UNITI

StaticElectricfields

Inthischapterwewilldiscussonthefollowings:

Coulomb'sLaw

ElectricField&ElectricFluxDensity

Gauss'sLawwithApplication

ElectrostaticPotential,EquipotentialSurfaces

BoundaryConditionsforStaticElectricFields

CapacitanceandCapacitors

ElectrostaticEnergy

Laplace'sandPoisson'sEquations

UniquenessofElectrostaticSolutions

MethodofImages

SolutionofBoundaryValueProblemsinDifferentCoordinateSystems.

IntroductionInthepreviouschapterwehavecoveredtheessentialmathematicaltoolsneededtostudyEM

fields.Wehavealreadymentionedinthepreviouschapterthatelectricchargeisa

fundamentalpropertyofmatterandchargeexistinintegralmultipleofelectroniccharge.

Electrostaticscanbedefinedasthestudyofelectricchargesatrest.Electricfieldshavetheir

sourcesinelectriccharges.

(Note:Almostallrealelectricfieldsvarytosomeextentwithtime.However,formany

problems,thefieldvariationisslowandthefieldmaybeconsideredasstatic.Forsomeother

casesspatialdistributionisnearlysameasforthestaticcaseeventhoughtheactualfieldmay

varywithtime.Suchcasesaretermedasquasistatic.)

Inthischapterwefirststudytwofundamentallawsgoverningtheelectrostaticfields,viz,(1)

Coulomb'sLawand(2)Gauss'sLaw.Boththeselawhaveexperimentalbasis.Coulomb's

lawisapplicableinfindingelectricfieldduetoanychargedistribution,Gauss'slawiseasier

tousewhenthedistributionissymmetrical.

Coulomb'sLaw

Coulomb'sLawstatesthattheforcebetweentwopointchargesQ1andQ2isdirectly

proportionaltotheproductofthechargesandinverselyproportionaltothesquareofthe

distancebetweenthem.

Pointchargeisahypotheticalchargelocatedatasinglepointinspace.Itisanidealized

modelofaparticlehavinganelectriccharge.

Mathematically, ,wherekistheproportionalityconstant.

InSIunits,Q1andQ2areexpressedinCoulombs(C)andRisinmeters.

ForceFisinNewtons(N)and , iscalledthepermittivityoffreespace.

(Weareassumingthechargesareinfreespace.Ifthechargesareanyotherdielectric

medium,wewilluse insteadwhere iscalledtherelativepermittivityorthe

dielectricconstantofthemedium).

Therefore .......................(1)

AsshownintheFigure1letthepositionvectorsofthepointchargesQ1andQ2aregivenby

and .Let representtheforceonQ1duetochargeQ2.

Fig1:Coulomb'sLaw

Thechargesareseparatedbyadistanceof .Wedefinetheunitvectorsas

and ..................................(2)

canbedefinedas .

SimilarlytheforceonQ1duetochargeQ2canbecalculatedandif representsthisforcethenwecan

write

Whenwehaveanumberofpointcharges,todeterminetheforceonaparticularcharge

duetoallothercharges,weapplyprincipleofsuperposition.IfwehaveNnumberof

chargesQ1,Q2,.........QNlocatedrespectivelyatthepointsrepresentedbytheposition

vectors , ,...... ,theforceexperiencedbyachargeQlocatedat isgivenby,

.................................(3)

ElectricField:

Theelectricfieldintensityortheelectricfieldstrengthatapointisdefinedastheforce

perunitcharge.Thatis

or, .......................................(4)

TheelectricfieldintensityEatapointr(observationpoint)dueapointchargeQlocated

at (sourcepoint)isgivenby:

..........................................(5)

ForacollectionofNpointchargesQ1,Q2,.........QNlocatedat , ,...... ,theelectric

fieldintensityatpoint isobtainedas

........................................(6)

Theexpression(6)canbemodifiedsuitablytocomputetheelectricfiledduetoa

continuousdistributionofcharges.

Infigure2weconsideracontinuousvolumedistributionofcharge(t)intheregion

denotedasthesourceregion.

