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Assignment4_1 - 2 1.5 2.1.13 2.1.23 IDENTIFY Apply...

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Unformatted text preview: 2 1.5. 2.1.13. 2.1.23. IDENTIFY: Apply Coulomb's law and calculate the net charge g on each sphere. SETUP: 'I'hernagnitude nfthecharge ufanelectrun is 2:11:11 3-: 10'” c. 2 Enema: F = 4L1}. This gives |q| = Jugs-1 = image? 3-: 111-” muses m)2 = 1.4-3): 10-” c. And :qu :- therefctre, the tntal number of electmns required is r.- = |q|re=(1.43xm'1‘ omens-10'“ Clelectrnn} =sso electrnns. Et-‘ALUATE: Each sphere has 391] excess electrons and each sphere has a net: negative charge. The turn like charges repel IDENTIFY: Apply Coulomb’s law. The turn forces on g; must have equal magnitudes and uppnsite directions. SET UP: Like charges repel and unlike charges attract. EXECUTE: Thefc-Ice 17"} that g! exertsnn gghas magnitude Fg=qu§3| andisinthe+xdirectinn 1—51 rnusthein 2 the —Idirectiun,se glunlsthepesitive. F1 =15; gives klfillfgl =quzqusl. r r r 2 EDDcrn 2 1 2 = —1 = .00 ' =tl.lr'5tl C. Isl elm (3 new m] n EFALUATE: The result for the magnitude of 431 doesn't depend en the magnitude of g; . IDENTIFY: ApplyCuulnmh’s lawtn calculatethefnrceexertednnune ufthechargeshyeachnfthe utherthree and then add these forces as vectors. (:1) SET UP: The charges are placed as shnwn in Figure 21.233. Q1=qz =gs=94=q Figure 21.13: Consider forces nn Q4. The free—body diagram is giveninFigure 21.1311. Take they-axis tuheparallel tethe diagnnalhetween q} and q.I andlet +3; heinthedirectiun awayfinrn 93. Then F1 is intlle +y-dlreclinn. 1-1} = +E nuns-15" =+EHE F3, =+F3sin45" = HQNE F = +P;cns45“ = +F3H'i Flglrelllfih me=s+d+d=n %=E,+s,+sy=wfi}ii+ 1 9‘3 ‘1: [“2451 43:51.3 4EQE=EK 3 a: 9'1 (1+2Ji15anmrmnrmnhmges Smart.“ EVALUATE: 1n generalthe resultant farcecn menf the charges isdirectedawajrficmthe apposite corner. The fercesareaflrepulsiwsincethechargesareaflthesame. Bysjrrmnetrythenetfcrcecnnnechargecanhavenc cmponent perpendicular In the diagonal of the square. E F _ SETUP: in: is Ward a negative charge and away finnl a pcsitive charge. EXECUTE: [a] The field is taward the negative charge sn is dcwnward. 21.111 IDENTIFY: Fur a paint charge, E = Ii: 1:" = [ESE-<11]! N-mIICIJM=432 are. {0.250 n1} 3: _ 2 2 (h) r: IE: (sssxisw mil: )[muxiw C} =l_5E|n1 E 12.0mm: EVALUATE: At different paints me electric field has difl'erent directions but it is always directed toward the negative paint charge. E I,.2 charge girtan electric field E experiencesa force fi=qfi SETUP: 'Iheelectriefieldofa negative chargeis directed toward the charge. PeintA is 0.1111] mfiom q: and [Hill mfiemql. PeintBis 0.1m mfi’nmgl and H.350 mfiemqg, EXECUTE: [a] The electric fields duete the charges atpoirltA are shot-min Figure 21 31:1. |q1| g I gfilfixlfl'gfl E =e—= 3.99 II] N- re — 1 [ x m ) (utilise)2 3:31 E2 [email protected]=[fl.99xlfl" N-mzflnjm r3! {dictum} Sineefiie twufieldsareinnppesite directidnswe mhlracttheirrmgnimdeste findflienetfield E=E1 —El =1l’..