2.1. Threepositivepointchargesofequalmagnitude 𝑞 arelocatedat 𝑥 =−2, 𝑦 =2,and 𝑦 =−√2.Find thecoordinatesofafourthpositivecharge,alsoofmagnitude 𝑞,thatwillyieldazeronetelectricfield attheorigin:Thefieldattheoriginthatarisesfromthethreechargescanbeexpressedas
Themagnitudeofthisfieldis
Tocounterthisfield,thefourthchargemustbepositionedalonga 45◦ lineinthefirstquadrant.Its distancefromtheoriginalongthislinewillbe
2=1.68.Thistranslatesinto equal 𝑥 and 𝑦 coordinatesof 21∕4 =1 19.Thereforethefourthchargeofpositivemagnitude 𝑞 is locatedat (1.19, 1.19)
2.2. Pointchargesof1nCand-2nCarelocatedat(0,0,0)and(1,1,1),respectively,infreespace.Determine thevectorforceactingoneachcharge.
First,theelectricfieldintensityassociatedwiththe1nCcharge,evalutatedatthe-2nCcharge locationis:
inwhichthedistancebetweenchargesis √3 m.Theforceonthe-2nCchargeisthen
Theforceonthe1nCchargeattheoriginisjusttheoppositeofthisresult,or
2.3. Pointchargesof50nCeacharelocatedat 𝐴(1, 0, 0),
, 0, 0), 𝐶(0, 1, 0),and 𝐷
infree space.Findthetotalforceonthechargeat 𝐴.
Theforcewillbe:
wheredistancesareinmeters.
2.4. Eightidenticalpointchargesof 𝑄 Ceacharelocatedatthecornersofacubeofsidelength 𝑎,with onechargeattheorigin,andwiththethreenearestchargesat (𝑎, 0, 0), (0,𝑎, 0),and (0, 0,𝑎).Findan expressionforthetotalvectorforceonthechargeat 𝑃 (𝑎,𝑎,𝑎),assumingfreespace:
Thetotalelectricfieldat 𝑃 (𝑎,𝑎,𝑎) thatproducesaforceonthechargetherewillbethesumofthe fieldsfromtheothersevencharges.Thisiswrittenbelow,wherethechargelocationsassociated witheachtermareindicated:
Theforceisnowtheproductofthisfieldandthechargeat (𝑎,𝑎,𝑎).Simplifying,weobtain
2.5. Apointchargeof3nCislocatedat(1,1,1)infreespace.Whatchargemustbelocatedat(1,3,2)to causethe 𝑦 componentof 𝐄 tobezeroattheorigin?
Fortwopointcharges,wemaywrite:
where 𝑞1 =3nC,andwhere 𝑞2 istobefound.With 𝑞1 locatedat(1,1,1), 𝐫′ 1 = 𝐚𝑥
.The positionvectorfor 𝑞
.Becausetheobservationpointisattheorigin,we have 𝐫 =0.Thefieldnowbecomes:
Forazero 𝑦 component,wethusfind 𝑞2 =−(14)3∕2∕33∕2 =−10 1nC
2.6. Twopointchargesofequalmagnitude 𝑞 arepositionedat 𝑧 =±𝑑∕2
a)findtheelectricfieldeverywhereonthe 𝑧 axis:Forapointchargeatanylocation,wehave
Inthecaseoftwocharges,wewouldthereforehave
Inthepresentcase,weassign
thechargepositionvectorsas
then
Substitutetheseresultsinto(1)toobtain:
b)findtheelectricfieldeverywhereonthe 𝑥𝑦 plane:Weproceedasinpart 𝑎,exceptthatnow 𝐫 lies inthe 𝑥𝑦 plane.Forsimplicity,wecanchoosethe 𝑥 axisonwhichtoevaluatethefield,sothat 𝐫 = 𝑥𝐚𝑥.Eq.(1)becomes
where
Therefore(3)becomes