Foranelementarycharge ,i.e.consideringthischargeaspointcharge,

wecanwritethefieldexpressionas:

.............(7)

Fig2:ContinuousVolumeDistributionofCharge

Whenthisexpressionisintegratedoverthesourceregion,wegettheelectricfieldatthepointPduetothisdistributionofcharges.Thustheexpressionfortheelectricfield

atPcanbewrittenas:

..........................................(8)

Similartechniquecanbeadoptedwhenthechargedistributionisintheformofaline

chargedensityorasurfacechargedensity.

........................................(9)

........................................(10)

Electricfluxdensity:

AsstatedearlierelectricfieldintensityorsimplyElectricfield'givesthestrengthofthe

fieldataparticularpoint.Theelectricfielddependsonthematerialmediainwhichthe

fieldisbeingconsidered.Thefluxdensityvectorisdefinedtobeindependentofthe

materialmedia(aswe'llseethatitrelatestothechargethatisproducingit).Foralinear

isotropicmediumunderconsiderationthefluxdensityvectorisdefinedas:

................................................(11)

Wedefinetheelectricfluxas

.....................................(12)

Gauss'sLaw:Gauss'slawisoneofthefundamentallawsofelectromagnetismandit

statesthatthetotalelectricfluxthroughaclosedsurfaceisequaltothetotalcharge

enclosedbythesurface.

Fig3:Gauss'sLaw

LetusconsiderapointchargeQlocatedinanisotropichomogeneousmediumof

dielectricconstant.Thefluxdensityatadistanceronasurfaceenclosingthechargeis

givenby

...............................................(13)

Ifweconsideranelementaryareads,theamountoffluxpassingthroughthe

elementaryareaisgivenby

.....................................(14)

But ,istheelementarysolidanglesubtendedbythearea atthe

locationofQ.Thereforewecanwrite

Foraclosedsurfaceenclosingthecharge,wecanwrite

whichcanseentobesameaswhatwehavestatedinthedefinitionofGauss'sLaw.

ApplicationofGauss'sLaw:

Gauss'slawisparticularlyusefulincomputing or wherethechargedistributionhassomesymmetry.WeshallillustratetheapplicationofGauss'sLawwithsome

examples.

1.Aninfinitelinecharge

AsthefirstexampleofillustrationofuseofGauss'slaw,letconsidertheproblemof

determinationoftheelectricfieldproducedbyaninfinitelinechargeofdensityLC/m.Let

usconsideralinechargepositionedalongthezaxisasshowninFig.4(a)(nextslide).

Sincethelinechargeisassumedtobeinfinitelylong,theelectricfieldwillbeoftheform

asshowninFig.4(b)(nextslide).

IfweconsideraclosecylindricalsurfaceasshowninFig.2.4(a),usingGauss'stheorm

wecanwrite,

.....................................(15)

ConsideringthefactthattheunitnormalvectortoareasS1andS3areperpendicularto

theelectricfield,thesurfaceintegralsforthetopandbottomsurfacesevaluatestozero.

Hencewecanwrite,

Fig4:InfiniteLineCharge

.....................................(16)

2.InfiniteSheetofCharge

AsasecondexampleofapplicationofGauss'stheorem,weconsideraninfinitecharged

sheetcoveringthexzplaneasshowninfigure5.Assumingasurfacechargedensityof

fortheinfinitesurfacecharge,ifweconsideracylindricalvolumehavingsidesplacedsymmetricallyasshowninfigure5,wecanwrite:

..............(17)

Fig5:InfiniteSheetofCharge

Itmaybenotedthattheelectricfieldstrengthisindependentofdistance.Thisistruefor

theinfiniteplaneofchargeelectriclinesofforceoneithersideofthechargewillbe

perpendiculartothesheetandextendtoinfinityasparallellines.Asnumberoflinesof

forceperunitareagivesthestrengthofthefield,thefieldbecomesindependentof

distance.Forafinitechargesheet,thefieldwillbeafunctionofdistance.

3.UniformlyChargedSphere

Letusconsiderasphereofradiusr0havingauniformvolumechargedensityofrv

C/m3.Todetermine everywhere,insideandoutsidethesphere,weconstructGaussiansurfacesofradiusrr0asshowninFig.6(a)andFig.6(b).

