homework 5 - PH 132 SPRING 2005 HOMEWORK # 5 Assigned:...

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Unformatted text preview: PH 132 SPRING 2005 HOMEWORK # 5 Assigned: 02/07/05 (Elm/5’; 92/1)”;an 63503562332) _‘ gg/‘C‘REJZZF 42245)Lé/F C In. 2 5' \ - A 1.1_. _._ . ._- ___.__________i. H. _ _ -w_._.__.._m.___..-.‘........d._..__....___ Q The two metal objects in Fig‘ 26-24 have met charges of +"0 . H 'pC and —7‘0 pC. which result in a 20 V potential differenccbetweén Q t' : 2 0 v E them. {3} “Mat is the capacitance “"—""' -f 2 Qlt+70PC3+701V0 C '. - — 2 ((1)z = - 70¢ - “70m ’6 of Lhe system? (b) If Eh: charpee I are changed {6' +200 pC and- ._ —200 pC. what does the capaci—.' \__, tance become? (c) What does thei' . potential difference become? f it a \_ _ EM! 0.) Capaa'kmce C Q. “’ C?" ‘* *ZOOPC Champ; 2 ~200PC, 3% C “W2 .bmew amahna 6’ 1 - ‘Q' Q2 C)N€w 9mm ba‘FérenCe Av‘ I R 9.. 3‘? r O .3 Km I ‘55: m 3; :5? n (h % V (‘3 D— < («3 f 8‘ N x G E; {‘3’ (Ln U} K "R. Q} g“ u 90 U‘} .0 .n E“ EU a 9‘: 313 b) Look a+ €7Mffbns 2c; *2, 20%; 26-17; ma’ 2%” 40452433) 5 ' flflfé Mai/52k Mad (waééwc afiepe’m/s can/j on 746 E and 1% 599mm)»; 01" #5:», W0 ob we 4% make, up % ' capam’an I70 6? Wfljes ) AV Wrfi‘dkd 6/3? 7%) keep 1%6 Cipdéfiéf/fce, F265 (65 W as Me earn: $.65 Md 1%; Jam/52,753” 011‘" M0 oégfédfi do 057‘ ' L. _. mm. i i 5 @111 Fig. 26-30! the battery -! as a potential difference of 10 V | | i and the five capacitors each have a capacitance of 10 HF. What is the charge on {a} capacitor 1 and VI) V1: V3; V4,\/5- (b) capacitor 2? If. M a", )u2JM3J Md); Cim uswafi Wo-fcfi'r'OW-J M315 545M om gAcH GNP/‘t-LJTOR PDT“. bJFF. Atoms (EACH I CAPACITOR Mb amazes) 97011513 B‘r‘ 5W CA?A¢JT02. fig 217mb EQUIVALEW CAP. :12 "g: a I = A. at, - = "23 52,3 C; if C; W Cu 59“}: 5%; = C1347 3‘ C2,}, W Gaga?! = If“ I ,1' f 2‘ a he; : “ff-'- : Tr C éto F “ "‘ {I C2345 Cam C5 23% M I . 935; = C: *Cams M Case = [EMF 3 gNDTg 519mg com EACH CAP. PLATE: . Paras coreMv—ECTED “I‘D T3199. 752M. 0F EATI’ELY Lou-4. BE 3’09 . PLAy-gs “amid—Eb TO He. N56,. 7.5M. «9F BAVEPJ-a’; V @1227; QM £62053, 3 FaWDWJk-vé M3615" 77363“. + TEAM. g=¢r+iz+2v FDWWIKJG c4.“ 1 : fr+25 Amw- = g; +g‘f 23:22 tail/TE VOLTAGE (Loo?) 56:09,: 6440955 we? $111. new man": A?“ Make»?- FMM — TERM. ow: 56m- In 51W §E¢TIOM. 5‘02. Loo? Q: F02. Loo? £22! \Y,“ VI :0 Vt'v‘f "V5 :0 Fa?— L—oo?@1 ‘Y/HVL : D FOE LOOPSDI- NEE: I . PLAvEg ALTERNATE; 5.3.1 915;»; am camafibzs m gait-35$, “a”; :VC. For. 5N.» QEAPJ : a 56W”. 