griffiths_li_needle

Charge density onaconducting needle David J.Griffiths and YeLi Physics Department, Reed College, Portland, Oregon 97202 (Received 9August 1995; accepted 7December 1995) ‘Weattempt todetermine thelinear charge density onafinite straight segment ofthin charged conducting wire. Several different methods arepresented, butnone yields entirely convincing results, anditappears thattheproblem itself may beill-posed. ©1996 American Association of Physics Teachers. LINTRODUCTION I.SOLID MODELS Imagineastraightsegmentofconducting wire,length2a,A,Ellipsoidal‘onwhichweplaceanelectric chargeQ(Fig.1).Question: intheabsence ofexternal fields, howwillthecharge distribute ‘Thecharge density onanellipsoidal conductor itself along thewire? That is,what istheequilibrium linear gy 2 charge density, 4(x)? Itseems obvious thatCoulomb repul- gtptacl 1) sion willpush charge outtoward theends, justasthecharge .onasolidconductor flowstothesurface. However, itisnot Withtotalcharge Qis* clearhowmuchofthechargegoestotheends;presumably Q{xyt22-2someofitisleftspreadoutalongthelengthofthe‘‘needle.”” o-alate] . (2.2) ‘Thequestion sounds simpleenough, anditmustsurly 14 arethetoesemivaxes (Fig.2).faveBeenconsidered ToneagebySommerteld, Smythe, “TocalculatethetotalchargedQonaTingofwidthdx,we ,ormaybeevenMaxwellhimself.However,wehave f foundnoreferencetoitintheliterature. ThereasonmaybeOtthatanelementofareadAonthesurfaceisrelatedtoits °au jectioninthexzplaneby thattheproblem asitstands isillposed: theanswer appar- PY entlydepends ontheparticular model usedtorepresent the dxdz=cos 6dA, (2.3) needle."Ifwethinkofitasasolidobject,weareobligedtowhere@istheanglebetweentheyaxisandtheunitvectorftspecify itsprofile: isit(say) anelongated ellipsoid (Sec. normal tothesurface:ILA),orisitpethapsathincircularcylinder(Sec.IIB)?As aawweshallsee,thesetwocasesleadtoradically different re- 0sO=j-f. (24) sults, andthedifferences seem topersist even inthelimit as Now, fcanbecalculated bytaking thegradient of thecrosssection goestozero.Alternatively, wemight model f(x.¥,2)=(x/a)?-+(y/6)?+(z/e)?, and.dividing offits thesystem asacollection ofcharged “‘beads”” onastring length:stretchedbetween—aand+a(Sec.IIIA).Wecansolvefor _fety?2x yztheirequilibriumpositions,andexaminethelimitasthei-(S+5+](2.2.4). (25)number ofbeads goes toinfinity (and thecharge oneach one goestozero). Orwemight putthebeads atfixedlocations Thus along thestring, andsolve fortheequilibrium partitioningof y thetotalcharge (Sec.IIIB).Willthesetwo‘“bead”” models 6080=ae (2.6)yieldthesameeffective linearchargedensity? Towhatsolid Bea 16+21 shape (ifany) dothey correspond? Unfortunately, formore Evidently thecharge onapatch ofsurface above dxdzis thanfourbeads itisprohibitively difficult toperform the 0b1calculations analytically, andwemust resort tonumerical odA= = axdz. en methods. [However, ifwesubstitute aHooke’s lawinterac- 4macytionforthetrueCoulombic one,theproblemisexactly Thetotalchargeontheringisfourtimesthechargeononesoluble (seetheAppendix).] quadrant: Inthispaper weexplore eachofthese models: solids in Ob peaSec.IlandbeadsinSec.II.Afterthat,asasortof“reality dQ-—— ax <dz. (2.8)check,”” itisinstructive tocompare thetwo-dimensional ana- mac Jo y log:thecharge density onaninfinite conducting “‘ribbon’” From Eq.(2.1) wehave (ofwidth 2aandinfinitesimal thickness). This problem has “oOthevirtuethatitiswelldefinedandexactlysoluble,sothey=bvt—(ala—GleP es)methodsappliedcarliertotheneedlecanbeputtorigorous Theintegralsimplifiesifweletu=2/[eV1—(x/a)"):test(Sec.IV).Intheconcluding section (Sec.V),wesum- _‘marizeourresults,andreturntothequestionofwhatthis ag-2axfatuoy 10)problem really means. wa JoJi-u? 2a 706Am.J.Phys.64(6),June1996 ©1996AmericanAssocationofPhysicsTeachers 706 y v a, a e “ ° z Fig.3.Conducting cylinder. Fig.1.Conducting“needle.” whereRistheradiusofthecylinder(Fig.3).