11k #1 one .41 7% ffiM‘mz a. L :33 L - 3 L .. 4.2.3. HP): 9_|fcji4727'2=)= ‘F'rgzfizgfi " j (“hf-+43: : lg— {hu “1—” "d -= ('5) NFL/32' 6‘ We “ 0‘ Z At what distance along the central axis ofa uniformly charged i _ ; ‘c disk of radius R is the magnitude of the electric field equal i One-half the magnitude of the field at the center of the surface the disk? 6M” =-cenkr of dl'sk is at” £90 ' R: rad/a5 oFdI‘Sk . Find: Ekoftfc fi’efl due- ‘1‘?) a. WWW dishibumh of aha/55: . - Charged Disk. Hedda field due 1'0 4 charge/disk! £62) €20 -— LJQQ 25'2‘0 _' 22 141 50 SHEETS n 22-!42 IOO SHEETS 22' I46 '200 SHEETS ‘ _ Eb): 55.0 “W) zeo We are bob}? For 5(2): 95(0): - 9; 3'; N I": I 31'“ ’2 I NI... Sci: At some instant the velocity components _of an electron mov- betwem two charged parallel plates an: vx = 1.5 X 105 mfs and v, = 3.0 x m: “3;; Suppose up: the elacu-ic field between , Q t :0 the plates is given by E 2 (120 N}C)j. (a) What is the acceleration 3 ‘\_ : 9f the electron? (b) What will be the velocity of the electron after 3 O _ its x coerdinate has changed by 7.0 cm? Vxh) }' 5 KIDS Mg U30)? 3.0 #03 “£5 ' a4 a ’d-kr -Hme “6’38” ’I’Cmro’rbm of {lecfflm changes by 0.02M. 433.24%)? b)“\7"@t- 22 I41 50 SHEETS 22-142 'IIINII SHEETS 22 144 200 SHEETS ® 3 5} 311; ‘6: 3 N '7" Eu‘ a a? 9... -— .rs 565" “é o J‘ V I} . .3 Nd“: 55"” 033 : aft) swcéméiznsm as;: b) 7716 mab'an beam: a! a. paer I'nsfanf in name, So M's Aer fiw‘ mm be. +=o and £64 ‘X/o)'= 0. 4+ a war Mm t, x/é) = 0.02 M, Look (a file Magma/n6 on“. V”: q...m..._.-.u._...mu.. u... .........l... a: j A 10.0 g_lglock with a charge of +3.90 X 10': C is placed In eotric field)? = (3.00 X 103)i - 600j, whereE is in newtons per coulomb. (a) What are the magnitude and direction of the force on Unblock? (b) Iflheblock is released fromrestatthe origin at M: ' m = 1. 00 1%?2 lea, ' (a r + 3x20'5c, Q :=0.whalwillbeitscoord.inatesatr=3.005? - Ex: XID3N/C. 153 .- 4900 ’% Find; a) maflnfiude 414,) Egg dIPeoflon or F ' b)c00rd:mfc of Offijed' Hz: 31005 auger mime W red- aé (0, o) i 6% 6L) Sf iEx" (9.00wo'5c)(5.00x203”’¢)= 0.240 N F3 = (£51 = (9.2.00 mo" (ax-(«co Me) = -% o. 0460 N .._a. F ‘= 0.2VON3 -- aotrs’ouj‘ . I - ‘ - WW +fi’%e=+an*'(%)=+an“(i%%§%) ' l I b) X: 0+ Ye waxy: _, £5)£z=1(€3.2V?~)(3.ws)¢:lo(gm ANS 108m,-21.<om ANS. 2214: so SHEETS 22.142 IOOSHEETS 22144 zoo SHEETS ® I . - charge of radius R discussod in Section 23—6. Show that trostatic force on the electron can cause it: to oscillate -:: center of the ring with an angular frequmcy "" ' _ If “I ‘9 41m0mR3 ’ ékah'c fieid due «f» a Wham "charge disMMM: charged nhj We are inJereskcl in small antpb‘Jude asa‘lbfi'dns = 2462. . .50,N£ will 002. #:e apmfimfim= a” 2‘2: R‘. “SP’JWS'H‘LH MW of F Makes 1%: ekdrm wave. with a. 56mph homo/.112, Mort. °' InFig.23-31.pointdaarges+I.0qand-2.0qamfixeda . a f . :liswncefidapan.(a)findEatpointsA.B.andC.{b}Sketchme e; Z eclric eidlincs. - .. 0 e; 2e 3* - fiml: 606.. ,28 E; - 535-5stw Eledh'c Edd 04:0: pain-l Change 44%. eledn'c field [inc pomé {n.‘ml's will I'VE as “It paper 0) and (-) Sf n5 (h #2 efmém 15(th Wake/m 52%. 22-14! 