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20091ee1_1_hw4-sol

20091ee1_1_hw4-sol -...

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Unformatted text preview: 4.1. 'l‘hevalueofEatP(p=2,¢=40°,s=3)isgivena.sE=10039—200344-11133 V/rn. Determine the incremental work nquimdtommampc drugoad'utsnceof6pm: a) in the direction of up: The incremental work '3 g'ven by W = -qE- dL, where in this case, dL =dpa, =6x10"a,. Thu M = —(20 x 10-8 C)(100 V/rn)(6 x 10-6 m) = —12 x 10-9 J = —12nJ b) inthedirectionofy:Inthiscma=2d¢q=6x104abandso M = —(20 x 10-°)(_200)(0 x 104) = 2.4 x 10-31 = 24 c) inthedirectionofagz Hero,dL=dsa; =6x10"a;,andso aw = _(20 x 10-“)(300)(5 x 10-') = -35 x 10-81 = -35 111 d) in the direction of E: Hero, dL = 5 x 10-853, where _ lOOa,—200a¢+MOa; _ _ a; — —[100‘2+2009+3009]1/9 _0.267a, 0.51*I!5a¢-I-0.1802aI Thus aw = —(20 x 10-‘)[100~ — 200g +300..]- [02675, — 0.535., +0.002 a,](0 x 10-“) = 44.9111 o) InthodiroctionofG=2a.—3ay+da¢:Inth'neuo,dL=6xlO"ac,whore 23. -3_a,+4a. .0 = [2’ +3, +4211” =o.371a. —0.557a, +0.743a. Sonow aw = —(20 x 10-°)[100a, — 200g + 3005,]. [0371 a, — 0557., +0.743..](5 x 10-“) = —(20 x 10-6) [37.1(., - a.) — 55.7(5, 1,) — 742m - a.) + 111.4(5, -a,) + 222.9] (0 x 104) where, at P, (a, - a.) = (3‘ - 8') = 009(40") = 0.766, (aF - 3,) = sin(40°) = 0.643, and (8‘ - a,) = — sin(40°) = —0.643. Substituting these results in aw = _(20 x 10-“)[254 _ 35.8 + 47.7 + 853 + mane x 10-“) = -41.an 4.3. If E = l20n,V/m, find the incrementd amount of work done in moving 3 50pm chuge e d'utenee of2 mm from: s) P(1,2,3) toward Q(2,1,4): The vector along this direction will be Q — P = (1,4,1) from which em = [3, -a,+a,]/\/§. We now write I = —(so x 10")(m) «a. .s.) - (an . a.» fin x 10") dW=-qE.dL=—(5ox10-°)[12on,.("—"‘gfl (2x 10-') At P, e = ten-1(2/1) = 63.4°. Thus (n, . s,) = eos(ea.4) = 0.447 and (e, . s,) = sin(62.4) = 0.894. substituting these, we obtsin N = 3.1 E- b) Q(2,1,4) toward P(l,2,3): A little thought is in order here: Note that the field has only aredielcomponentenddoesnotdependonéorz. NotedeothatPendQueatthe same redim (J6) from the: axis, buthsve different esnds coordinates. we could just unwell position the two points at theeeme zlocetionendthe problem would not change. Ifthiswereso,then movingalongastnightlinebetweenpendeouldtlns involve movingalongachordofecirclewhoeendiusiefi.Halfweyelongthieline'uapointof symmetryinthefield(mnbeeehetachtoaeethh). 'I'hismeensthntwhensuutingfrom eithetpoint,theinifinlforcewillhetheenme. Thustheanswetide=3tl£asinput a. Th'nisehofoundbygoingthmughtheumeptocedureesinpena,butwiththe direction (rolesofPandQ) revemed. 