(Thefunctional Astonishingly,theeffectivelinechargeM(x)=dQ/dx isformofthesingularity atthe“‘corners”” isdictatedbya Astonishing, BeN(x)=d0, generaltheorem:itgoeslike5"where3sthedistance fromtheedge.Smythe’sansatzsimplyincorporates this Mx)=O)2a. (2.11) structure inapower seriesexpansion.) Anapproximate so- Since thisresult isindependent ofbandc,itholdsinthe lution isobtained bytruncating thesumsatn=randn=s, limitb,c—+0, whentheellipsoid collapses toalinesegment _Tespectively. Smythe provides analgorithm fordetermining along thexaxis,Conclusion: Iftheneedle isthelimiting theCoefficients A,andB,andacriterionforchoosing thecaseofanellipsoid, thenthelinear charge density iscon- optimal values ofrands,foraprescribed degree ofaccu- stant.Inthiscasethetapering oftheendsexactly cancels the TY:Heapplies thetechnique toparticular values ofthetendency forcharge topushouttoward theextremities. aspect ratiob=a/R ranging from{upto4,buthedoesnotexplore thelarge-b régime that concems ushere. Intruth,Smythe’s method isnotwellsuited toourprob- B.Cylindrical lem.Itispretty clearthatweneedonlyonetermintheB ‘Theexact charge distribution onaconducting cylinder is series, andoff-hand onewould suppose thatthelarger ris notknown. Inhispioneering study, Smythe* remarks that thebetter. However, inpractice thecalculation iswildly un- “the literature isblank onthissubject;"” Taylor? adds that stable, andmany ofSmythe’s ownnumbers (obtained with theproblem “mustberegardedasintractablefromthepointtheaidofadeskcalculator)areincorrect.Taylor’refined ‘ofview ofconventional methods.” Anumber ofauthors Smythe’s method, using aslightly different series that con- haveextended andimproved Smythe’s preliminary results,° verges _more rapidly. Applying Taylor’s technique, withbutthelimitingcaseofzeroradiusremainselusive. b=1000andr=10,weobtainedthelinearchargedensitySmythe begins byexpressing thesurface charge density plotted inFig.4.Thisgraph appears toconfirm ourintuition ‘ontheends(¢,)andonthesides(@,)intheformoftwothataportion ofthechargepushes outtothetwoends,series: leavingthedensityrelatively flattowardthemiddle.« However, Djordjevic,’ drawing onextensive numerical olz)= a? x2y"-13, studies using aquite different approach, reports thatthe(2)=BeAga?x2)", (2.12)Chargedensityonalongconducting cylinderisessentially . constant [\(x)=Q/2a] over theentire rod, except inthe odr)=ZeBAR=P)", (2.13) y oa oa i os —<p SZ ws xa Fig.4.Linearchargedensityonaconducting cyindr,using@=1,Q=1, 707Am.3.Phys,Vol.64,No.6June1996 D5.Grifittsand¥.Li 707 29+ +++ 1 0 «0 x oo oa Fig. 5.Four equal charges onafinite wire immediate vicinity ofthetwoends; thelength dofthere-. . gionsoverwhichitdeviatessignificantly isproportional toR Te oesort (Specifically, d~SR), andthetotal charge onthese twoend caps isproportional toR/a. IfDjordjevié isright, then the charge density inthelimit R—0 isconstant fortheentire oa cylinder, asitisfortheellipsoid. ILLBEADMODELS? SSS TS A.Fixed charge ‘Supposeweplace2nequalpointcharges(q=Q/2n)on thelinefromx=~atox=a.Ourtaskistofindtheirequi- ylibrium positions. Ifn=1, thetwocharges willobviously repel outtotheends. However, forn=2 theproblem is os already nontrivial. The outermost pair will beat+a; letx ‘denote thepositions oftheother two(Fig. 5).Theforce on . . thecharge at+xis . “ . po 1it L) a1 re oe©Ware \(etay™ (2x) (ax) 6D Setting Fequal tozero"® yields aquartic equation: oa (a?-x?)