Thisresultcanbegeneralizedtoapplyanywhereinthe 𝑥𝑦 planebynotingthattheproblem exhibitscylindricalsymmetry–anyrotationofthe 𝑥 axisaboutthe 𝑧 axiswillproducenochange. Therefore,wemayusecylindricalcoordinates,andreplacethe 𝑥 variablebytheradialvariable 𝜌,anduse 𝐚𝜌 insteadof 𝐚𝑥.Thefieldthenbecomes:
2.7. Twopointchargesofequalmagnitudebutofoppositesignarepositionedwithcharge +𝑞 at 𝑧 = 𝑑∕2, andcharge 𝑞 at 𝑧 =−𝑑∕2.Thepairforman electricdipole
a)Findtheelectricfieldintensityeverywhereonthe 𝑧 axis:Forthetwochargeswewouldwritein general:
.Usingthesesubstitutions,wefind:
b)Evaluateyourpart 𝑎 resultattheorigin:Weset 𝑧 =0 intheaboveresulttoobtain
c)Findtheelectricfieldintensityeverywhereonthe 𝑥𝑦 plane,expressingyourresultasafunction ofradius 𝜌 incylindricalcoordinates:Thiswillbeginwiththesameinitialsetupasinpart 𝑎, exceptnow 𝐫 = 𝜌𝐚𝜌 describestheobservationpointinthe 𝑥𝑦 plane.Withthischange,wehave
𝐚𝜌 +(𝑑∕2)
d)Evaluateyourpart 𝑐 resultattheorigin:Setting 𝜌 =0 inthepart 𝑐 expression,wefind: 𝐄(𝜌
asinpart 𝑏, −andasexpected
e)Simplifyyourpart 𝑐 resultforthecaseinwhich 𝜌>>𝑑:Withthisrequirement,wefind 𝐄(𝜌>>𝑑) = 𝑞𝑑 𝐚𝑧 4𝜋𝜖0𝜌3
2.8. Acrudedeviceformeasuringchargeconsistsoftwosmallinsulatingspheresofradius 𝑎,oneofwhich isfixedinposition.Theotherismovablealongthe 𝑥 axis,andissubjecttoarestrainingforce 𝑘𝑥, where 𝑘 isaspringconstant.Theunchargedspheresarecenteredat 𝑥 =0 and 𝑥 = 𝑑,thelatterfixed. Ifthespheresaregivenequalandoppositechargesof 𝑄 coulombs:
a)Obtaintheexpressionbywhich 𝑄 maybefoundasafunctionof 𝑥:Thesphereswillattract,and sothemovablesphereat 𝑥 =0 willmovetowardtheotheruntilthespringandCoulombforces balance.Thiswilloccuratlocation 𝑥 forthemovablesphere.Withequalandoppositeforces, wehave 𝑄2 4𝜋𝜖0(𝑑 𝑥)2 = 𝑘𝑥 fromwhich 𝑄 =2(𝑑 𝑥)√𝜋𝜖0𝑘𝑥.
2.8
b)Determinethemaximumchargethatcanbemeasuredintermsof 𝜖0, 𝑘,and 𝑑,andstatethe separationofthespheresthen:
Withincreasingcharge,thespheresmovetowardeachotheruntiltheyjusttouchat
Usingthepart 𝑎 result,wefindthemaximummeasurablecharge: 𝑄𝑚𝑎𝑥
)
Presumablysomeformofstopmechanismisplacedat 𝑥 = 𝑥 𝑚𝑎𝑥 topreventthespheresfrom actuallytouching.
c)Whathappensifalargerchargeisapplied?Nofurthermotionispossible,sonothinghappens.
2.9. A100nCpointchargeislocatedat 𝐴(−1, 1, 3) infreespace.
a)Findthelocusofallpoints 𝑃 (𝑥,𝑦,𝑧) atwhich 𝐸
=500 V/m:Thetotalfieldat 𝑃 willbe:
The 𝑥 componentofthefieldwillbe
Andsoourconditionbecomes:
b)Find 𝑦1 if 𝑃 (0,𝑦1, 3) liesonthatlocus:Atpoint 𝑃,theconditionofpart 𝑎 becomes
2.10. Aconfigurationofpointchargesconsistsofasinglechargeofvalue −2𝑞 attheorigin,andtwocharges ofvalue +𝑞 atlocations 𝑧 =−𝑑 and +𝑑.Thechargesaspositionedforman electricquadrupole, equivalenttotwodipolesofoppositeorientationthatareseparatedbydistance 𝑑 alongthe 𝑧 axis.