Fortheregion thetotalenclosedchargewillbe

.........................(18)

Fig6:UniformlyChargedSphere

ByapplyingGauss'stheorem,

...............(19)

Therefore

..............................................(20)

Fortheregion thetotalenclosedchargewillbe

...........................................................(21)

ByapplyingGauss'stheorem,

.......................................(22)

ElectrostaticPotentialandEquipotentialSurfaces

Intheprevioussectionswehaveseenhowtheelectricfieldintensityduetoachargeor

achargedistributioncanbefoundusingCoulomb'slaworGauss'slaw.Sinceacharge

placedinthevicinityofanothercharge(orinotherwordsinthefieldofothercharge)

experiencesaforce,themovementofthechargerepresentsenergyexchange.

Electrostaticpotentialisrelatedtotheworkdoneincarryingachargefromonepointto

theotherinthepresenceofanelectricfield.Letussupposethatwewishtomovea

positivetestcharge fromapointPtoanotherpointQasshownintheFig.8.Theforceatanypointalongitspathwouldcausetheparticletoaccelerateandmoveitout

oftheregionifunconstrained.Sincewearedealingwithanelectrostaticcase,aforce

equaltothenegativeofthatactingonthechargeistobeappliedwhile movesfrom

PtoQ.Theworkdonebythisexternalagentinmovingthechargebyadistance isgivenby:

.............................(23)

Fig8:MovementofTestChargeinElectricField

Thenegativesignaccountsforthefactthatworkisdoneonthesystembytheexternal

agent.

.....................................(24)ThepotentialdifferencebetweentwopointsPandQ,VPQ,isdefinedasthework

doneperunitcharge,i.e.

...............................(25)Itmaybenotedthatinmovingachargefromtheinitialpointtothefinalpointifthe

potentialdifferenceispositive,thereisagaininpotentialenergyinthemovement,

externalagentperformstheworkagainstthefield.Ifthesignofthepotentialdifference

isnegative,workisdonebythefield.

Wewillseethattheelectrostaticsystemisconservativeinthatnonetenergyis

exchangedifthetestchargeismovedaboutaclosedpath,i.e.returningtoitsinitial

position.Further,thepotentialdifferencebetweentwopointsinanelectrostaticfieldisa

pointfunctionitisindependentofthepathtaken.Thepotentialdifferenceismeasured

inJoules/CoulombwhichisreferredtoasVolts.

LetusconsiderapointchargeQasshownintheFig.9.

Fig9:ElectrostaticPotentialcalculationforapointcharge

FurtherconsiderthetwopointsAandBasshownintheFig.9.Consideringthe

movementofaunitpositivetestchargefromBtoA,wecanwriteanexpressionforthe

potentialdifferenceas:

...................(26)Itiscustomarytochoosethepotentialtobezeroatinfinity.Thuspotentialatanypoint(

rA=r)duetoapointchargeQcanbewrittenastheamountofworkdoneinbringinga

unitpositivechargefrominfinitytothatpoint(i.e.rB=0).

..................................(27)

Or,inotherwords,

..................................(28)LetusnowconsiderasituationwherethepointchargeQisnotlocatedattheoriginas

showninFig.10.

Fig10:ElectrostaticPotentialdueaDisplacedCharge

ThepotentialatapointPbecomes

..................................(29)Sofarwehaveconsideredthepotentialduetopointchargesonly.Asanyothertypeof

chargedistributioncanbeconsideredtobeconsistingofpointcharges,thesamebasic

ideasnowcanbeextendedtoothertypesofchargedistributionalso.Letusfirst

considerNpointchargesQ1,Q2,.....QNlocatedatpointswithpositionvectors , ,

....... .Thepotentialatapointhavingpositionvector canbewrittenas:

..................................(30a)OR

...................................(30b)

Forcontinuouschargedistribution,wereplacepointchargesQnbycorresponding

chargeelements or or dependingonwhetherthechargedistributionislinear,surfaceoravolumechargedistributionandthesummationisreplacedbyan

integral.Withthesemodificationswecanwrite:

Forlinecharge, (31)

Forsurfacecharge, .................................(32)

Forvolumecharge, .................................(33)

Itmaybenotedherethattheprimedcoordinatesrepresentthesourcecoordinatesand

theunprimedcoordinatesrepresentfieldpoint.