2 = 3E9 “' F "L any. ‘11: V'EJJ iZ-zVLCL ‘i V FDR. a} =V3Ej}, *— cAgF'. 2‘7 “M191, E is -‘ “55:15, AM 34.39th varnas HA:st 355»: MAW-Eb WITH A GUFVED WWN . .._ -.____~__.__.___.__m._.__# _p:’~._, 50 SHEETS 22-142 100 SHEETS 22-]44 200 SHEETS 22-141 “9 (AM PR 3) “x._/ 1' i r WE" an» smug“ F2972. V FIE—s7”: —__________L___.____# \/'-\fl :0 .W V1: MEX}: ’70w/E F051 ‘2 .' it :V'C' W at : NEKT/ 901,415 F0}?— ZS I A. NM fl...- )/ Zrz’sz M fig : éD/LLC gag/v5“ F0; U4: is _= V5 C; M V; 2 éHDV NEKT, $02,425 FOR v.2: QT}. . V: “Vq 'Vs- =0 Ana— V.t = ‘fiov (3R! FOP-— Z '. i E f 3 i E l r 4 i E E 50 SHEETS 2244! I00 SHEETS / 22-144 200 SHEETS 22«14T Kama-AD ("ii (2%“: Wm" 9 Cid-9W7 . 624.9%, C,=39;F,C,,*¢QM i Egg“ a) Si {5 chased: charge on each , i ' “Ln Fig. 26—331 barlcry B supplies 12 V. Find the charg: on j_ "fi'capacimr (a) first whan only switch SI is closed and (h) later . .211 switch 82 is also closad. Take CI = 1.0 pF, C2 = 2.0 1.113, u 3.0 ,LLF, and C4 = 4.0 HF. TWO DIFFERENT APPfloA-cefES 72> mas PROBLEM A126" PREséNTEb Rs ' b ' .Sowncms. )5! “(52 J" 6:3: 5" 50mm 1 C’qodai’m :‘n Sena; aux» awe/fling? 32mm mm . . C; - first wn'k. eymfikns dcscn‘b’nj 14%. Circuit C, and C3 arc I}? sank and C2 dfldf’ydre in seats. This mam: “‘ J ‘ ' . ~ cbw: L h?“ ' .F_ -o 3 9am}. 501mg firch . VEDA/q — 5:: a __.., QB: gig, _3 Q: Qqu: @699“qu :(8+I)Qq ~—>> C95,: HMS. Camer in Wsandzh fim/é/"égm’mém‘ (Wm! W C E5? 6~ 5 c a) f——{. SW41 '5 EEE W ' afié of“‘iq_ CMmfl G1 ' v I _J 1 - me 1 1 1 cc -- — 4- ~ 3 z " --"> C = a...“ q C”, C; H5 C43 CI+C3 Czq C2. +C~{ 2‘; Cal-Cg .C; a“ 'jflJ-Jlnr’SJ-Sc' 9'3‘7Q‘ =03' Al‘keWISfi) qu: CR1: Gas-“(9: r"6-33,: C13V= __.'_C,.3_>V =4AB. h CI+C3 Oz? :01: on! g CZ‘IV 1662f: )v =AUS. b C‘ C’ 2 4 C12 634 > 5W“? H 3 Me (21+ —Cq Cw * V- VI:VZ: V12, V3=V4:V3g’ <52: =ctw=qw¢ ,ozscaug czvn 50; (3.3”??st flew/q wag We can 0W9? 14% Wfiu if we knew M2 and _CQrz - CP _ .. Viz' 5; JVSLJ‘ “5:3 3 Qz‘qu >th/I1‘LVH -: : g} 03 a 012. (9,1, .. _J_ v Viz-“jg”! Cu. + — H Kg! — Q2[Ciz+&:f] C34 ("2 i E Y CM: CI+C2 é”,de Q chzc3+cq l I m_ mm;m_w__mfiwwwmmw_;4_. I a 22 141 50 SHEETS '22 142 100 SHEETS 22 144 200 SHEETS AD 0: szn W Q‘CSVEQ: fag/C ANS’ _J Wm _-.__.‘. _.___,__ _._,__-_-M_d._c.._. ..,.-___m.__._‘____.__ _m— P» - ‘ r _ :26 A paraIIeI-plate air—filled capacitor having area 40 cm3 and GI 2 V: i are spacing 1,0 mm is charged to a potential difference of '----.._....._. ' -3 ' 600 V. Find (a) the capacitance, (b) the magnitude of the charge d : 3 /X{0 M on each plate, (c) the SEC}er energy. (11] the electric field between ' mlhe plates. and (e) the energyifiensity hem-ecu the plates. A :- 2 = x/D'fi/mz End: a.) a 006‘ . b)l®] e)u, c) u 50 SHEETS I 00 SHEETS 12-] 44 200 SHEETS Qua/U pm: 64/046453 51‘3"“! 