Electr Magnetic for Eng 4_2

8/13/2019 Electr Magnetic for Eng 4_2

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92 CHAPTER ELECTROST

Figure 4-17: A dielectric medium polarizedby an exter-nal electric field Es11.

an electric dipole consisting of charge *q at the cen-ter of the nucleus and charge -q at the center of theelectroncloud [Fig. a-16(c)]. Each such dipole setsup a small electric field, pointing from the positivelychargednucleus to the center of the equally but neg-atively charged electron cloud. This induced electricfield, called a polarizntionfield, is weaker than and op-posite n direction to Esx1.Consequently, he net elec-tric field present in the dielectric material is smallerttran E"*,. At the microscopic level, each dipole ex-hibits a dipole moment similar to that described in Ex-ample4-7. Within the dielectric material, the dipolesalign themselves n a linear arrangement,as shown inFig. 4-17. Along the upper and lower edges of the ma-terial, he dipole arrangementexhibits a positivesurfacechargedensity on the upper surfaceand a negativeden-sity on the lower surface.The relatively simple picture described n Figs. 4-16and 4-17 pertainsto nonpolnr materials in which themolecules do not have permanent dipole moments.Nonpolar molecules become polarized only when anextemalelectric field is applied, and when the field is

terminated, he molecules eturn to their originalpolarized state. In somematerials, such as watcqmolccrrlarstnrcftrc is suchthat thc molcculcsbuilt-in permanentdipole moments hat areorientedn the absence f anappliedelectric field-terials composed f permanent ipolesarecallodmaterials.Owing to their randomorientations,polesof polar materialsproduceno net dipolemacroscopicallyatthe macroscopic cale,eachthe material representsa small volume containingsandsof molecules). Under the influence of anfield, the permanentdipoles tend to alignto someextent along the direction of the electicin an arrangement omewhatsimilar to thatFig.4-17 or nonpolarmaterials.Whereas D and E are related by e6 in freepresenceof these microscopic dipoles in amaterial alters that relationship in that material to

D : e o E * P ,where P, called the electric polarizationfield ,for the polarization properties of the material.larization field is produced by the electric field Edepends on the material properties.A dielectric medium s said to be linear If tlrcnitude of the inducedpolarization ield is directlyportional to the magnitude of E, and it is saidisotropic if the polarization field and E are in thedirection. n somecrystals, heperiodicstructurematerial allows more polarizationto take placecertaindirections, uch as the crystal axes, hanothers. n suchanisotropicdielectrics,E and Dhavedifferentdirections.A medium s said omogeneousf its constitutive arameterse, pa,are constant throughout the medium. Our presentment will be limited to media that are linear, iand homogeneous.For suchmedia theis directly proportional o E and is expressedrelationship

P: aoXeE,


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r.8 DIELECTRICS 93

he

E

tod

D

by

Petoleum oilPolystyreneGlassQuartzBakeliteMica

2 .12.64.5-10

3.8-555.44

aJ122025403020200

Table 4-2: Relativepermittivity (dielectricconstant)and dielectricstrengthof commonmaterials.

I : er6oand es : 8.854x 10-12 F/m.

-E-texe is called the electric susceptibility of the ma-:rl. Inserting q. (4.8a) nto Eq. (4.83),we have

D: eoE e6X"E: eo(l* X")E rE, (4.85)

--',-h efineshe permittivitye of the materialase : e o ( l + X e ) . (4.86)

rr-; was mentioned earlier, it is often convenient--haracterize the permittivity of a material relativetrat of free space, eo; this is accommodated byrelative permittivity r : t/to. Values of e, arein Table L2 for a few common materials. androger list is given in Appendix B. In free space= l, and or mostconductors r : l. The dielectric:rant of air is approximately1.0006at sea evel,rt decreases oward unity with increasing altitude.::pt in some special circumstances,such as when:rlating electromagneticwave refraction (bending)the atmosphere over long distances, air is:-',d thesameas ree space.ae dielectricpolarizationmodelpresentedhusfar:lacedno restriction n theupperendof the strength

te applied electric field E. In reality, if E exceedsa

certain critical value, known as the dielectric strength ofthe material, it will free the electronscompletely fromthe moleculesand cause hem to accelerate hrough thematerial in the form of a conduction current.When thishappens,sparking can occur, and the dielectric materialcan sustainpermanentdamagedue to electron collisionwith the molecular structure.This abruptchange n be-havior is called a dielectric breakdown. The dielectricstrength 86, is the highest magnitudeof E that the ma-terial can sustain without breakdown.Dielectric break-down can occur in gas, liquid, and solid dielectrics.The associated ield strength dependson the materialcomposition, as well as other factors such as temper-ature and humidity. The dielectric strength for air is 3(MV/m); for glass, t is 25 to 40 (MV/m); and for mica,it is 200 (MV/m) [seeTable 4-2].A charged hundercloud with an electric potential V,relative to the ground, nducesan electric field E : V/din the air medium between the ground and the cloud,where d is the height of the cloud base above theground's surface. If 7 is sufficiently large so that Eexceeds he dielectric strength of air, ionization occursand discharge (lightning) follows. The breakdownvoltage V61of a parallel-platecapacitor is discussed nExample4-11.


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94 CHAPTER 4 ELECTROSTA

Figure 4-18: nterface etweenwo dielectricmedia.

REVIEWUESTI(}NSQ4.20 What is a polarmaterial?A nonpolarmaterial?Q4.21 Do D andE alwayspoint in the samedirection?If not, when do theynot?Q4.22 What happenswhen dielecfic breakdown oc-curs?