Further,inourdiscussionsofarwehaveusedthereferenceorzeropotentialatinfinity.

Ifanyotherpointischosenasreference,wecanwrite:

.................................(34)

whereCisaconstant.Inthesamemannerwhenpotentialiscomputedfromaknown

electricfieldwecanwrite:

..(35)Thepotentialdifferenceishoweverindependentofthechoiceofreference.

.......................(36)

Wehavementionedthatelectrostaticfieldisaconservativefieldtheworkdonein

movingachargefromonepointtotheotherisindependentofthepath.Letusconsider

movingachargefrompointP1toP2inonepathandthenfrompointP2backtoP1

overadifferentpath.Iftheworkdoneonthetwopathsweredifferent,anetpositiveor

negativeamountofworkwouldhavebeendonewhenthebodyreturnstoitsoriginal

positionP1.Inaconservativefieldthereisnomechanismfordissipatingenergy

correspondingtoanypositiveworkneitheranysourceispresentfromwhichenergy

couldbeabsorbedinthecaseofnegativework.Hencethequestionofdifferentworks

intwopathsisuntenable,theworkmusthavetobeindependentofpathanddepends

ontheinitialandfinalpositions.

Sincethepotentialdifferenceisindependentofthepathstaken,VAB=VBA,andover

aclosedpath,

.................................(37)ApplyingStokes'stheorem,wecanwrite:

............................(38)fromwhichitfollowsthatforelectrostaticfield,

......................(39)Anyvectorfieldthatsatisfiesiscalledanirrotationalfield.

Fromourdefinitionofpotential,wecanwrite

.................................(40)fromwhichweobtain,

..........................................(41)

Fromtheforegoingdiscussionsweobservethattheelectricfieldstrengthatanypointis

thenegativeofthepotentialgradientatanypoint,negativesignshowsthat is

directedfromhighertolowervaluesof .Thisgivesusanothermethodofcomputingtheelectricfield,i.e.ifweknowthepotentialfunction,theelectricfieldmaybe

computed.Wemaynoteherethatthatonescalarfunction containallthe

informationthatthreecomponentsof carry,thesameispossiblebecauseofthefact

thatthreecomponentsof areinterrelatedbytherelation .

EquipotentialSurfaces

An equipotential surface refers to a surface where the potential is constant. The

intersection of an equipotential surface with an plane surface results into a path called

an equipotential line. No work is done in moving a charge from one point to the other

alonganequipotentiallineorsurface.

In figure 12, the dashes lines show the equipotential lines for a positive point charge. By

symmetry, the equipotential surfaces are spherical surfaces and the equipotential lines

arecircles.Thesolidlinesshowthefluxlinesorelectriclinesofforce.

Fig12:EquipotentialLinesforaPositivePointCharge

MichaelFaradayasawayofvisualizingelectricfieldsintroducedfluxlines.Itmaybe

seenthattheelectricfluxlinesandtheequipotentiallinesarenormaltoeachother.In

ordertoplottheequipotentiallinesforanelectricdipole,weobservethatforagivenQ

andd,aconstantVrequiresthat isaconstant.Fromthiswecanwrite

tobetheequationforanequipotentialsurfaceandafamilyofsurfacescanbegeneratedforvariousvaluesofcv.Whenplottedin2Dthiswouldgiveequipotential

lines.

To determine the equation for the electric field lines, we note that field lines represent

thedirectionof inspace.Therefore,

,kisaconstant.............................................(42)

.................(43)

Forthedipoleunderconsideration =0,andthereforewecanwrite,

..................................(44)ElectrostaticEnergyandEnergyDensity:

We have stated that the electric potential at a point in an electric field is the amount of

work required to bring a unit positive charge from infinity (reference of zero potential) to

that point. To determine the energy that is present in an assembly of charges, let us first

determine the amount of work required to assemble them. Let us consider a number of

discrete charges Q1, Q2,......., QN are brought from infinity to their present position one

by one. Since initially there is no field present, the amount of work done in bring Q1 is

zero. Q2 is brought in the presence of the field of Q1, the work done W1= Q2V21 where