50 SHEETS 22-142 "30 SHEETS 22-444 200 SHEETS - _ We wit! «(50 00901442 expression 5r 1% ekdn'c fiela’ ab 0 pom-f charge: 61‘ng . @ JIMMD 60.1? is 'lb do pad“ 6) firs!" 50 wt know who?) dirt?de ' ' Haw @Mmmdmp 1.20mnindiamem-ismpcnded ' calm air owing to a downwani—directed aunosphefic electric fieldE=462NK1 (a)Whmisfl1cweightufdmdrop?(b)l-low many excess electrons does it have? ("given l“ ' =1’? : odfamdcrafdrop, b = mic/WM ‘ I | __ “‘3 Nofe= 5m «3 :3 «game, moire? A W . warm; .4 Coulomb Force. :n 4h diacfi'm 99m in 2’. Oqanfi’aafi’on d1arge,Newfon‘s 2"" Law 1 a) w=mj ,so we, need -Hwe mass 0F Me drop we are and 3mm fifit—mwzwae do know {am we dm’fi at” WA-leg/p’ ‘5 ’0 kym3./=v So M30 . 22 I“ 50 SHEETS 22-142 100 SHEEYS 22-14i 200 SHEEI'S .m__..._......-— ................_.. _.... _.__ .__ M.m———___._._. b) Llij Ninn'S Z'dLaw: -"-‘ mag W-QE=O (41m drop is smpmded) ' W*nlelE’ I ‘ ANs. lelg 0-bKIO’"¢)(4(:2“/c,) PH 132 Suggested Problems 66 (a) In Fig. 234?. two fixed POWChaTE3591= “54811092 = +24 are separated by distance d. locate the point (or points) at which the net electric field due to the two charges is zero. 0:) Sketch the net electric field lines qmli- tatively. At points left of g; (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'lézl in this region. In the region between the charges (0 < .1: < at) both fieldspoint leftward and there is no possibility of cancellation. At points to the right of Q2 (where z > :1), E1 points leftward and E3 points rightward so the net field in this range is ' fine, = — 13.1] in thei direction. Although lqll > 92 there is the possibility of Em = 0 since these points are closer to 92 than to ql. Thus, we look for the zero net field point in the a: > d region: g, lei = lei __ 1 E 2 1 Q2 .. 4st.) 1'3 41mg (:1: — (1)2 which leads to $;d=‘/E=¢g_ Thus, we obtain 2 = W as 2.7d. A sketch of the field lines is shown below. 22 I4! 50 SHEETS 22-!42 IOO SHEETS 22446 200 SHEETS 3 - AMMD .f" _‘__‘___ @ In Fig. 23-23, what is the ' @m 6”“ ‘2; ‘ *“Z - d _ Qt: 93‘ 33f? \ias-W 2" ‘ f m; . has an em. m 4+ Poi/9f P ‘ E}: I More: The dam, field dc'flcfim élac as: {he Pow awaits) mm! a. - 'codek 5am haw: be”? daé/ea/ 1’0 Me d’djré/tc. ' Elednt field; due 40 $5.714 charges . . disk of radius 2.5 cm has a Stu-face charge density of .3 pain-12 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its team] axis at distanpe z = 12 cm from the disk? '” e ..——--' -- “M g o 2-" g 2600 W2) R 0 5w: c. (= 5233:“) Ki. ,6 c 'h - 53"“) 4" (V cm“ 3;”; 12M 410’— "- '7" , z mum m7. (0'ng +(.o75.~m . 6°: Kgflonflcfl 5' u 'Z‘fimo 30— ‘7'“0 M _.1 IV?) (GI-14%.???) ‘0 M .. 5’s} 1?, " 1‘0 “a: ( l 50 SHEETS 22-142 100 SHEETS 22-144 200 SHEETS 22-14! '® _mmn smfaceoflhenegafivdychm’gedplateandsflikesthemnfaoeof deopposite plate, lflqnawayjnatimelj X10"s.(a)Wha1 ismespdedofflndmsfikgmcmplawflmm ismemagnimdeoflhe ‘ fieflflfi - "ll" &=D.Dzm 5:534: 2 2'5 5°“57—«)- .. fr; Car-557T; So ____ rm: Acwmrrm 1‘5.¢P”"-- -‘7‘Hu$ am: am 056 CpNé'r‘._ Am. EGZMS , _ - ...
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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 2 - PH 132 SPRING 2005 HOMEWORK # 2 Assigned:...

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