4-g ElectricoundaryonditionsAn electric field is said to be spatially continuous f itdoes not exhibit abrupt changes n either its magnitudeor direction as a function of spatial position. Eventhough the electric field may be continuous in eachof two dissimilar media, it may be discontinuous atthe boundary betweenthem if surface charge existsalong that boundary.Boundary conditions specify howthe tangential and normal components of the field inone medium arerelated o the componentsof the fieldacross he boundaryn anothermedium. We will derivea generalset of boundaryconditions, applicableat theinterface betweenany two dissimilar media, be theytwo different dielectricsor a conductorand a dielectric.

Also, any of the dielectrics may be free space.Ethough these boundary conditions will be derivedelectrostatic conditions, they will be equally rnfor time-varying electric fields. Figure 4-18 showsinterface between medium 1 with permittivity e1 rmedium 2 with permittivity e2. In the generalcase,boundary may have a surfacechargedensity p..To derive the boundary conditions for the tangeilcomponentsof E and D, we begin by constructingclosedrectangular oop abcda shown in Fig. 4-18, rthen we apply the conservative property of the eloclfield given by Eq. (4.40), which states hat the lincitegral of the electrostatic field around a closed pdalwaysequal to zero. By letting A4* 0, the coltions to the line integral by segmenisbc anddazero. Hence.

f *. or:l"uEz. t f"o 1dt o,where E1 and &z are he electric fields in media Irespectively. In terms of the tangential and normCrectionsshown n Fig. 4-18,

Er : Er t * Ern , (Ez:Ezt * Ezn. (


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y9 ELECTRICBOUNDARY CONDHONS 95

go

lJttr segment b, E7 anddl have he samedirection,lul oversegment d, Ey anddl are n oppositedirec-u. Conscquently, q. (4.87)givesEy Al - E11A/= 0, (4.8e)

Er: Ezt ff/m). (4.90)t'rordingly, the tangential componentof the electricr'r.i is continuousacross he boundarybetweenany tw,o-, -;:, i .SinceDt : e1811 nd Dz, : t2E2t, he bound-r. conditionon the tangentialcomponentof theelectricftu densitv is

Dta 1

Dzt2 (4.e1)

\ext we apply Gauss's law, as expressed by\. G.29), to determinethe boundaryconditionson therurmal componentsof E and D. According to Gauss'slrr. the total outward flux of D through the three sur-lx of the small cylinder shown in Fig. 4-18 mustrFrel ths total charge enclosed n the cylinder. By let-rng the cylinder's height L,h + 0, the contribution totu total flux by the side surface goes to zero. Also,rm if each of the two media happens o have free orn:rmdvolume chargedensities, he only charge emain-n,_:n the collapsed cylinder is that distributed on theI:undary. Thus, Q : p, As, and

where f1and f2are theoutwardnormalunit vectorsofthebottomand op surfaces,espectively.t is importantto rcmember tnt thenormalunit vectorat the surfaceofanymcdiumsalwaysdefnedto be n theoutwarddirec_tionawayfromthatmediuru.ince ir : -ff2, W.9.92)simplifiestoff2.@r Dz) :A (Clm2). (4.93.)

With D1n and D2, defined as the normal componentsof D1 and D2 along ff2, we have

Dn - D2n: p, (C/m2). (4.94)

Thus, the normalcomponent f D changes bruptlt,at ctchargedboundary befween wo dffirent media,and theatnountof change s equal to the surfacechargedensity.The correspondingboundary condition for E isrEn- zEzn ps. (4.95)

In summary,1) theconservative ropertyof E,V x E-0


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CHAPTER 4

Table 4-3: Boundaryconditions or theclectric fields.

0 ApplicationfBoundaryonditionsThe x-y plane s a charge-freeoundary eparatingtwo dielectricmedia with permittivit iese1 and 12,asshownn Fig. 4-19. f theelectric ield in medium isEr : iEr, * I Ery* iEp, find(a) heelectric ieldE2in medium2 and(b) theanglesl and02.

Solution: a) Let Ez : i'Eu * 9Ezy+ 2822.Our task

r-y plane

Figurt 4-19: Application of boundary conditions at theinterfacebetween wo dielectric media (Example 4-10).

is to find the componentsof E2 in terms of thecomponentsof E1. The normal to the boundaryHence, he x and y componentsof the fields aretial to the boundaryand the z componentsarethe boundary.At a charge-free nterface, thecomponentsof E and the normal componentsofcontinuous.Consequently,

E z t : E k , E z y : E b ,and

D z r : D k szEzz stEb.HenG,

E z : i E r , * | E r y + 2 ! E u .(b) The tangential omponenrsi ", and l- l l l - _ _ _ - - l - t - - - ] - - - - ] -En: tlEi, a Ef, andE21:J EL + E y.Tllr;91 and 02 are then given by

^ Et ,tan Ar :' E u

tanl.t:E"

:- Ez,

r--7---------;_\f Ef, + EiyE u 'r.-:-_^\\fE;r + E;yEzz $,+ no(e1/e)Ep

TangentialETirngentialDNormal ENormalD

En= ExD*/et: Dzt/ezff.(erEr -tzBz)* p"

f l . ( o r - D z ) : p ,

E*= ExDt/er: Dzr/ezstEln - e2B2n- ps

Dy1 D21t: 4


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(o

H.ECTRIC BOUNDARY CONDMONS

6e two angles are related bytan02 (4.ee)tan01

4.16 With referenceo Fig. 4-19,find E1 if8z : 8eo. n - i3 + i3 (V/m),dr = 20, ndtheboundary o be charge ree.Nrrrrrrrrrrrrrrrr .r : i2 * y3 + 212(Ylm). (See G;ffiOSE 4.17 RepeatExercise4.16 for a boundaryuddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddhargedensity ps:3.54 x l0-ll (C/m2)..r"ns-Er : *2 * i3 + iI4 (Ylm). (See0;