V21 is the potential at the location of Q2 due to Q1. Proceeding in this manner, we can

write, the total work done ....................(45)

Hadthechargesbeenbroughtinthereverseorder,

................(46)

Therefore,

................(47)

HereVIJrepresentvoltageattheIthchargelocationduetoJthcharge.Therefore,

Or, ................(48)Ifinsteadofdiscretecharges,wenowhaveadistributionofchargesoveravolumev

thenwecanwrite, ................(49)

where isthevolumechargedensityandVrepresentsthepotentialfunction.

Since, ,wecanwrite

.......................................(50)

Usingthevectoridentity,,wecanwrite

................(51)

Intheexpression ,forpointcharges,sinceVvariesas andDvariesas

,thetermV variesas whiletheareavariesasr2.Hencetheintegralterm

variesatleastas andtheassurfacebecomeslarge(i.e. )theintegraltermtendstozero.

ThustheequationforWreducesto

................(52)

,iscalledtheenergydensityintheelectrostaticfield.PoissonsandLaplacesEquations

Forelectrostaticfield,wehaveseenthat

................................................................(53)Formtheabovetwoequationswecanwrite

................................................(54)

Usingvectoridentitywecanwrite, ................(55)

Forasimplehomogeneousmedium, isconstantand .Therefore,

................(56)ThisequationisknownasPoissonsequation.Herewehaveintroducedanewoperator

,(delsquare),calledtheLaplacianoperator.InCartesiancoordinates,

...............(57)Therefore,inCartesiancoordinates,Poissonequationcanbewrittenas:

...............(58)Incylindricalcoordinates,

...............(59)Insphericalpolarcoordinatesystem,

...............(60)

Atpointsinsimplemedia,wherenofreechargeispresent,Poissonsequationreducesto

...................................(61)

whichisknownasLaplacesequation.

Laplaces and Poissons equation are very useful for solving many practical electrostatic field

problems where only the electrostatic conditions (potential and charge) at some boundaries are

known and solution of electric field and potential is to be found hroughout the volume. We shall

considersuchapplicationsinthesectionwherewedealwithboundaryvalueproblems.

Conventionandconductioncurrent:

CapacitanceandCapacitors

We have already stated that a conductor in an electrostatic field is an Equipotential body and any

charge given to such conductor will distribute themselves in such a manner that electric field

inside the conductor vanishes. If an additional amount of charge is supplied to an isolated

conductor at a given potential, this additional charge will increase the surface charge density

. Since the potential of the conductor is given by , the potential of the

conductor will also increase maintaining the ratio same . Thus we can write where

the constant of proportionality C is called the capacitance of the isolated conductor. SI unit of

capacitance is Coulomb/ Volt also called Farad denoted by F. It can It can be seen that if V=1, C

= Q. Thus capacity of an isolated conductor can also be defined as the amount of charge in

Coulombrequiredtoraisethepotentialoftheconductorby1Volt.

Of considerable interest in practice is a capacitor that consists of two (or more) conductors

carrying equal and opposite charges and separated by some dielectric media or free space. The

conductorsmayhavearbitraryshapes.Atwoconductorcapacitorisshowninfigurebelow.

Fig:CapacitanceandCapacitors

Whenadcvoltagesourceisconnectedbetweentheconductors,achargetransferoccurswhich

resultsintoapositivechargeononeconductorandnegativechargeontheotherconductor.The

conductorsareequipotentialsurfacesandthefieldlinesareperpendiculartotheconductor

surface.IfVisthemeanpotentialdifferencebetweentheconductors,thecapacitanceisgivenby

.Capacitanceofacapacitordependsonthegeometryoftheconductorandthe

permittivityofthemediumbetweenthemanddoesnotdependonthechargeorpotential

differencebetweenconductors.ThecapacitancecanbecomputedbyassumingQ(atthesame

timeQontheotherconductor),firstdetermining usingGaussstheoremandthen

determining .Weillustratethisprocedurebytakingtheexampleofaparallelplate

capacitor.