a)Findtheelectricfieldintensity 𝐄 everywhereinthe 𝑥𝑦 plane,expressingyourresultasafunction ofcylindricalradius 𝜌:Webeginbyapplyingthegeneralformulaforthepointchargefield,where thethreetermsapplytothethreecharges:
Thepositionvectorswillbe
substitutions,thefieldexpressionbecomes:
2.10
b)Specializeyourpart 𝑎 resultforlargedistances, 𝜌>>𝑑:Underthiscondition,wemayusethe expansion:
Carryingouttheproductandneglectingtheterminvolving 𝑑4∕𝜌4,wefind:
fromwhich
2.11. Acharge 𝑄0 locatedattheorigininfreespaceproducesafieldforwhich 𝐸𝑧 =1 kV/matpoint 𝑃 (−2, 1, −1)
a)Find 𝑄0:Thefieldat 𝑃 willbe
Sincethe 𝑧 componentisofvalue1kV/m,wefind 𝑄0 =−4𝜋𝜖061
b)Find 𝐄 at 𝑀(1, 6, 5) incartesiancoordinates:Thisfieldwillbe:
𝑧 [1+36+25]1 5 ] or 𝐄𝑀 =−30.11𝐚𝑥 −180.63𝐚𝑦 −150.53𝐚
c)Find 𝐄 at 𝑀(1, 6, 5) incylindricalcoordinates:At 𝑀, 𝜌 = √1+36=6 08, 𝜙 =tan−1(6∕1)= 80 54◦ ,and 𝑧 =5.Now
𝜌 = 𝐄𝑀 ⋅ 𝐚𝜌 =−30 11cos 𝜙 −180 63sin 𝜙 =−183 12 𝐸𝜙 = 𝐄𝑀 ⋅ 𝐚𝜙 =−30 11(−sin 𝜙)−180 63cos 𝜙 =0(asexpected) sothat 𝐄𝑀 =−183.12𝐚𝜌 −150.53𝐚𝑧.
d)Find 𝐄 at 𝑀(1, 6, 5) insphericalcoordinates:At 𝑀, 𝑟 = √1+36+25=7 87, 𝜙 =80 54◦ (as before),and 𝜃 =cos−1(5∕7 87)=50 58◦ .Now,sincethechargeisattheorigin,weexpectto obtainonlyaradialcomponentof 𝐄𝑀.Thiswillbe:
𝑟 = 𝐄𝑀 ⋅ 𝐚𝑟 =−30 11sin 𝜃 cos 𝜙 −180 63sin 𝜃 sin 𝜙 −150 53cos 𝜃 =−237 1
2.12. Electronsareinrandommotioninafixedregioninspace.Duringany1𝜇sinterval,theprobabilityof findinganelectroninasubregionofvolume 10−15 m2 is0.27.Whatvolumechargedensity,appropriate forsuchtimedurations,shouldbeassignedtothatsubregion?
Thefiniteprobabiltyeffectivelyreducesthenetchargequantitybytheprobabilityfraction.With 𝑒 =−1.602×10−19 C,thedensitybecomes
2.13. Auniformvolumechargedensityof0.2 𝜇C∕m3 ispresentthroughoutthesphericalshellextending from 𝑟 =3 cmto 𝑟 =5 cm.If 𝜌𝑣 =0 elsewhere:
a)findthetotalchargepresentthroughouttheshell:Thiswillbe
b)find 𝑟1 ifhalfthetotalchargeislocatedintheregion 3cm <𝑟<𝑟1:Iftheintegralover 𝑟 inpart 𝑎 istakento 𝑟1,wewouldobtain
2.14. Theelectronbeaminacertaincathoderaytubepossessescylindricalsymmetry,andthechargedensityisrepresentedby 𝜌𝑣 =−0
for 𝜌> 3×10−4 m.
a)Findthetotalchargepermeteralongthelengthofthebeam:Weintegratethechargedensity overthecylindricalvolumehavingradius 3×10−4 m,andlength1m.