+9.1 Dielectric-ConductorBoundaryf,:'nsiderhecasewhenmedium1 n Fig.4-18 s a die-s-tric and medium2 is aperfectconductor.n aperfect:L'nductor, = D : 0 everywheren the conductor.Fience, z = Dz: 0, which requireshe angential nd

normal componentsof E2 and D2 to be zero. Conse-quently, from Eq. (4.90) and Eq. (4.94), the fields in thedielectric medium, at the boundary with the conductor.are given by_ t , I 1

E n : D r t - O ,D ln g1E6: p" .

(4.100a).(4.100b)

Ps=stEo

Conductingsla$---..+.

tllre &20: When a conductingslab s placed n an externalelectric field &, chargeshat accumulate n the conductor

These wo boundary onditions an be combinedntoDr : erBr : ffp, (atconductor urface), 4.101)

where ff is a unit vector directed normally outward fromthe conductingsurface.This means hat theelectricfieldlinespoint directly awayftom the conductorsurfacewhenp, is positive and directly toward the conductor surfacewhenp, is negative.Figure 4-20 showsan infinitely long conducting slabplaced n a uniform electric field &. The medium aboveand below the slab has a permittivity e1. Because E6points away from the upper surface, t induces a pos-

itive charge density ps : rllllol on the upper surface


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Figure 4-2lz Metal sphereplaced n an externalelectricfieldFa.

of the slab. On the bottom surface, E6 points towardthe surface,and therefore the induced charge density is-pr. The presenceof these surface charges nduces anelectric ield E1 n the conductor, esulting in a total fieldE : Fo * Ei. To satisfy the condition that E must beeverywherezero in the conductor,Ei must equal -S.If we place a metallic sphere n an electrostatic field,as shown n Fig. 4-2I,negatwe chargeswill accumulateon the lower hemisphereand positive charges will ac-cumulateon the upper hemisphere.The presenceof thesphere auses he field lines to bend to satisfy the con-ditiongivenby Eq. (a.101); hat s, E is alwav-s ormalto thesurfaceat the conductor boundary.4-9.2Conductor-Gonductoroundarylffe now examine the general case of the boundary be-tween two media neither of which are perfect dielectricsor perfect conductors. Depicted in Fig. 4-22, medium 1haspermittivity e1 and conductivity o1, medium 2 hass2 and 02, nn.d he interface between hem has a surfacechargedensitypr. For the electric fields, Eqs.(4.90) and(4.95)give

E*: Ezt, srErn sz0zn- ps. 9.102)

CHAPTER4 ELECTROST

Since we are dealing with conducting media, theric fields give rise to current densitiesJr and Jz,Jr being proportional to El and Jz beingto E2. From J - oE, we haveJrn Jzn, t l 01 62

The tangential components .Irt and ,I21 epresentflowing in the two media n i direction parallelbboundary,and henceno transferof charge s ibetween hem. This is not the case or the normalponents. f Jh +,/2n, then a different amountofarrives at the boundary than leaves it. Hence, p,remain constant with time. which violates thetion of electrostaticsequiring all fields andremainconstant.Consequently,he nonnalof J has o be continuous cross he boundarytwo dffirent mediaunderelectrostatic onditions.settingJn: Jzn n Eq.(4.103),we have

Jn Jz,o1 o2


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100 CHAPTER4 ELECTROSTA

where is the integrationpath from conductor 2 to con-ductor 1. To avoid making sign errors when applyingEq. (4.109), t is important to remember that surface,Sis the *0 surfaceand P1 s on S. BecauseE appearsnboth the numeratorand denominatorof Eq. (4.109), thevalueof C obtainedforany specificcapacitor configura-tion s always ndependent f E. In fact, C dependsonlyon the capacitor geometry (sizes, shapesand relativepositionsof the two conductors)and the permittivity ofthe insulatingmaterial.If the material between the conductors s not a per-fect dielectric (i.e., if it has a small conductivity o),then current can flow through the material between theconductors,and the material will exhibit a resistanceR.The generalexpression or R for a resistor of arbitraryshapes givenby Eq. (4.71):

I en.asC : "s ^ (F), (4.109)- J,n ' t

- [ n . a tR: v t , ( ( .2 ) .J ,oE 'ds

wherepointsP1 and Pz areany two pointson conduc-tors and2, respectively. ubstituting qs. 4.107)and(4.108)nto Eq. (4.105)givesCrpacitancendEreakdownoltaeParallel-Plateapacilor

Obtainanexpressionor the capacitanceC of apclel-platecapacitor omprisedof rwo parallel platesaof surfaceare,a and separated y a distance .lcapacitor s filled with a dielectric material with pmittivity e. Also, determine he breakdownvoltapd : 1 cm and the dielectricmaterial s quartz.Solution: n Fig. 4-24, weplace he lowerplateof Icapacitorn the .r-y planeand the upperplate n Iplanez : d. Because f the appliedvoltageditenceV, charge O accumulatesniformly on thclplateand Q accumulatesniformlyon he owerflIn thedielectricmediumbetweerlthe lates,he ch{inducea uniformelectric ield n the -2-direction lpositive to negative charges). n addition, someingfield lines will exist near the edges,but theirmay be ignored f the dimensionsof the platesarelarger than the separationd between them,that case he bulk of the electric field lines will exithe medium between he plates.The chargethe upperplate s p' : QlA.Hence,

g : - i E ,andfrom Eq. (4.106), he magnitude f E at tlrcductor-dielectric boundary s E : ptle : QltA-Eq. (4.108), the voltage difference is

and the capacitances

where usewas made of the relation B : QleA.From V - Ed, as given by Eq. (4.112), Vwhen E : Eds, he dielectric strength of the

v - - lr' " dr: Ioo-rr, .idz Ed,For a medium with uniform o and e, the product ofEqs. 4.109) nd (4.110)gives

R C : I6 (4 .111 )

(4 .1 l0)

if C is known.his simple relation allows us to find Ror vice versa.