*= 16ax?, (3.2) towhich the numerical solution is SSR ST Ss TES x=0.36148a. 3) Ingeneral, the 2ncharges areat*(x1X2,.0%q—15 x,=a), andtheforce ontheithcharge is y 2/8 a q¢1 1 Fr=f—|> —+d —— os -> G4) N “4it a) Setting F,=0 (fori=1,2,...,.2—1) yields n—1coupled ne equations fortheequilibrium positions ofthecharges. For example, with n=6 (12charges), wefind on 1=0.100102, 15=0299130, Sa as TE x=0.494410, Fig.6.Chargedensitiesfr(a)n=5,(6)n=10,and(6)n=100. x4=0.68241a, +50.85692a. graphsappeartoconvergeasnincreases, suggesting thatHowever, what wereally want isnotsomuch theposition there isawell-defined limit function asn—»2:. Indeed, an ofeach charge astheeffective charge density, h(x). Tothis expression oftheform endwecompute 2 a MW=A+ A 0)M==——, G3) oe Fini (inspiredbySmythe’streatmentofthefinitecylinder)fitsthe whereresults quite well (see Fig. 7),ifA=0.384985 and Flot 2.6) 3=0.083684. However, wehavenotheoretical justification FR ta) forthisform, noabinitio means forcalculating theparam- isthecenter oftheinterval. Theresulting plots (using Q~1 etersAandB,and,infact,norealassurancethatthegraphs anda=1)forn=5,10,and100areshowninFig.6.Theconvergeatall 708Am.1.Phys,Vol64,No.6,June1996 DJ.GriffithsandY.Li 708 y ou eaoag ues 7 Fig.9.The2nevenlyspacedchargesonafinitewie. “ a(24 mg iZ uoa i(3aoe ®eye jt (G-i*d Fig.7.GraphofEg.(3.7)superimposed onFi.6) Inequilibrium F/=0, so 3 yw ywqa? =0, B.Fixedposition 2%wae’ Ep3,Ga3.14)‘Theprevious modelisalgebraically cumbersome because on,thevariablesweseek(thepositionsofthecharges)appear wherei=1,2,3,....n—1 (qq,ofcourse,issubjecttotheextraquadratically, andinthedenominator. Evenforthesimplest constraining forceofthewire). Meanwhile, thetotalcharge (nontrivial) case(n=2) thisledtoaquartic equation (3.2). isQ: ‘Analternative model, inwhich wefixthepositions ofthe . particles andlettheircharges varyyieldsasystemoflinear y2 (G15)‘equations. a2 .Forexample, with four evenly spaced beads there aretwo distinet charges: q;at+a/3, andq3ata(Fig. 8).Inequi- Taken together, (3.14) and (3.15) provide nsimultaneous librium, thenetelectrical force onq;(ata/3) iszero—the linear equations forthenunknowns. The corresponding force t0theleft(due toq2ata)balances theforce tothe charge density is right (due tog2at—aandq,at—a/3): qi _n-12 1(22) 1(2%+n) 7 RMaD=Gain Pde (3.16)Grey\@}”tre,\Qayea?!” G8)wherex,=(i-1/2)d. Theresultingplotsforn=5,10,and where d=2a/3 istheseparation between thecharges. Itfol- 100areshown inFig.10.Again thegraphs appear tocon- lows that verge asnincreases, andtothesame function asbefore 1 (compare Figs.6and10)nents 39) ‘Aninteresting variantonthisapproach issuggested bythej work ofRoss.'” Instead ofadjusting thecharge distribution ontheotherhand,thesumofallthecharges isQ,so eee aeneeneoncgi. chars debate ata=ON. (3.10) thatthepotentials beequal atpoints midway between the charges. Thepotential atapointd/2totheleftofq,is Solving, we find ; 1{2 aie 4 =n, =i 3.11 =—— sy nerd,are eww wal Tama, j-THa Similarly, for n=3weobtain : 3725 1980 ans +>oe @.17) 4"764% BTA 4Tyg]P12) mae . Letting $476 Vd,where Visthecommon potential, we Ingeneral, for2ncharges adistance d=2a/(2n—1) obtainnlinearequations, apart,therearemunknowns, 43,42»--.sdy (Fig.9).Theforce . oncharge 4;is 7 gw © 4 "44% 43ty %(i+j-3/2)%(i-j-12) 2(j-i+12) =50, i= em (3.18) ® qgeti4, %& which,together with(3.15),determine then+1_unknownsGide). Forexample,whensn=2wefindg,=(2/9)Oand’q,=(5/18)Q. InFig. 11wegraph thecharge densities -0 0/3/38 forn=5, 10,and100;theagreement withprevious results is excellent, forlarge n,suggesting again convergence toa Fig.8.Fourequally spaced charges onafinite wie common’ shape. 