6’7599 : ._ Eon; 37-35x/0"2F4n)(t!ox/0‘W) 6%) Eva para/M p/ak (apaafir, C ' a“ (egzo-e i7.932)- . = 5_5X/0"”F rams. lo) Q=QV = (3,5r/0‘”F)(000V)=AA5_ a) Mai/2w? = é(3.5xxo"'F)((aoov)2=A/us. 22—] 4! 22mm 9 (e5, 29129.63?) _ . d) was! v-=> (400v ;. 62.0x/05V/mAfl/s. 8) w 2205‘ = :7."(f-ng/o"2%a)(é-owow/“023ANS «3.2923 9.931) : and (e) the stored energy for each capacitor. Assume the numericai values of Exercise 10, with V = 100 V. t ‘ I i ' .r' w 1 Fig. 26-26‘ find (a) the charge. (b) the. potential difference, V C'I CS C5: WO/xF % (22 | Md (2.) 62;, Q1, 62:, FIGURE 26-26 Exercise 15 and Problem 47. VI ; Va. )' V5 - ' ' CM, Uz, U3 Capaah‘ars m finks «WM; Dam/M “M a. ( @amflel-plale air—filled capacitor has a capacitance of 50 pF. [I (ajl If each of iL“. plates has an area of 0.35 m2, what is the sepa— 50 SHEETS 22-142 100 SHEETS 22-144 200 SHEETS 22-141 r . > c d l? V' Mb“ I Capacr'hane mm a. dIE/{a‘r‘rt a) gut; -: 2 F b) C’= ICC (3529—27 9034) C“: (5.0)(5‘DX/0“"F> = m5 fl=0.35mz {£1.41 add b) new Capaaknce , C’) W71 WliH/l 0- didmb tc=5.(o a) C = 8%: A d: €064: (€.£/5&’o"“mz)(0-3sz) :IANS‘ i i —----—-—--- --—‘--ww ----—v—---‘—v—v-———‘——__1____‘ { given: C = SOPF = gape/0‘”): hi 4J1 @2951 ("Lia/wag A1__ MC 7 f c; ’: waafmmg mm; M 0“ 51gb ‘thbéfivkzd bJQJw‘QéM f"— “ ‘ “g mefie; _ CL- 3E Aft all: arpet d day y hand h "“““"""—--—- ——-_-.—._..- . er youw overac ona ry , our comes ‘ 1 Gwen: R=25mrazsm close to a metal doorknob and a 5 mm spark results. Such a spark AV = VCSpherg) ~ vacuole) = r6 w means that there must have been a potential difference of possiny l5 kV between you and the doorknob. Assuming this potential difference, how much charge did you accumulate in walking over the carpet? For this extremely rough calculation, assume that your body can be represented by a uniformly charged conducting . Fffldt Q sphere 25 em in radius and electrically isolated from its ._._..-—- it surroundings. J | fl 2 fl Ill HI III I III III at . : I z i 1.") {H U! D O O In a o '- P1 - cw e E E E! a’a rim-i . t“ I"! N . /" C nMMn Saluo'on Capaa'faflce :fsolakd Sphere. _ .1 E O V(Sphcre)-—\/C .ob)=l5 kV - ii C= 4776012 (35. 