O O e A- E d - d (4 .113)


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r-10 CAPACITANCE l0r

ad.

ofindithc

ex i

theA .

(4 -

A -r l -

Dielectric e

ConductingplateFigure4-2A:A d-cvoltagesource onnectedo aparallel-plateapacitorExample -l l).

to TableL2, E6s= 30 (MV/m) for quartz.

iIlII

, thebreakdownvoltage sl'- : Easd: 30 x 106 10-2= 3 x 105V. I

Gapacitancel Goarialine'lbtain an expression or the capacitanceof the coax-me shown nFig.4-25.

Solution: For a given voltage V across he capacitor(Fig.4-25), hargeQ and Q will accumulaten hesurfaces f theouterand nnerconductors,espectively.We assumehat hese harges reuniformlydistributedalong he engthof theconductors ith line charge en-sityp; : Q/ I on heouterconductor nd pt onthe n-ner conductor.gnoring ringingfieldsnear he endsofthecoaxial ine,we canconstruct cylindricalGaussiansurface n the dielectric, around the inner conductor.with radius such hatc < r < b. The nnerconductoris a line charge imilar o that of Example4-6, except

-Pt

Outer conductorFigure4-25: Coaxialcapacitorilledwith insulatingmaterialof perminivitye @xample -12).


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CapacitiveensorsTo sense is to respond o a stimulus. SeeResistiwSercors.)A iruf::ftflrehionga sensor f ttle stinulus4{d ,&WlgSgeometry usually he,$Bgirigl:,$G$itsfr,lt$conductivelementrs orffidigt ficproperiles'of the insulatingmaterial*ated betweenhem.Capacitiveensorsaie uSd'' n"a mullitude fapplications.fewexamplesollow.''Fluid augeThe two metalelectrodesn (A), usually odsorplates,orma capacitorwhosecapacitances di-rectly roportionaltohepermittivityf the materialbetweenhem. f he luid ectionsof heightHy and

TECHNOLOGYBRIEF: CAPACITTVEthe heightof theemptyspaceabove t is (I/ - flthen heoverallcapacitancesequivalento rvopacitors n parallel,ori , , : _ ^ ( w H t ) , ^ ( H - H iC z = C f * C o : e f= J ! * t . T

uhere .w is the elec{rodeplate width, d isspaoing etweenelectrodes, nd eyand s4arsperrnittiWlesf the fluidand air, respectively.rangingheexpressionsa linearequationCz: kHf a Co, l

where he constarit t : (e1 e)w/d andCs scapacitance f thetank when otallyempty.the inear quation,he fluidheight an beminedby measuringC2,whichcan be realizedusinga bridge ircuit B).Theoutput oltage{proportionaltohedeviationetweenC1andC2-settingCr : Co afixedcapacitor) ndbying he ankelectrodeso the bridge ircuitoC2,Vo." ecomes roportionaltohe luidheigtrt


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Y BRIEF: CAPACITTVESENSORS 103

e

H

s ditySensor

.

metalelectrodeshaped n an interdigi-pattern to enhance he ratioA/d) are Iabri-on a siliconsubstrateC).The spacingbe-digits s typically n heorderof 0.2pm.Theof hematerialseparatingheelectrodes

varieswith he relativehumidityof the surroundingenvironmentHence,he capacitorbecomesa hu-miditysensor.PressureensorA flexiblemetaldiaphragm eparates n oil-filledchamberwithreference ressurep6 roma secondchamber xposedo the gasor fluidwhosepres-sureP is to be measuredy hesensor D1). hemembranessandwiched,ut electricallystfiated,betweenwoconductivearallel urfaces,ormingtwo capacitorsn series(D2).When p > ps,themembraneends nthedirection f the owerplate(D3).Consequently,1 ncreasesndd2decreasesand, n turn,C1decreases ndC2 ncreases. heconverse appens henP < po.With he useof acapacitanceridge ircuit, uchas the one n (B),thesensor an becalibratedo measurehe pres-sureP withgoodprecision.


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104that ttre line chargeof the innerconductor s negative.With a minussignadded o theexpressionor E givenby Eq.(4.33),we have

(4.r14\

CHAPTER4 ELECTROST4-11 ElectrostaticPotentialnergyrilhen a s{xrrues conncctcd o a caryitm, itenergy in charging up the caprcitor. If thcplatesare madeof a good conductorwithzero resistance ndif thc dielectricseparatinghcconductors asnegligibleconductivity,henno realrent can flow through the dielcctric, and no ohmicoccur anywheren the capacitor.Wherethendoescharging-upenergy go? The energyendsupstored n the dielectricmedium n the fqm ofstnticpotentialenergy.Theamountof storedenerg:tis related o Q, C, andV.Under he influenceof the electric ield in thelectric medium between the two conductors.accumulateson one of the conductors,andan equaloppositechargeaccumulateson the other conductc,effect, charge q has been transfened from one ofconductors o the other.The voltage u across heitor is related to q by

From the basicdefinition of the electric potential V,amountof work dW" required o transferanincrementalamountof chargedg isd W s : u d q : t O n .

n ^ P t ^ OE : - f - : - | . -Zner - ZrerlThepotentialdifferenceV between heouterand nnerconductorss

v - I" 'r.dt:-" ' u#) .rur,y: * " ( : )

The capacitanceC is then given by

C _ ov 2nelln(b/a) (4 .116)The capacitanceper unit length of the coaxial

2treln(b/a) (F/m). I

REVIEWUESTIONSQ4.25 How is the capacitanceof a two-conductorstructurerelated to the resistanceof the insulatins ma-terial between he conductors?Q4.26 What are fringing fields and when may they beignored?