$r'4::cll}|'l M‘C, toflieright. (h)The elechicfielthatpeimsfiare shuwniriFigure 21.311}. El [email protected]=(sssxin9 N-nflfizjm=ifi]9xlflj we rm {flJIIIIIm} E2 = FIE =[399x1fl9 N-nflCEJ—lz'jflw C T}: (0 35‘“ m}: Emceflmfiekisaremfimsarmdfiecfimwaddflaeirnflgninflesmfindfilemtfifld E=E1+E, 4.543;“? rec, to fileright. mam, E=BJ4x1fl3NICJDtteright Thefercennapmtenplaeedatthisputntwmfldhe F=gE=fLfiflx1lTE 01(334xm3 NIQ=l_-1flxlfl’"N, tuflseright EVALUATE: AmmmmmmmrmmthMdeeemmmmu 21.31. IDENTIFY: Furapcintcharge,E=k .IhenetfieldisthevecmrsimlnffllefielikproducedbyeachchargeA = 2.511 X103 me = 1.124): lfl‘ MC =9_1Txlflz NEG the field t—EIJSI} m—H—fldm m—r s—{lJflDm—H—fllfifl 111—? El 3 u-r—o—in- i 3 = I i in E1 A £1 #2 p] I {1'1 {a} {bi Figure 11.1] 21.38. IDENTIFY: Applycuflstant aceelerafinfl equatienstethe mefiendfthepretnn. E=Ff|q|_ SETUP: Apremnhasmassmp=l_d?xlfl'fikgandeharge+clet+xbeinfl1edirecfinnnfmefinnnffl1epmmn EXECUTE: [a] vfll=fl_ a=—_ x—Ifl=vhf+%afgives Jr—er=%allt2 =é—F. Behring forEgives m P E = 203mm) was? xlfl'” kg} (Lanna-13 C)[l.5fl>c:lt‘l" s}? (h) v: =vm+axt =£t=113xlfl4 this. 37: F EVALUATE: The electric field is directed fi'om the PDSifiVElj" charged plate toward the negatively charged plate aridtheforce ontheprcrtrmis alscrin this direction. = 143 we. 21.54. 21.63. [a] IDENTIFY: The field is caused by a finite imiformljr charged wire. SETUP: 'I'hefieldfor suchavvireadistancer fi'omitsmidpoirtt is 1 1 1 a E =——=2 —. 29W} Liam)” +1 [4?Ffii]x1f(rfa)z +1 (isotcitiEr N-mitcfifitsxiot can} son cm 1 +1 4.25 cm (11} IDENTJEY: The field is caused by a uniformlyr charged circular vvire. SETUP: 'I'hefieldforsuchatvireadistancerfi‘omitsmidpoirttisE=%%.Wefirstfindfiaeradius Ken 1: +0 ' Enema: E = = 3.113 x 111‘1 we, directed upward (ovens m} ofthe circle using Ecru-=1. EXECUTE: Solving for r given :- = Ma: = (3.50 unfit: = 1.353 cm The charge on this circle is Q = all = {1T5 nCJmXflflESfl m) = 14.83 1:0 The electric field is 1 flflflxlflgN-mzfcz 14.33xlfl'9fl’m cocoon: 9x _ E t E = __ _ 4m {f +a1 )3” [(tlflfitltlm)’ +{ti.o1353m)1]”2 E=14s x IO‘MEmeartL EVALUETE: hbofimses,mefiddsmeofmesameorduofuagnimdemdflmvahflmediflumthmanseme chargehasheenbentintodifl'erentshapes. [a] IDEXICIEY and SET UP: Use Eq.{21.14} to relate the dipole moment to the charge magnihide and the separation d of the two charges The direction is fiom the negative charge toward the positive charge. EXECUTE: p=qd=(4.5xlfl'9 Cjfilxlfl'a m}=1.4>¢1l‘.}|'11 lIii-m; Thedirectionofji isfiemq. towardqz. (1}ijva andSET L'P: Use Eq. {21.15) to relate themagnitmles ofthe torqueand field. EXECUTE: r = pEsinpfi, with 435 as defined in Figure 21.63, so I pains = sexist“ N-m {1.4XID‘11 C-m)sin36.9” E: = 3151] NFC Figure 21.53 EUALUATE: Eq.(21.15} gives the torque about an axis through the center of the dipole. But the forces on the two charges fcsma couple {Problemllfl} andthe torqueis thesamefor anvasisparalleltothis one.Theforceon eaehchargeis |q|E andthemaairmminiomentarmforanaaisatthecenteris dfl, sothemaairmmitorqueis 2flg1E){df2)=1.2xlfl4N-n1'I'hetorquefortheorientatinnofthedipoleintheproblemislessthanthis maximum. ...
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