20% pass) “H C) = CVcsphm) Q: (#fl6¢R>Vts r3 = EX: (O'ZSMXI5X/OLV)= ' N5. PM 5 k 8,.quIOq/umpé; A 1 CHM “q I _ _ _ ________%_ 9E. T h fal ' f 'l h the , — _ tion ofwlffismcritf aid :EgcTJSe oafvfo fling: $153253: Sheds have "HR 59, 1 12 V. (a) Calculate the area of each sheet. The separation is now decreased by 0.10 mm with the: charge held constant. (b) What is J (196? I lhc new capacitance? (c) By how much does the potential diffcr- ' 3. = 41 i \_.. once changc? Explain how a microphone might be constructed - C F- ;qo F using this principle. v: v 1-:- H - d=fl0mm dowo 3M £7" . 1mm Find: 50A .47 ' - b)d’= d- OJHM +5116! Hw ii A new Capacflance , C’ c) AV= v’ -v’ ' 4-1.- 9) aA L Fora Farah! Flak Mpacflbg =a— Z6-7P'G’32) . Q8 1 —3> A : C31,: (fox/a"‘F)(/.6>qo‘3w»\ _ /,__{x/D"5m" d 5» MEMO-'chm, ANS. '9) Qt; 6"“! : C 35.9.}; I ; 6.!”- éfi": 5.0L,“ P l AHA d-aa" d a” ' C” (fl-eax/o"”%.»)(uxxo*3m'~) : LIX/0""F: NPF (flax/0‘3," - onxo' m) $5 On :‘n “1e, 0F WK W305 CUIbMflMS) e 1%?— separafmn 33.5%an Wfle +900 capda'fvrsm’ezfs. Th5 resulfi’rg change; r}: 1446 MM afilCé/mce arc Mum/ed as sound blj Speakers. 50 SHEETS 22-142 100 SHEETS 22-144 200 SHEETS 22-14] in Amman QMZS—flit’ 221’. (a) A potential difference of 300 V is applied to a series connection of two capacitors, of capacitance CI = 2.0 11F and given: v {Mug}: 300v capacitance C2 = 8.0 ,uF. What are the charge on and the pcten- C 3 Z 0 F- _ tial difference across each capacitor? (b) The charged capacitors i ‘ fl ' are disconnected from each other and from the battery. They are ' _ F- d'ien reconnected. positive plate to positive plate and negative 6 z " plate to negative plate, with no external voltage being applied. What are the charge and the potential difference for each now? (c) Suppose the charged capacitors in (a) were reconnected with plates of opposite sign tugether. What then would be the steady- state charge and potential difference for each? Mra)QlQ1-JV'7UZ b)New tram-.0,'}Q;',U,‘,U&’ ‘ b . . () 9n”: (C ‘31)," + " + — i Cl C. '1" “I l - .9 E CZ Q3”... 01" Vz‘ E l 1 ) Since. C, and C; are «in asides ,(see 9-6934) ' ' 1 Also, 6e p. 634) A 1/2 = v-v: We have unknowns = (Q) L}J ’t/z attde 75w g/gafiw, .so we need some more. Infirmafi‘m: :8 =Q:_Q—:Civ :clvf =.£!'—-\[—“2‘{0 I I Cl V... w. V-I'x. CZ v—v. “"4 we V f VN‘VJW @toscigcnms. E lb) New C, and (a are in Wile/J5me, a poknfiéz/drep app/fde fidr embihafibn would (duff in if)! 5M: pawl/72d fi'firme 45135.5 ad! 1 6‘10th (903") W‘VL’ @379 0; but Q‘wzkconcpgzo? ' 1' Combim film in» rehnmshliw ' (cmgmmm ‘99 “"345 i VII: c, “"Vzlz % *5? Q5: (Cl/ca) 9; 4159* QEZQ‘Q; ‘ r I. ; 2(5) -4 (C'WJ=Z<9~©’ ezswzfisoeoemfli «aw-e; i to) Cheri/gas of opposfie ai'gn can now How Mum inc (amazing. ‘ 1 Since Hae i’hi‘fih/ 0111135 on 4414 capaa'y‘are are gym/i C} and (z are (fascia/yd filler“ t1”;t_t"_‘l__-—.r" o I I _. _ . .