(4.rIf we start with an uncharged capacitor andup from zero chargeuntil a final chargeO hasreached,henthe total amountof work performedr

Using C = Q/V, where V is the final voltage,also be written aswu: |cvz (J).

For the parallel-plate apacitordiscussednple 4-11, its capacitances given by Eq. (4.113)

(4 .115)

line is(4.rr7)

(4.rU : _C

*":Ionon:f (D (4.wc


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: IMAGE METHOD 105

i \t lt ,I t

? _ .4Charge Q above grounded planeF;ue 1'26zBy image heory a chargeQ above grounded erfectlyconductingplane s equivalent o e andits mage- er"c de groundplane emoved.

- tA/d, where A is the surface areaof eachof itsand d is the separationbetweenthem. Also, thesc V acrosshe capacitors related o themagni---nfheelectric ield,E, in thedielectric v V - Ed.6ese wo expressionsn Eq. (4.121) ives

REVIEWUESTIONSQ4.27 To bring a charge4 from infinity to a givenpoint n space, certainamount f work llz sexpended.Wheredoes heenergycorrespondingo $r go?

EXERCISE.18 The adii of the nnerandouterconduc-torsof a coaxialcableare 2 cm and5 cm,respectively,and he nsulatingmaterialbetweenhemhasa relativepermittivityof 4. The chargedensityon theoutercon-ductor s pl - 10-4 (C/m).Use the expressionor Ederived n Example4-12to calculatehe total energystoredn a 20-cm engthof thecable.Ans. We:4.1 J. (See S1

4-12 lmage ethodConsidera point charge Q at a distanced above a per-fectly conductingplane, as shown in the left-hand sec-tion of Fig. 4-26.We want to determineV,E, and D atany point in the spaceabove he groundedconductor,aswell as he distribution of surfacechargeon the conduct-ing plate.Three different methodshavebeen ntroduced

, e Ar== i jeaY: lefi1eg - |eE2v, (4.122)(ttlEe v" Ad is the volume of the capacitor.Ib cbctrostatic energy density u.r" s defined as the

s

ic potentialenergy y" peJunit volume:

though this expressionwas derived for a,el-platecapacitor,t is equally valid for any di-mediumn an electric ieldE. Furthermore.orrolume v containing a dielectric e, the total elec-potential energy stored in y isw e - (J). (4.124))1,, 'a,

(4.r23)


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in this chapter for finding E. The first, basedon Cou-lomb's law, requiresknowledge of the magnitudesandlocationsof all the chargescontributing to E at a givenpoint in space. n the presentcase, he chargeQ will in-ducean unknown and nonuniform distribution of chargeon the surface of the conductor.Hence,we cannot uti-lize Coulomb's method.The secondmethod s basedonthe applicationof Gauss's aw, and it is equallydifficultto use because t is not clear how one would constructa Gaussiansurface such that E is always totally tan-gential or totally normal at every point on that surface.ln the third method, the electric field is found fromE - -Vy after solving Poisson'sor Laplace's equa-tion for V, subject o the availableboundaryconditions;that is, V :0 at any point on the groundedconductingsurface and at infinity. Although such an approach sfeasible in principle, the solution is quite complicatedmathematically. Alternatively, the problem at hand canbe solved with great ease using image theory, whichstates hat cuttt ivenchargeconfigurationabovean inf-nite,perfectlyconducting lane s electricallyequivalentto thecombinationof thegivenchargeconfigurationandits image configuration, with the conducting plane re-moved.The image-methodequivalentof the chargeeabove a conducting plane is shown in the right-handsectionof Fig. 4-26. It consistsof the charge p itselfand an image charge -Q at a distance 2d from e,

CHAPTER4 ELECTROSTA

with nothing else betweenthem. The electric fieldto the two isolatedchargescan now be easily foundany point (x,y,z) by applyingCoulomb'smethod,demonstratedy Example -13.Thecombination ftwo chargeswill alwaysproducea potentialV = 0everypoint in the planewhere he conductinghad been. If the charge s in the presenceof moreone groundedplane, it is necessary o establishof the charge relative to each of the planes andto establishmages f eachof thosemagesagainsremainingplanes.The processs continueduntilcondition V : O is satisfiedat all pointson allgrounded lanes.The imagemethodappliesnotto point charges,but also to any distributionsofsuch as the line and volume distributions depictodFig. 4-77.

Erample-13 lmageMethodorGhargeabove onduclinglaneUse image heory to determine7 andE at obitrary point P(.r, ), z) in the regionz > 0 ducchargeQ in freespace t a distance aboveaconductingplane.