709 ——_Am.J. Phys, Vol. 64,No,6,June 1996 DJ. Grifiths nd¥,Li700 oa oa Sa ae TES et STS TE at 0 Sa ae ass et IST Tes Passos oe ee et SS TR st Fig.10.Chargedensitiesusingthefixed-position method:(a)n=5,(b)Fig.11.ChargedensitiesusingRoss’method:(a)n=5,(b)n=10,and(e)n=10,and(6)n=100. a0, IV.INFINITE CONDUCTING RIBBON A A.Exactsolution VO"Treo" 42) ‘Theanalogous caseofaninfinite conducting ribbon, of Equipotentialsareellipses,withsemi-axesacosh(V/Vo) and width2aandinfinitesimal thickness(Fig.12),canbesolved @Sinh(V/Vo). AsV-+0theycollapsetoaline,~a<x<exactly. Forthisisastrictly two-dimensional problem (the +4,along thexaxis.(Inthisproblem itisconvenient toset potential isplainly independent ofz),andhence accessible to _thepotential equal to0ontheconductor.) the methodofconformalmapping.IfAisthenetchargeper_Thechargedensityontheribbonisdetermined bythe unitlength (inthezdirection), thepotential V(x,y) isgiven discontinuity inthenormal derivative ofV:(implicitly) bytheequation’? MVsove MVecw 2 2 a(x)=—€|S|- 5(4.3) soe ee ap %bo%0 acoah’(VIVo) ©a”sink’(VIVo) Forverysmally,ontheinterval—a<x<a(whereVisalso wherevery small), Eq.(4.1) reduces to 710 Am. 4.Phys, Vol. 64,No.6,June 1996 DJ. GrifithsndY.Li710 ‘ ‘ Sf . “ 4 Fig.14.Arayofpaalelwireswithequalcharges. Fi.12Infiniteconductingribbon 7 M(S 1 fo 40 ~— = (hat - ES Very)==esbh 44)Tre(3ataBtme a) Ca) 0 Settingf,equalto0(fori=1,2,....zn—1)yieldsn—1equa- MVaveve MVyretow Yo tionsfortheequilibrium positionsofthewires.Forexample,re er ar 45)ifn=2wefindx;=a/V5=0.447a;ifn=3,thenx,vo ve =(a/y3)(1—2/V7)"7=0.285a andx=(a/Y3)(1 andhence!? +2)47)"2=0.765a. A ‘Weareinterested intheeffective surface charge o(x), in o()= (4.6) thelimit n—. Asbefore, max ‘ ‘Theistheexactcharge density onaconducting ribbon—the ok)= > (48)analogtotheformulathathasremainedsoelusiveinthecase ioftheneedle.ItisplottedinFig.13(forA=1anda=1).AswithX=A/2nand expected, thechargei towardtheedgesofthe z=l peste,thechargeirepelledot 1edges EmKereytx). 9)InFig.15weplotnumerical solutionsforn=5,10,and100.‘Theresults fittheexact solution (4.6) very well. B.Wire models ry 1.Fixed charge 2.Fixed position Consider anarrayof2ninfinite wiresrunning parallel © Suppose nowthatthewiresareevenly spaced, adistancethe2axisinthexzplane, freetomove inthexdirection _g—2a/(2n—1) apart,buttheitcharges(X;.hg,-.-.,) are between—aand+a(Fig.14).Ifeachwirecarriesthesamevariable(Fig.16).Theforceperunitlengthontheithwireis linearcharge density A,whataretheirequilibrium positions, . i" *(X1,X25.49%,—19¥,™4)? Theforceperunitlengthonthe i(» Sey2 iti ash eE+E S- HI- ithwireis Wp (PiAAR Hi)BesFH (4.10) Atequilibrium f,=0, from which itfollows that = X M=Qi-1P7D inn.ou CWE Gage (41) os Thisgives us 1equations innunknowns—the remaining constraint is oe oA2%Maz: (4.12) oa For example, ifm=2, we obtain =A/6=0.167A, 2=A/3=0.333A; if'n=3, —dy=(31/290)A=0.107A, 2,=(18/145)A=0.124A, d3=(39/145)A=0.2694. InFig.17 Sasss easo'so-75 1theresultingsurfacechargeo(x)isplottedforn=5,10,and 100;again,thegraphsareincloseagreement withtheexact Fig.13.Chargedensityonribbon, answer. TLL Am.1.Phys,Vol.64,No.6,June1996 DJ.GrifandY.Li 711 °| : IS Es TEs ot Creme ECC TEs at F » a 0.4 a a 0.4 7: °.| I as TE iS Eas TES ee °.| |od os on oa a a as TE et I ST TE ot Fig.15.Surfacechargebyconstant chargemethod:(a)n=5,(b)m=10,and Fig.17.Surfacechargebyconstant spacingmethod: (a)n=5,(b)n=10,(n=100, and(6)n=100. V.CONCLUSION function isfairlyflatinthecenter, with“rabbit ears’”atthe ‘What, then, isthecharge density onaconducting needle? twoends. And yettheellipsoid model andDjordjevic’s Allournumerical studies support one’s intuition thatthe analysis ofthecylinder model indicate thatthecharge den- sityshould beconstant, inthelimit ofinfinitesimal cross section. Isitconceivable that wehave been