Fromintegraltables,thisevaluatesas
b)iftheelectronvelocityis 5×107 m/s,andwithoneamperedefinedas1C/s,findthebeamcurrent:
2.15. Asphericalvolumehavinga2 𝜇mradiuscontainsauniformvolumechargedensityof 105 C∕m3
a)Whattotalchargeisenclosedinthesphericalvolume?
Thiswillbe 𝑄 =(4∕3)𝜋(2×10−6)3 ×105 =3 35×10−12 C
b)Nowassumethatalargeregioncontainsoneoftheselittlespheresateverycornerofacubical grid3mmonaside,andthatthereisnochargebetweenspheres.Whatistheaveragevolume chargedensitythroughoutthislargeregion?Eachcubewillcontaintheequivalentofonelittle sphere.Neglectingthelittlespherevolume,theaveragedensitybecomes
2.16. Withinaregionoffreespace,chargedensityisgivenas 𝜌
C∕m3,where 𝜌
and 𝑎 are constants.Findthetotalchargelyingwithin:
a)thesphere, 𝑟 ≤ 𝑎:Thiswillbe
Itistheintegralover 𝜃,performedfirst,thatgivesthezeroresult.
b)thecone,
2.17. Alength 𝑑 oflinechargeliesonthe 𝑧 axisinfreespace.Thechargedensityonthelineis
𝜌𝐿 =+𝜌0 C/m(0 <𝑧<𝑑∕2)and 𝜌𝐿 =−𝜌0 C/m( 𝑑∕2 <𝑧< 0),where 𝜌0 isapositiveconstant.
a)Findtheelectricfieldintensity 𝐄 everywhereinthe 𝑥𝑦 plane,expressingyourresultasafunction ofcylindricalradius 𝜌:Beginbyconstructingthedifferentialfieldatradius 𝜌 inthe 𝑥𝑦 planethat arisesfromapointcharge 𝑑𝑞 = 𝜌𝐿𝑑𝑧 onthe 𝑧 axis.Todothis,usethegeneralexpression:
whereinthiscase, 𝐫 =
and
𝐚𝑧.Withthesesubstitutions,wefind
wherethepositivesignappliestotheregion 𝑧> 0;thenegativesignto 𝑧< 0.Thetotalfieldat radius 𝜌 isthenfoundbyintegrating 𝑑𝐄 overthetotalchargelength:
2.17. a)(continued)Notethattheradialcomponentwillintegratetozerothroughoddparity,leavingonly the 𝑧 component,aswouldbeexpected.Theintegralsimplifiesto:
b)Simplifyyourpart 𝑎 resultforthecaseinwhichradius 𝜌>>𝑑,andexpressthisresultinterms ofcharge 𝑞 = 𝜌0𝑑∕2:Atlargeradii,wecanusethebinomialexpansiontothefirsttwoterms:
withwhich
where 𝑞
2.18. a)Find 𝐄 intheplane 𝑧 =0 thatisproducedbyauniformlinecharge, 𝜌𝐿,extendingalongthe 𝑧 axisovertherange 𝐿<𝑧<𝐿 inacylindricalcoordinatesystem:Wefind 𝐄 through
wheretheobservationpointpositionvectoris 𝐫 = 𝜌𝐚𝜌 (anywhereinthe 𝑥-𝑦 plane),andwhere thepositionvectorthatlocatesanydifferentialchargeelementonthe 𝑧 axisis 𝐫′ = 𝑧𝐚𝑧.So
.Theserelationsaresubstitutedintotheintegral toyield:
Notethatthesecondtermintheleft-handintegral(involving
)haseffectivelyvanishedbecauseitproducesequalandoppositesigncontributionswhentheintegralistakenoversymmetric limits(oddparity).Evaluatingtheintegralresultsin
Notethatas 𝐿 → ∞,theexpressionreducestotheexpectedfieldoftheinfinitelinechargein freespace, 𝜌𝐿
)
b)ifthefinitelinechargeisapproximatedbyaninfinitelinecharge(𝐿 → ∞),bywhatpercentage is 𝐸𝜌 inerrorif 𝜌 =0 5𝐿?Thepercenterrorinthissituationwillbe %error= [1− 1
For 𝜌 =0 5𝐿,thisbecomes %error=10 6% c)repeat 𝑏 with 𝜌 =0.1𝐿.Forthisvalue,obtain %error=0.496%.