4.5. Compute the value ofjfc -dL for G = 21m, with A(l, -1,2) and P(2,1,2) using the path: a) straight-line segments A(1, —1,2) to B(1,1,2) to P(2,1,2): In general we would have [Mm/I’m ThechengeinzoccuuwhenmovingheweenBendP,duringwhichy=l.Thu P P 2 fG-JL=/ 2ydz=/ 2(l)dt=g A B 1 b) stratum-line segment: 40,—1.2) to C(2,—l,2) to P(2, 1,2): In th'a case the change in zooms whenmoving fromA toC,duringwhichy= -1. This LPG-a.=[2ydx=[2(—1)a=—_2 4.10. Exp-ea the potential field of an infinite line charge a) with mo refierence at. p = m: We write in general: Vt(p)=- —dP+01=-—ln(p)+01=0atp=po heap Therefore =2,“ LMPO) and finally v.0»)= fi-flnm) w>1=— 2—; (a b) withV=Voatp=poz Usingthereuoningofpnrta,wehmve V¢(Po)=Vo= 2—‘mln(po)+ca=>ca= -Vo+ 2:0an WV)= fihgflgm c) Can the zero reference be placed at infinity? Why? Answer: &. becnuse we would have a potential that 'n proportional to the undefined ln(oo/p). and finally 4.12. In spherical coordinates, E = 2r/(ra + 09),” V/m. Find the potential at Any point, wins the rderence a) V=Oatinfinity: Wowfitoingmal 2rdr l —(19+09)9 ”'6’ —r9 +a2 Withamo mfeteme atr—too,C=0 andthowfomV(r) = Z1r9+a9!. b) V=Oatr=0: Usingthogmudexprubmwfind V(0)=%+C=0 : C=_i V(1‘)=— +0 Therefore 1 _.2 V(r)= 1 - = m :5 056541!) c) V=100Vatr=a: Honwefind V(a)=i,+0=1oo => cane-i 2a 202 Therefore 1 1 09-73 4.16. The potential at. any point in space is given in cylindrical coordinate by V = (la/p9) coca») V/m, where I: and b are constants. a) Where '3 the mo refinance for potential? This will occur at. 2—» 00, or whenever anew) =0, which gives ¢= (2m— 1)1r/2b,whaem = 1,2,3... b) Find the vector electric field intemity at any point (p,¢,z). We use 8V 1 W I: . 309"”) = ‘W = _3—P —;w% = 7 [2°“(W)&p+bm(b¢)l¢] 4.21. Int V = 229994-31“? +2y° +33”) V in free space. Evaluate each ofthe following quantities at P(3, 2, -1): a) V: Substitute P directly to obtain: V = —15.0V h) |V|. This will bejustm c) E: We have as 3' a 12!! E|p=‘wlp=‘[(2” ‘ +29+2y9+3:9)"+ (“”3 +29+2y°+339) " 18: + (621:2? + m) 8.] P = 7.18. + 22.3! — 71.183 VIII! 421d) |E|pz taking the magnitude of the part c result, we find IEIp = 75.0Vlm. e) a”: By definition, this will be E aglp _ -W _ -009», — ”0‘2: +0348... t) D: This is 0|? = «OEL = 628., +2023“ — 629a. pC/m9. 4.32. Using Eq. (36), a) find the enetgy stored in the dipole field in the region r > a: Westartwith d E(r,0)= 41:0r3[2coe0a,+ein08o] Thentheenergywillbe _ l _ 2' (1")2 a - a - Wg—fuflmE-Edu—A [faaaeors [4co: its? 9],?anodrdau on + —2!r 2 1 00 I . 2 1r =—32‘(:2 F. [a [300320+1]91n0d0=—48?:20a3 [—cos'o_eoao]o 4 _ («0’ J -121r£oa3 b) Whycanwenot let aapptoachwoualimit? Fromthe abovereeult, aeingularityinthe energy 00cm: asaa 0. More importantly, a cannot betoo small, or the original fabfield aesumptionusedtoderiveEq. (38) (a >>d)willnothold,andeothe fieldempmion will not bevalid. ...
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