fooled byvery ¥ slow convergence ofthegraphs, andthat, actually, asngoes toinfinity, Figs. 6,10,and11would allflatten outto(x) =0.5?Thiswouldresolve theawkward paradox—and yet,in sthecaseoftheribbon theanalogous graphs (Figs. 15and17) ene > giveareasonably accurate representation oftheexact an-On d2ALA An ‘wer,evenforquitesmalln.Moreover, ifthechargedensitywere constant, onatruly one-dimensional needle, how could theforce onanoff-center point bezero, considering thatthe charge outtothenear endisexactly balanced, leaving, in- Fig.16.Arrayof2nparallelwireswithequalspacing. evitably, someextrauncanceled charge atthefarend? T2——_AmJ. Phys,Vol64,No.6June1996 DA.GrifhitsandYL 712 SOCCES xfreryae'=fvnta’)de'afA(x)dx" lrtd taf ncean'=0 as SOCNow re artiic [Lawrereo as isthetotal “‘charge.”” And since theconfiguration issym- metrical about theorigin, A(x) must beaneven function, so Itisembarrassing toconclude thatwestilldonotknow -whatthechargedensityonaconductingneedleis.Wesus-iix'X(2")dx"=0. (a6) pect thattheproblem isillposed (inthesense thatthean- - swer depends onthemodel adopted), andeven forthose Therefore, (“bead”) models that seem tobeapproaching acommon limit we cannot tell what the functional form ofthat fimit 2af*xcae’= mightbe!S—norcanweabsolutely excludethecounterintui- (x+a)Q-2a J2Me')dx'=0. (antiibilitythatitisinfactaconstant. ivepossibility thatitsinfactac Differentiating withrespecttox,wefindthat Q-2ar(x)=0, (A8) APPENDIX: HOOKE’S LAW ANALOG or Weconsiderhereanexactlysoluble“toy”?model,in AG)=O?a. (9)which thetrue Coulombic interaction isreplaced bya Again, the“‘charge" density isuniform ontheinterval —a Hooke’s law force. Specifically, wesuppose there are2n <x<a (and 0otherwise).beadsspreadoutalongthexaxis,andeachoneisconnected ‘ThevirtueoftheHooke’slawmodelisthatwecansolve toeveryotheronebyaspringofforceconstantkandnatural itanalytically, inboththediscreteandcontinuum regimes.length a(Fig. 18).Ourproblem istodetermine theequilib- However, itishardly arealistic model, anditisnotclear rium spacing ofthebeads. ‘what, ifanything, ittells usabout thetrueCoulombic case. Clearly, they will arrange themselves symmetrically about, thecenter, atpositions wemayaswellcall—x,,...,—X9, WLEI =,X45X5-eonkn- Theforceontheithbeadis ‘ACKNOWLEDGMENTS , 4 Wethank RayMayer formany useful discussions, and ‘ MattKlebanforextensive computer exploration ofthe Fi=—k>»(etxj-a) +B(x-x)-a) Smytheapproach(Sec.IIB)andfordrawingmanyofthe, , figures. Wethank ananonymous referee forsupplying Ref. x 3.Above all,wethank A.R.Djordjevié ofBelgrade Univer- ->@j-x-a) sityforilluminating e-mail correspondence regarding the soe charge density onaconducting cylinder; wehave notbegun ==K2nx;-(24-1)a). (a1) 0dojustice (inSec.IB)tohisresults andhisinsights,which wehope hewillseefittopublish. Atequilibrium thisforceiszero,so "Thisisleeadyasomewhatsurprisingconclusion.Thechargedensityona(2i-1) thinconductingdis,forexample,canbeobtainedbyavarityoflimiting=e (A2)proceduresalofwcyhsamert[reprayingensmethod,seR.Friedberg,“Theelectrostaticsandmagactotaticsof& Evidentlythebeadsareevenlyspaced,adistancea/napart.tameacypathologyintetectiontncoedeenaseta Inthelimitn+the‘‘charge”” densityisuniformomthedoesnotinfecthereductionfromthreetotwo.intervalfrom—ato+a. WilliamR.Smythe,StaticandDynamicElecricity,Sede,revisedpintInthis“‘toy’*modelwecanalsohandlethecontinuum ng(Hemisphere, NewYork,1989),Sec.5.02. lineardensity\(x).Thebeadatxwillexperience forces iclagntism200 zsNewYork,1949);ourdeefrombothsides;atequilibrium theforceduetoallthebeads{jag“wedonJereyCima'sunpublishedReedCollegeseniosthesistoitsleftwillbalance theforce ductoallthebeads toits “W.R’Smythe, “Charged RightCircular Cylinder,” J.App.Phys.27,right:"® 917-920(1956).SeesloRef.2,Sec.539.