Example:Parallelplatecapacitor

Fig:ParallelPlateCapacitor

Fortheparallelplatecapacitorshowninthefigureabout,leteachplatehasareaAandadistance

hseparatestheplates.Adielectricofpermittivity fillstheregionbetweentheplates.The

electricfieldlinesareconfinedbetweentheplates.Weignorethefluxfringingattheedgesof

theplatesandchargesareassumedtobeuniformlydistributedovertheconductingplateswith

densities and , .

ByGaussstheoremwecanwrite, .......................(1)

Aswehaveassumed tobeuniformandfringingoffieldisneglected,weseethatEis

constantintheregionbetweentheplatesandtherefore,wecanwrite .Thus,fora

parallelplatecapacitorwehave,

........................(2)

SeriesandparallelConnectionofcapacitors

Capacitorsareconnectedinvariousmannersinelectricalcircuitsseriesandparallelconnections

arethetwobasicwaysofconnectingcapacitors.Wecomputetheequivalentcapacitanceforsuch

connections.

SeriesCase:Seriesconnectionoftwocapacitorsisshowninthefigure1.Forthiscasewecan

write,

.......................(1)

Fig1.:SeriesConnectionofCapacitors

Fig2:ParallelConnectionofCapacitors

Thesameapproachmaybeextendedtomorethantwocapacitorsconnectedinseries.

ParallelCase:Fortheparallelcase,thevoltagesacrossthecapacitorsarethesame.

Thetotalcharge

Therefore, .......................(2)

ContinuityEquationandKirchhoffsCurrentLaw

Let us consider a volume V bounded by a surface S. A net charge Q exists within this region. If a

net current I flows across the surface out of this region, from the principle of conservation of

charge this current can be equated to the time rate of decrease of charge within this volume.

Similarly, if a net current flows into the region, the charge in the volume must increase at a rate

equaltothecurrent.Thuswecanwrite,

.....................................(3)

or, ......................(4)

Applyingdivergencetheoremwecanwrite,

.....................(5)

It may be noted that, since in general may be a function of space and time, partial derivatives

are used. Further, the equation holds regardless of the choice of volume V , the integrands must

beequal.

Thereforewecanwrite,

................(6)

The equation (6) is called the continuity equation, which relates the divergence of current density

vectortotherateofchangeofchargedensityatapoint.

Forsteadycurrentflowinginaregion,wehave

......................(7)

Consideringaregionboundedbyaclosedsurface,

..................(8)

whichcanbewrittenas,

......................(9)

whenweconsidertheclosesurfaceessentiallyenclosesajunctionofanelectricalcircuit.

The above equation is the Kirchhoffs current law of circuit theory, which states that algebraic

sumofallthecurrentsflowingoutofajunctioninanelectriccircuit,iszero.

Questionbank:

1stunit

Bits:

1. Displacementcurrentinaconductorisgreaterthanconductioncurrent(yes/no)2. Electricdipolemomentisavector (yes/no)3. Electricsusceptibilityhastheunitofpermittivity(yes/no)4. Capacitancedependsondielectricmaterialbetweentheconductors(yes/no)5. TheunitofpotentialisJoule/coulomb(yes/no)

6. (yes/no)

7. Coulombsforceisproportionalto 8. Theunitofelectricfluxiscoulombs

9. Theelectricfieldonxaxisduetoalinechargeextendingfrom 10.Potentialatallpointsonthesurfaceofaconductoristhesame11.Laplaceequationhasonlyonesolution12.Exampleofnonpolartypeofdielectricisoxygen13.Theelectricsusceptabilittyofadielectricis4,itsrelativepermittivityis514.BoundaryconditionforthenormalcomponentofEontheboundaryofadielectricis

= 15.Potentialduetoachargeatapointsituatedatinfinityis0

16.Relationtimeis 17.Theforcemagnitudeb/wQ1=1CandQ2=1Cwhentheyareseparatedby1minfree

spaceis9*109N

18. =0isinpointform =019.Directionofdipolemomentisindirectionofappliedelectricfield20. Ifaforce,F=4ax+ay+2aZmoves1 Cchargethroughadisplacementof4ax+2ay

6aZtheresultantworkdoneis6 J