2.19. Alinehavingchargedensity 𝜌0|𝑧| C/mandoflength 𝓁 isorientedalongthe 𝑧 axisat 𝓁∕2 <𝑧< 𝓁∕2
a)Findtheelectricfieldintensity 𝐄 everywhereinthe 𝑥𝑦 plane,expressingyourresultincylindrical coordinates:Astheproblemexhibitscylindricalsymmetry,wemaywritethepositionvectorfor theobservationpointinthe 𝑥𝑦 planeas
.Then,with
,wemaywrite
Notethatthesecondtermintheintegrand(the 𝑧 component)iszero,becauseofoddparity. Weareleftwith
Evaluatingthelimitsthefinalresultcanbewrittenas
b)Evaluateyourresultofpart 𝑎 inthelimitas 𝓁 (not 𝑧)approachesinfinity:Inthislimit,thesecond terminthebrackettendstozero,andwehave
thusexhibitingnoradialvariation!
2.20. Alinechargeofuniformchargedensity 𝜌0 C∕m andoflength 𝓁,isorientedalongthe 𝑧 axisat
∕2 <𝑧< 𝓁∕2.
a)Findtheelectricfieldstrength, 𝐄,inmagnitudeanddirectionatanypositionalongthe 𝑥 axis: ThisfollowsthemethodinProblem2.18.Wefind 𝐄 through
wheretheobservationpointpositionvectoris 𝐫 = 𝑥𝐚𝑥 (anywhereonthe 𝑥 axis),andwhere thepositionvectorthatlocatesanydifferentialchargeelementonthe
axisis
.So
.Theserelationsaresubstitutedintotheintegral toyield:
Notethatthesecondtermintheleft-handintegral(involving
)haseffectivelyvanishedbecauseitproducesequalandoppositesigncontributionswhentheintegralistakenoversymmetric limits(oddparity).Evaluatingtheintegralresultsin
2.20. b)withthegivenlinechargeinposition,findtheforceactingonanidenticallinechargethatis orientedalongthe 𝑥 axisat 𝓁∕2 <𝑥< 3𝓁∕2:Thedifferentialforceonanelementofthe 𝑥directedlinechargewillbe 𝑑𝐅 = 𝑑𝑞𝐄 =(𝜌
𝑑𝑥)𝐄,where 𝐄 isthefieldasdeterminedinpart 𝑎.Thenetforceisthentheintegralofthedifferentialforceoverthelengthofthehorizontalline charge,or
Thiscanbere-writtenandthenevaluatedusingintegraltablesas
2.21. Achargedfilamentformsacircleofradius 𝑎 inthe 𝑥𝑦 planewithcenterattheorigin.Thefilament carriesuniformlinechargedensity +𝜌0 C/mfor 𝜋
Findtheelectricfieldintensityattheorigin: Thefieldattheoriginarisingfromadifferentiallength 𝑎𝑑𝜙 ofchargeontheringwillbe
wherethepositivesignappliestothenegativechargecontribution,thenegativesigntothepositive chargecontribution.Nowthetotalfieldwillbethepiecewiseintegralofthedifferentialfieldoverthe twosemi-circles.Using
(asisrequiredtoincludeall 𝜙 dependence),wehave
Notingthatthetwotermsintheaboveexpressionareequaltoeachother,wemayevaluatethefirst integralandintroduceafactorof2:
2.22. Twoidenticaluniformsheetchargeswith 𝜌𝑠 =100nC∕m2 arelocatedinfreespaceat 𝑧 =±2.0 cm. Whatforceperunitareadoeseachsheetexertontheother?
Thefieldfromthetopsheetis 𝐄 =−𝜌𝑠∕(2𝜖
) 𝐚𝑧 V/m.Thedifferentialforceproducedbythis fieldonthebottomsheetisthechargedensityonthebottomsheettimesthedifferentialarea there,multipliedbytheelectricfieldfromthetopsheet: 𝑑
.Theforceperunitareais thenjust 𝐅
2.23. Adiskofradius 𝑎 inthe 𝑥𝑦 planecarriessurfacechargeofdensity 𝜌𝑠 = 𝜌0∕𝜌,where 𝜌0 isaconstant. Findtheelectricfieldstrength, 𝐄,everywhereonthe 𝑧 axis.