“7.Taylor,“BlectiPolarzabiltyofaShortRightCircularConducting eo nan Gylnder,"J.Res.NBSB64,135-143(1960).Taylorisconcernedwith (xx!—a)d(x')dx'= |“(xe-2—a)A(x")de'. theproblemofanunchargedcylinderinanexternalclericfield,butiis weys straightforward 1adapthismethodtothecaseof«chargedcylinderwith(A3) ‘noextemalfield.SeealsoT.T.Taylor,“Magnetic Polarizability ofa ShorRightCircularConducting Cylinder,” J,Res.NBSB64,199-210 Combining terms, (1960), 713 —_Am.J. Phys, Vol. 64,No.6, June 1996, DJ. GriffithsandYL713, OW.R.Smytine,“ChargedRightCiculaCylinder,”J.ApplPhys.38,usedthepotaialaeachcharge(duetoalltheothers),butforopen2366-2967(1962);PC.Waterman,“MatixMethodiaPotentialTheorysegmentsitisbettetousethemidpoint,sincetheendchargesaresubject‘andElectromagnetic Scattering,”J.Appl.Phys.50,4550-4566 (1979);B.tononelectrostatic forces. D.Popovig,M.B.Dragovié,andA.R.Djordjevic, AnalysisandSynthesis “Smythe,Ref.2,Sec.422. {aio ofBecoFesSureudagneCrearCynara fsabn!tyeringheitthening onacon"aSheemendingFieCreatCyloeclongatedellipsoid.WithA=Q/2candc—2,Eqs,(2.1)and(2.2) R.Djordjevic, “Comments on“Calculation ofElectostatic Fields Sor-roundingFiniteCircularCylindrical Conductors,’ *”IEEETrans.Anten- AL2,(eT?nasPropag.33,683-684(1985);P.K.Wang,C.H.Chuang,andN.L-ooo4{ee-e4(4)]. Miller, “Electrostatic, Thermal andVapor Density Fields Surrounding ysihenetsurfacechargedensity(bothsides)forainfinitecylinder AWSayiteRee2paanAmisS342,2371-25791985).epiccrocoonItem0fecesbon,indne ONReieecal recoverEq,(6)[Intheesec=,heanalogttetheoreminSee.A °MatcrialinthissectionibasedonYeLi'sunpublishedReedCollege“Satestatthechargpeunilength(inte=direction)onastipofwid seniorthesis(1994), dxisAdx/(Va"—2°)forallb—but(unlikethefiniteellipsoid)itisnot "Theelectricalforceontheoutermostcharges(at+)isnov220,of,jindependentofx} course,becausetheyaresubjecttotheextraconstraining forceholdingthe“Itiseasytocheckthatthetotallinearchargedensity(o(x)dz} isA. ‘chargesonthewire.Bytheway,onecan(equivalently)calculatethetotal_‘°Wehavepushedthefixed-positionbeadmodelupton=300withbarely potentialenergyoftheconfiguration,aodminimizeit10determinethedetectablechangesinthecurve;thebetfitoftheform(3.7)cursfor positionsofthecharges. A=0.384049,B=0.088285. NJ,B,Ross,“‘Plotingthechargedistribution ofaclosed-loop conducting "ThismethodcannotbeappliedtotheCoulombproblem,ofcourse,be-‘iteusing2microcomputer" Am.J.Phys.5S,948-950(10987).Rosscauseofthenastysingularityintheintegrand. Electric field line diagrams don’t work ‘Alan Wolf” Department ofPhysics,TheCooperUnion,NewYork,NewYork10003, Stephen J.VanHook,” EricR.WeeksCenterforNonlinear Dynamics,Department ofPhysics,TheUniversity ofTexas,Austin,Texas78712 (Received 27July 1995; accepted 10December 1995) Electric fields produced bycoplanar point charges have often been represented byfield line diagrams thatdepict two-dimensional slices ofthethree-dimensional field. Serious problems with these “conventional” field linediagrams (CFLDs) have been overlooked. Two ofthese problems, “equatorial clumping” and ‘false monopole moment,” occur because atwo-dimensional slice Jacks information vital totheaccurate representation ofaninherently three-dimensional field. Equatorial clumping causes most CFLDs toexhibit unphysical behavior such asirregular spacing between field lines terminating onnegative charges. CFLDs can also mistakenly indicate that a neutral charge distribution hasasignificant monopole moment. Such phenomena make thevisual estimation oflocal field strengths impossible andrender CFLDs oflittle utility forrepresenting three-dimensional fields. While these “‘projection’” problems can beavoided byusing two-dimensional field linediagrams torepresent two-dimensional (1/r) electric fields, orbyusing three-dimensional field linediagrams torepresent three-dimensional fields, other formsofdistortion generally remain. ©1996 American Association ofPhysics Teachers. LINTRODUCTION veys information about thelocal direction oftheelectricfield,butnotaboutthefield’smagnitude. ThelatterpropertyElementary physicstextbooks generally attemptsomerequires consideration ofthelocaldensityoffieldlines.Fig-form oftwo-dimensional graphical representation ofthe ure1(a),forexample, illustrates theCFLD forasimple di- three-dimensional electric field produced bycoplanar point —_pole. Field lines nearthedipole aregenerally spaced close charges. Themostcommon approach, theconventional field_together, whichisthought toindicate ahighfieldstrength,linediagram (CFLD), employs continuous electric field whilemoredistantlinesarespacedfartherapart.Theten-lines,or“linesofforce,"whichareeverywhere tangenttodencyofthefieldlinedensitytodecreasewithdistancefrom theelectricfield.Eachfieldlineistracedfromapositive thedipoleisapparently consistent withtheasymptotic 1/r*chargeuntilitterminates onanegative chargeoratinfinity decayofthefieldstrength. Fieldlinesare,ofcourse,crea-(ie.,“*far’”fromallcharges). Anindividual fieldlinecon-_turesoffiction.! However CFLDs suchasFig.1(a)aregen- 714 Amu2Phys.64(6),June1996 ©1996AmericanAstociationofPhysicsTeachers 714 NOTES AND DISCUSSIONS Comment on“Charge density onaconducting needle,” byDavid J. Griffiths and YeLi[Am. J.Phys. 64(6), 706-714 (1996)] R.H.Good Physics Department, California State University, Hayward, California 94542 (Received 14June 1996; accepted 25July 1996) Griffiths andLi'suggest with some reluctance thatthe cally onto thex-y plane. This gives asurface density linear charge density onaconducting needle might be‘in a(x,y) onthedisk... .””Referring toFig. 1,forapoint onthe fact aconstant.”” The purpose ofthisComment istosupport disk, thatconjecture with ageometrical demonstration. ‘Weshallbeginwiththewell-known proofthattheelec- baled Q) trostatic field inside auniformly charged spherical shell is non zero. Pick anarbitrary point, anddraw anarrow cone (in _(imilar triangles). So, from Eq,(1bothdirections). ThechargeQisproportional tothearea Similar triangles). So,fromEq,(1), subtended, which isproportional tothesquare ofthedistance Q1_Q7;80forthetwoareasatopposite endsofthecone: aoe 3) 2, and, ingeneral, thexandycomponents oftheEfieldcancel ma (Q)outasabove,whichmeansthatthediskisanequipotential, 17asrequired foraconductor inelectrostatics. which isthesameasthecondition thatthe£fieldsfromthe Theexpression forthesurface charge density follows im- twocharges areequal andopposite atthepoint. (Itworks mediately. Foro(x,y) wecanwrite op), withp=yx7Fy? only forasphere because anarbitrary straight linehitsop- (seeFig.2).Theprojection from thetwohemispheres tothe posite sides atthesame angle tothespherical surface.) The diskgives o(p)=2a/sin 8,where cos0=pla;so whole sphere can bemapped into such mutually canceling pairs foranyinterior point. Therefore, theEfield iszero - ainsidethesphere, O10)200s 4 Buildingonthissoneepweapproach theconducting cir- a . cular disk. R.Friedberg’ suggests (p.1087): “Imagine a(Which increases without limitattheedgeofthedisk).sphereofradius@centeredattheorigin,andonitssurface_Letusapplyasimilarprocedure totheconducting needle,place auniform surface charge density o)=Q/47a?. Now Fig.3.Now, instead ofprojecting down toaplane, weare collapse thezdirection sothatthecharge isprojected verti-. Projecting thecharge ontoalinealong thexaxis.Asbefore, theEfield component along thelineiszero, sotheresulting 2 Disk 2 =op) Fig.1.Equipotential disk. Fig.2.Projection ontoadisk. 