Solution: In Fig. 4-28, chargeQ is at (0,0, d)image -O is at (0,0, -d) in Cartesiancoordi


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HIGHLIGHTS

Eq. (4.19), he electric ield at P(x,y, z)

of

4neoL[x2+ y2 + (z - dyzltlzi r +yy+i (z+d)[ x 2 1 y z * ( z * d ) 2 ] 3 / 2

4.19 Use heresultof Example4-13to findnrface charge density p, on the surface of the con_plane.

ry=44/[2n(x2 + y2+ d21ttzt. (Seer$

n 6e trvo chargess givenbyORr -ORz\oI*-{)i r + iy +t(z- d)

OUESTIONSWhat s thefundamental remise f the mage

$ Givena certainchargedistribution,what are,inous approachesdescribed in this chapter for:uringthe electric field E at a given point in space?

CHAPTERIGHLIGHTS 'o Maxwell's equationsarethefundamentalenetsofelectrornagneticheory.o Understaticconditions,Maxwell's equations parate nto two uncoupledpairs,with onepair per_

taining to electrostatics ndthe other o magneto_statics.o Coulomb'saw providesanexplicitexpressionortheelectricfield due o a specifiedchargedistribu_tion.o Gauss'saw stateshat the total electric ield fluxthrougha closedsurfaces equal o thenetchargeenclosed y thesurface.o TheelectrostaticieldE at a point s relatedo theelectricpotential y at rhat point by E - _Vy,with V being eferencedo zeropotentialat infin_ity.o Becausemost metalshaveconductivities n theorderof 106 S/m), heyaretreatedn practiceasperfectconductors. y the same oken, nsulatorswith conductivities maller han l0-r0 (S/m)aretreatedasperfectdielectrics.o Boundary onditionsat the interface etweenwomaterialsspecifytherelationsbetweenhenormalandtangential omponents f D, E, andJ in oneof thematerials o the conesponding omponen$in the other.r The capacitance f a two-conductorbody andre_sistance f themediumbetweenhemcanbe com-puted rom knowledge f the electric ield in thatmedium.o Theelectrostatic nergydensitystoredn a dielec_tric medium s Lc," )eZz 1ltm3S.o Whena chargeconfiguration xistsabovean in_finite, perfectly conducting plane, the inducedfield E is the sameas that due to the configura_tion itselfand ts imagewith theconducting laneremoved.

point

; (o fA LI

4Teso

ar d


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108NoncontactensorsPrecisionpositioningb q"4ritilgt,qtg .qf:#mFconductordevice abrbation,as wll as tlp gpera-tion andcontrolofmanymechanicalyste'ms. on-contactcapacitive ensorsare U$edO$erF theposition fsiliconwafers uringhedeposifu, etch-ing,andcuttingprocesses,without oming n directcontactwithhewabrs.Theyarealsousedtosenseandcontrolobotarms nequipment anufacturingand oposition arddiscdrives, hotocopierollers,printing resses, nd othersimilar ystems.Basic rincipleThe concentric latecapacitorA1)consists f twometalplates, haring he sameplane,but electri-cally solatedromeachotherby an insulating a-terial.When onnectedo a voltage ource, hargesof opposite olaritywill ormon the twoplates, e-sultingnthecreation f electric-fieldinesbetweenthem.Thesameprinciple pplieso the adjacent-plates apacitorn (A2). n bothcases,he capaci-tance sdetermined y he shapes ndsizesof theconductive lementsand by the permittivity f thedielectricmedium ontaininghe electricield inesbetween hem.Otten, he capacitor urfaces cov-ered by a thin ilm of nonconductive aterial,hepurposeofwhich s to keep heplatesurfaces leanand dust ree.The ntroduction fan elternalobjectinto he proximity f the capacitor A3)will perturbtheelectricield ines,modifyinghecharge istribu-tionon theplates, s wellas modifyinghevalueofthecapacitances wouldbe measured y a capac-itancemeteror bridge ircuit.Hence, he capacitorbecomes proximityensor, nd ts sensitivity e-pends, n part,on howdifferenthe permittivityfthe object s from hat of the unperturbed ediumandon whethert is or is not madeof a conductivematerial.

TECHNOI-OCIYBRIEF: NONCONTACT S

Fingerprintmager.An interestingxtension f noncontactthe development f a fingerprint mageringof a two-dimensionalrrayof capacitivecells.constructedo recordan electricaltation f a fingerprint81 andB2).Eachis composed f an adjacent-platesapacitornected o a capacitancemeasurement ircuil


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Y BRIEF: NONCONIACT SENSORS

entiresurfaceof the imager s coveredby aayerof nonconductivexide.Whenhe ingerson heoxide urface,tperturbshe ield inesindividualensor ells o varying egrees,e-on the distancebetween he ridgesandof the inge/ssurfacerom hesensor ells.that he dimensions f an individualensorfi theorderof 65 pm on the side, he imager sof recording fingerprintmagat a reso-correspondingo 400dotsper inchor better.

ser

* Courtesy f Dr.M. Tartagni, niversity f Bologna,taly


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PROBLEMSSeclionsl-Z:Charge ndCurrent islribulions4.1* A cube 2 m on a side is located in the first octantin a Cartesian coordinate system, with one of its cornersat the origin. Find the total charge contained in the cubeif the chargedensity s given by p, : xy2e-22(mC/m3).4.2 Find the total charge contained in a cylindricalvolume defined by r < 2 m and 0 < z < 3 m ifpv: l0rz (mC/m3).4.3* Find the total charge contained in a cone de-fined by R < 2 m and 0 < e < r/4, given thatpu :20R2 cosz (mC/m3).4.4 If the line charge density is given by pr : I2y2(mC/m), find the total chargedistributed on the y-axisf r o m y : - 5 t o 1 l : 5 . s4.5* Find the total charge on a circular disk definedby r < a and :0 i f :(a) p, - pso ind (Clm2)(b) p. - pr6sin2Q (Ctmz)

O (c) p, : pso-r(C/m2)(d) p, : pso-'sin24 1C/m2;where ps6 s a constant.