Wefindthefieldthrough
wheretheintegralistakenoverthesurfaceofthedisk,andwhere 𝐫
and
.The integralthenbecomes
Inevaluatingthisintegral,weneedtointroducethe 𝜙 dependencein 𝐚𝜌 bywritingitas
cos
+sin
,where
and
areinvariantintheirorientationas 𝜙 varies.Sotheintegral nowsimplifiesto
2.24. a)Findtheelectricfieldonthe 𝑧 axisproducedbyanannularringofuniformsurfacechargedensity 𝜌𝑠 infreespace.Theringoccupiestheregion
incylindrical coordinates:Wefindthefieldthrough
wheretheintegralistakenoverthesurfaceoftheannularring,andwhere
Theintegralthenbecomes
Inevaluatingthisintegral,wefirstnotethattheterminvolving
integratestozerooverthe 𝜙 integrationrangeof 0 to 2𝜋.Thisisbecauseweneedtointroducethe 𝜙 dependencein 𝐚𝜌 bywritingitas
areinvariantintheirorientationas 𝜙 varies.Sotheintegralnowsimplifiesto
b)fromyourpart 𝑎 result,obtainthefieldofaninfiniteuniformsheetchargebytakingappropriate limits.Theinfinitesheetisobtainedbyletting
and
,inwhichcase
asexpected.
2.25. Adiskofradius 𝑎 inthe 𝑥𝑦 planecarriessurfacechargeofdensity 𝜌𝑠1 =+𝜌𝑠0∕𝜌 C∕m2 for 0 <𝜙<𝜋, and 𝜌𝑠2 =−𝜌𝑠0∕𝜌 C∕m2 for 𝜋<𝜙< 2𝜋,where 𝜌𝑠0 isaconstant.
a)Findtheelectricfieldintensity 𝐄 everywhereonthe 𝑧 axis:
Withthesetupshownatright,thefieldcanbefound throughthegeneralrelation 𝐄 =
where 𝐫 = 𝑧
and 𝐫
.Becausethechargeis 𝜙-dependent,weneedtoincludethe 𝜙 dependence inallterms.Thismeansthatwemustuse 𝐚𝜌
.Then,with
andwiththepositiveandnegativechargesaccounted for,theintegralisperformedpiecewiseoverthetwo halvesofthedisk:
Notingthatthe 𝑧 componentswillcancel,andperformingthe
integrationfirst,wefind:
b)Specializeyourpart 𝑎 resultfordistances 𝑧>>𝑎:Withthisconditionwehave
Notetheinversedistancecubeddependencethatischaracteristicofadipolefield.
2.26
a)Findtheelectricfieldintensityonthe𝑧axisproducedbyaconesurfacethatcarrieschargedensity
𝜌𝑠(𝑟)= 𝜌0∕𝑟 C∕m2 infreespace.Theconehasitsvertexattheoriginandoccupiestheregion
𝜃 = 𝛼, 0 <𝑟<𝑎,and 0 <𝜙< 2𝜋 insphericalcoordinates.Differentialareainspherical coordinatesisgivenas 𝑑𝑎 = 𝑟 sin 𝛼𝑑𝑟𝑑𝜙: 𝐄 isfoundusingthegeneralsurfaceintegral
whereinthiscase
where
and
hasbeenused.Theintegralnowbecomes:
Itisnecessarytoincludethe 𝜙 dependencein
. Wesubstitute
.Thefirsttwotermsofthis,involving sin 𝜙and cos 𝜙,willintegratetozerowhentaking𝜙from0to 2𝜋.Onlythe𝑧componentsurvives, andtheintegralbecomes,afterperformingthe 𝜙 integration:
Thetwointegralsevaluateas
Combiningthese,cancellingterms,andrearranging,finallyleadsto
b)Findthetotalchargeonthecone:Thiswillbe
c)Specializeyourresultofpart 𝑎 tothecaseinwhich 𝛼 =90◦ ,atwhichtheconeflattenstoadisk inthe 𝑥𝑦 plane.ComparethisresulttotheanswertoProblem2.23. When 𝛼 =90◦ , cos 𝛼 =0,andthefieldexpressioninpart 𝑎 canbeexpressedas:
whichisthesameresultasfoundinProblem2.23.