—_-.:~u==-L.'—"' PH 132 Suggested Problems 50 SHEETS 22-142 100 SHEETS 22-]44 200 SHEETS 22-]4‘! 371' AM FAD Check this formula for limiting cases. (Him: Can you justify this arrangement as being two capacitors in parallel?) (b) “(we CAPAc/UOR am BE WATED M ‘ A2 Two ck?x\orroms 3.24.: ‘PAMLUEL. EACH erH AREA 13. Mb "PL.ng 95?. buff. 6‘2 2- —-- FMEZ) can“)?! DIE¢£¢IFRJC WTEN'kLg 142/ brad-£44. wmqmws ’K, M31: K; Rafigcflvggfi I} m 0 )- 7/45: 715-9057 FOR. A ?AML£€L.T’LATEJ CA? parallel-plate capacitor of plate area A is fillbd with two _——- ___Jfi_wg_bxz_ 5 0 SHEETS 22-142 100 SHEETS 22-] '14 200 SHEETS 22-44] AMan a). m [a ,f a. C dielectrics as in Fig. 26—3523, Show that the capacitance is _ 230’dl KlKl d Ki‘sz l 1' Check this formula for limiting ‘ cases. (Him: Can you justify this i mgcmem as being two capac- [ itors in series?) 5591' —-: $2 ((53/4547072; CAN BE WATEZ} Two CA? AQJTOR-Ss :CN 'Ex‘iit‘gi A9- gfiw WITH AREA A Nu) MAW: SEE Mgr. .21 Ftuflgp “2/ hrEwar. MATERIALS, 1A, M75 IR; Wear I vat/7‘ : NOTE‘. IF fifz’yka THE—M C: &A K & 7W5 flag-var 510,2 A P Am MEL H hire CAP, SO SHEETS 22-14? It") SHEETS 22-144 200 SHEEYS 22441 6*) AMflnD . A slab of copper of thickness 1; is thrust into a parailo‘l-platc capacitor of plate area A. as shown in Fig. 263?; it is exactly halfway between the: plates. (a) What is the capacitance after tho slab is introduced? {b} If a charge q is maintained on the plates, what is the ratio of the Stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? Is the slab sucked in or must it be pushed in? *4. éfi - “l C'JI‘UM = q: c and“. flag].- a) new Capaa'z'arte J C’ b) Wa’ G)W¢m Slab 1's W<O or is M170? “ d f H—‘i FIGURE 26-37 Problem 61. Paraik! Flak Capaa’fir‘ waa a, o’fekdrfc. 62’) COPPW f5 0— Cmdudbq 60 Me ar'r space, befiuem Hie Copam‘or pkh‘es is reduced (93 Hit Copper flak. The Copper pick beams pm {2.94 0 arm mom 5% A! (I Befimffieslabbfnserwf:ct%a MU IPaf-G “’6'. M46! 1% sick :3 :hserfcdr AN5. 2 ' ' - LE: 159’}? pu'zc .01 QZMAANS (1’ch C, akin/{,3 d—b ' c) w=au= w—u (a; #1,.) 12—3.: f1,-_t>=1;‘[d_-_b .. awake—j 2c" 2c 2 C G 3 2.11 in z 6% 24A ANS L T The work is NEC'pHTIl/E) 50 Hi: slab is _“5uckcd~zh.“ ...
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homework 5 - PH 132 SPRING 2005 HOMEWORK # 5 Assigned:...

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