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Assignment4_1

# 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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