2.26 d)Showthatyourpart 𝑎 resultbecomesapointchargefieldwhen 𝑧>>𝑎. Usingtheexpressionforthetotalchargeasfoundinpart 𝑏,andrearrangingterms,thefieldof part 𝑎 takestheform:
wherewerecognizethesecondequalityasthepointchargefield.
e)Showthatyourpart 𝑎 resultbecomesaninverse-𝑧-dependent 𝐄 fieldwhen 𝑧<<𝑎 If 𝑧<<𝑎,wehave [1−(2
2.27. Giventheelectricfield
a)theequationofthestreamlinethatpassesthroughthepoint 𝑃 (2, 3, −4):Wewrite
Evaluatingat 𝑃 (2, 3, −4),obtain:
Finally,at 𝑃,therequestedequationis
b)aunitvectorspecifyingthedirectionof 𝐄
2.28 Anelectricdipole(discussedindetailinSec.4.7)consistsoftwopointchargesofequalandopposite magnitude ±𝑄 spacedbydistance 𝑑.Withthechargesalongthe 𝑧 axisatpositions 𝑧 =±𝑑∕2 (with thepositivechargeatthepositive 𝑧 location),theelectricfieldinsphericalcoordinatesisgivenby 𝐄(𝑟,𝜃)= [𝑄𝑑∕(4𝜋𝜖0𝑟3)][2cos 𝜃𝐚𝑟 +sin 𝜃𝐚𝜃],where 𝑟>>𝑑.Usingrectangularcoordinates,determineexpressionsforthevectorforceonapointchargeofmagnitude 𝑞:
a)at (0, 0,𝑧):Here, 𝜃 =0, 𝐚𝑟 = 𝐚𝑧,and 𝑟 = 𝑧.Therefore 𝐅(0, 0,𝑧)= 𝑞𝑄𝑑
b)at (0,𝑦, 0):Here, 𝜃 =90◦ , 𝐚𝜃 =−𝐚𝑧,and 𝑟 = 𝑦.Theforceis 𝐅(0,𝑦, 0)= 𝑞𝑄𝑑 𝐚𝑧 4𝜋𝜖0
2.29. If 𝐄 =20𝑒−5𝑦 (cos5𝑥𝐚𝑥 −sin5𝑥𝐚𝑦),find:
a) |𝐄| at 𝑃 (𝜋∕6, 0.1, 2):Substitutingthispoint,weobtain 𝐄𝑃 =−10.6𝐚𝑥 −6.1𝐚𝑦,andso |𝐄𝑃 | = 12.2.
b)aunitvectorinthedirectionof 𝐄𝑃 :Theunitvectorassociatedwith 𝐄 is (cos5𝑥𝐚𝑥 −sin5𝑥𝐚𝑦), whichevaluatedat 𝑃 becomes 𝐚𝐸 =−0 87𝐚𝑥 −0 50𝐚𝑦
c)theequationofthedirectionlinepassingthrough 𝑃:Use 𝑑𝑦 𝑑𝑥 = −sin5𝑥 cos5𝑥 =−tan5𝑥 ⇒
=−tan5𝑥𝑑𝑥 Thus 𝑦 = 1 5 lncos5𝑥 + 𝐶.Evaluatingat 𝑃,wefind 𝐶 =0 13,andso 𝑦 = 1 5 lncos5𝑥 +0.13
2.30. Forfieldsthatdonotvarywith 𝑧 incylindricalcoordinates,theequationsofthestreamlinesareobtainedbysolvingthedifferentialequation
).Findtheequationofthelinepassing throughthepoint (2, 30◦ , 0) forthefield 𝐄 = 𝜌 cos2𝜙 𝐚𝜌 𝜌 sin2𝜙 𝐚𝜙:
Integratetoobtain
Atthegivenpoint,wehave 4= 𝐶∕sin(60
𝐶
◦ =2√3.Finally,theequationfor thestreamlineis 𝜌2 =2