4.6 If J : jLxz (Nm2;, find the current / flow-ing througha squarewith cornersat (0, 0,0), (2, 0, 0) ,(2,0,2) , and (0,0, 2) .4.7* ffJ: frZS1n(A/m2), find 1 through he surfaceR : 5 m .4.8 An electron beam shaped ike a circular cylinderof radius r0 carries a chargedensity given by

CHAPTER4wherep6 is a positiveconstantand the bam'scoincidentwith the e-axis.(a) Delermhe th totd chargeconAincd inof thebeam.(b) If the electronsaremoving n the*zdirectio

uniform speedu, determine hemagnituderectionof the eurrentcrossing he z-plane.

Section-3:Coulomb'saw4.9* A squarewith sidesof 2 m hasacharge fat each of its four corners. Determine the electrbat a point 5 m above the center of the square.4.10 Threepoint charges, achwith 4 - 3located at the corners of a triangle in the .r-ywith onecornerat the origin, anotherat (2 cm,and the third at (0,2 cm,0). Find the forcethe charge located at the origin.4.11* Chargel:4 pC is ocated t (1 cm,and chargeq2 is locatedat (0,0,4 cm). Whatq2 be so that E at (O,2 cm, 0) has no

4.12 A line of charge with uniformfi :4 (pClm) exists n air along the z-axisz :0 andz :5 cm.FindE at (0,10 m,0).4.13* Electric charge is distributed along ancated n the -r-y plane and definedby r :2O < 0 S n 14. lf p1- 5 (pClm), find E at (0, 0,then evaluate it at:(a) Theorigin.(b) z :5 cm( c ) : - 5 c m

":(#) (c/m3)*Answer(s) available n Appendix D.

(D Solution available n CD-ROM.


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PROBLEMS l 1 l

C

4'l{ A line of chargewith uniformdensityO extendsbcrween : -L/2 and z : L/2 along the z-axis.Apply Coulomb's aw to obtain an expressionor theic field at anypoint P(r, @,0)on the x-y plane.that your result reduceso the expressiongiven(4.33)as he engthZ is extendedo infinity.

of radiusa, but in thepresent ase,assumehe:acechargedensity to vary with r asps: psor2 (Clm2\

pr6 s a constant.'r Multiple chargesat different locations are said to- equilibrium if the force acting on anyoneof them srical n magnitudeand direction to the forceactingonrf the others.Supposewe have wo negativecharges,.rrcated t heorigin and carrying charge 9e, and helocatedon the positivex-axis at a distanced from

15' RepeatExample4-5 for the circular disk of

Figure 4-29:Kite-shapedrrangmentf linechargesorProblem .17.

your result to the special case where d is infinite andcompare t with E4. @.25).Seclion -4:Gauss's aw

4,20 Given the electric flux densityD : *2(x * y) + y(3x - 2y) (C/m2)

determine(a) pu by applying Eq. @.26).(b) The total charge Q enclosed n a cube 2 m on aside, located in the first octant with three of itssides coincident with the x-, 1,-, and z-axes andone of its cornersat the origin.(c) The total chargeQ inthe cube,obtainedby apply-

ing Eq. (4.29).

:-rrt oneandcarryingcharge 36e, Determinehecon. polarity and magnitudeof a third chargewhose:mentwouldbring theentire system nto equilibrium.;-' Three nfinite lines of charge,all parallelto the. are ocatedat thethreecornersof the kite-shapedgement hown n Fig. 4-29.lt the two right trian-ue symmetrical ndof equalcorrespondingides,hdratheelectric ield is zeroat theorigin.Three infinite lines of charge, p4 = 5 (nC/m),= -5 (nC/m),andp4 = 5 (nC/m),areall parallel:-axis. fthey passhrough he respective oints-:,. (0,0), and (0,D) in the .r-y plane, ind the: field at (a,0,0). Evaluateyour result for a :: : d b : 1 c m ..{ horizontal strip lying in the .r-y plane is of; in the y-direction and infinitely long in the:ron. f the strip is in air and has a uniform chargeps,useCoulomb'saw to obtainanexplicitfor the electric ield at a point P locatedat

h above the centerline of the strip. Extend


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4.21* Repeat Problem 4.20 for D : f;xy2zi (Clm2).4.22 Charge Q1 is uniformly disributed over a rhinspherical shell of radius a, and charge Q2 is uniformlydisnibuted over a second spherical shell of radius D,with D > c. Apply Gauss's aw to find E in the regionsR < a , a < R < D , a n d > b .4.23* The electric flux density inside a dielectricsphereof radius a centeredat the origin is given by

D: fi.psR (Clm2)where po is a constant.Find the total charse nside thesphere.4.24 In a certain region of space, he chargedensityis given in cylindrical coordinatesby the function:

P, :20re-' (C/m3)Apply Gauss'saw to find D.4.25* An infinitely long cylindrical shell extendingbetween r : I m and r : 3 m contains a uniformchargedensity puo.Apply Gauss's aw to find D in allregions.4.26 If the chargedensity increases inearly with dis-tance from the origin such that pv : 0 at the originandpu: l0 C/m3 at R : 2 m, find the correspondingvariation of D.Seclion -5:Electric otential4.27 A square n the -r-y plane in free spacehas apoint chargeof *Q at corner (a/2,a/2), the same atcorner(a/2, -a/2), and a point chargeof -Q at eachof the other two corners.(a) Find the electric potential at any point P along the-r-axis.(b ) Eva luateVatx :a /2 .