155 Am.J.Phys. 65(2),February 1997 (©1997American Assocation ofPhysics Teachers 15S som 80 20 mmpp o2» Wx 0710 a fs ay Fig.4Apportionment oflinecharge. c Fortheexamplegiven,C=1/10~1/30=2/30=0.0667. Then Needlea) wwefind thatx,=8 corresponds tox,=17.1, andx,=9 to x)=22.5. Taking averages ofthese distances, ifthetwoseg- ‘ments cancel out then S41 5 peas© 0.0138=0.0138, (19) sothat checks pretty well. Foranygiven segment onthe right oftheorigin, thecorresponding oneontheleftpro- ducesafieldthatisequalandopposite. Mis,3.Projection onto«line. Andthat’sallveryplausible, exceptforthefollowingparadox, alluded toinGriffiths and Li’s conclusion. Refer- ting toFig. 4,symmetry would seem todictate that theE linear charge density \isappropriate foraconductor. In fieldattheorigin 0duetothecharge ontherightshould be order tofindA,weproject aringofchargefromthespherical €actlyequaltothefieldduetothechargejustuptothe ‘surface ontothecorresponding lineelement oftheneedle: 10”markontheleft,andsotheremaining charge onthe leftshouldprovideanetfieldpointingtotheright(forposi- oo2ma sin6adO=had@ sin6. (5) tiveA),incontradiction toourconclusion thatthere isnoSothelinearchargedensityturnsouttobeaconstant, fieldcomponent alongtheneedle.Thismightbecalledanearchargedensityturnsourtobeaconstan ‘paradoxofinfinity””whichproduceswhatGriffithsandLiey 2 (6)Tlertoasa“deeppathology inthereduction toaone-aan. dimensional object thatdoes notinfect thereduction from ne to“ , ability" threetotwo.”TheEfieldandscalarpotential Vareinfinitecorresponding to“thecounterintuitive possibility” sug-fortheneedle,andthecapacitance iszero,noneofwhichis gested byGriffiths and Li.|. trueforthedisk (except forEattheveryedge).Inparticular, TesalwayspossibletocarveupauniformlinechargeinjnFig”4,thefieldattheorginduetothechargeontheright suchawayastoshow thatthefieldatanarbitrary point iginfinite, andthefieldfromtheleftisalsoinfinite; soadd- within thelineisero. Anumerical example willprobably be ngsfinitefieldduetothecharge from10to-30ontheleftmostunderstandable, andmostconvincing, Fig,4.Forclat- willmakeonlyanegligible difference tothesituation atthatity,wemeasurex,fromtheorigintowardtheright,andx2point.Thisispresumably asatisfactory approximation foratoward theleft.Whatweneedisthat,foranarbitrary origin Pyically realizable thinneedle©,anygivensegment dxiscanceled bythecorresponding So,inviewoftheabove andRef.1,itappears thatthe segment dx2,interms offieldattheorigin: linear charge density isuniform foraneedle ofinfinitesimal dx)_dry thickness andforathinprolatespheroid; andthat,foraieeea (7)cylinderoffinitegirth,thereisanexcessofchargeatthe Po ends. sincedxisproportional tocharge. Thesolution is °D,.Osis and 11,“Charge dem on»coadocig wee” Am, 14 4.Phys. 64,705~714 (1996). == +c. (8) 7%Friedberg, “The clectostatics andmagnetostatcs of conductingces) disk”Am.J.Phys.61,1084—1096(1993). RADIO ENGINEERING ‘Wally[Selove]wasasuperbstudentinhighschool,andtheUniversity ofChicagoofferedhimatuition scholarship. Theuniversity awakened hisintellect. Hehadthought hewanted tobea radio engineer butChicago hasnosuch major. Hewasadvised thatphysics wastheclosest area. Wally sought outextra work todoinhisclasses because heneeded anAaverage tomaintain his scholarship. Hefound physics tobemuch more interesting thanhehadimagined, andhewas fascinated bymarvelous courses intheclassics andinhistory. FayAjzenberg-Selove, AMaterofChoicesMemoirs ofaFemalePhysicist(RutgersUniversityres,NewBrunswick,Nev Jersey, 1994), p.113 156 Am.J. Phys, Vol.65,No.2,February 1997 Notes andDiscussions 156