CHAPTER44.28 Thecirculardisk of radiusc shown n Fighasuniform chargedensitypsacrossts surface.(a) Obtainan exprcssion or the electricat a pointP(0,0,2) on thez-axis.(b) Useyour result to find E and thenevaluatc

z = h. Compareyour final expressionwittr (whichwasobtained n thebasisof Coulomb'r4.29* A circular ing of charge f radiusa lies r.r-y planeand s centeredat the origin. Assumethe ring is in air and carriesa uniform densityp1.(a) Show hat the electricalpotentialat (0, 0, z) isby V - p1a/[2es(a2 z2)t/2].(b) Find the correspondingelectric field E.

4.30 Show that the electric potential differenceV12tweentwo points in air at radial distances11&nd12an nfinite line of chargewith densityp1alongthezis Vrz- @t/2nedln(r2/r).4.31* Find the electric potential V at a location atance b from the origin in the x-y plane due to achargewith charge density p1 andof length ,. Thccharge is coincident with the z-axis and extendsz : - l / 2 t o z : l / 2 .4.32 For the electric dipole shown in Fig.d : I cmand El : 2 (mV/m)at R - I mandgF i n d E a t R : 2 m a n d : 9 0 .4.33 For each of the distributions of the electrktentialV shown n Fig. 4-30,sketch hedistribution fE (in all cases,heverticalaxis s inand he horizontal xis s in meters).4.34 Given heelectric ield

s

u:R# tv/m)find the electricpotentialof point A withpointB whereA is at *2 m andB at -4 m, bothz-axis.


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ttOBLEMS 1 1 3

rt

a

Fsure 4-30:Electricpotentialdistributions f Problem] ! t - -

An infinitely long line of charge with uniform': pt :6 (nCVm)ies in the ;-y planeparallel or-erisat x : 2 m. Find thepotentialVanat pointn- 0.4 m) in Cartesian oordinates ith respecto8,0,0, 0) byapplyinghe esult f Problem .30.

4.36 The r-y planecontainsa uniform sheetof chargewith pr, : 0.2 (nC/m2).A secondsheetwith ps2 -9.,(nClm) occupies he plane e : 6 m. Fnd VAu Vac, illdVrc for A(0,0,6 m), B(0,0,0), andC(0, -Z m,2 m).Sectionl-7:Conductorc4.37* A cylindrical bar of silicon hasa radiusof 2 mmandalengthof5 cm.If avoltageof5 V is appliedbetweentheendsof thebar Md p": 0.13 1m2lV.s), n : 0.05(m2lV.s),N" : 1.5 x 1016 lectrons/m3, nd Nn : N",find the following:(a) The conductivity of silicon.(b) The current.l flowing in the bar.(c) The drift velocitiesu. andu6.(d) The resistanceof the bar.(e) The powerdissipated n the bar.

4.38 Repeat Problem 4.37 for a bar of germaniumwith i"r, : 0.4 (m2lV.s), trr1 : 0.2 (m2N.s), andN" : Nn :2.4 x 101e lectronsor holes/m3.4.39 A 100-m-longconductorofuniformcross-sectionhasa voltagedropof 2 V between ts ends. f the densityofthecurrent lowing through it is 7 x ld (A/m2), identifythe material of the conductor.4.40 A coaxial resistorof length / consistsof two con-centric cylinders. The inner cylinder has radius a andis made of a material with conductivity o1, and theoutercylinder,extendingbetween : aandr = S,is made of a material with conductivity o2. lf thetwo ends of the resistor are capped with conductingplates, show that the resistancebetween the two endsis R : I [n(opz * o2(b2 a\)].4.41" Apply theresultof Problem4.40 to find the resis-tanceof a l0-cm-long hollow cylinder (Fig.4-31) madeofcarbon with o : 3 x 104 S/m).4.42 A2 x 10-3-mm-thicksquaresheetof aluminumhas 0 cm x 10cm faces.Find the followine:


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Figure4-31:Cross-section fhollow cylinder ofproblem4.41.

(a) Theresistance etweenoppositeedgeson a squareface.(b) The resistance etween he two square aces.(SeeAppendixB for the electrical constantsof materi_

als.)Seclion-9:Boundaryonditions4.43* With reference to Fig. 4-lg, find E1 ifEz : i3 - j,z+i{(V/m), r : 2eo, 2 : 18e0,and the boundary has a surface charge densityps = 7.08 x l0-rr (C/mz). What angle does E2 maklwith thez-axis?4.44 An infinitely long dielectriccylinder with e1, : {anddescribed y r = l0 cm is surroundedby a materialwith e2, 8. If Er - ?r2 sinQ+ 6Zr2cos@ i3 (V/m)in thecylinder region, find E2 and D2 in the surround_ing region.Assume hat no free chargesexist along thecylinder'sboundary.4.45* A 2-cmdielectricspherewith e11 3 is embed_dedinamediumwithe2, 9.I fEz : fr3cosg-63sing(V/m) in the surrounding egion, find E1 and D1 in thesphere.4.46 If E : ftSO ylm) ar the surfaceof a 5-cm con-ductingspherecenteredat the origin, what is the totalchargep on thesphere'ssurface?

CHAPTER4

4.47* Figure4-32showshree lanar ielectricequal hickness ut with differentdielectricE6 in airmakesan angleof 45. with respect o thefind the angleof E in eachof the other avers.Seclions-10 nd -11: apacitancendElectrical4.48 Determine the force of attraction in acapacitorwith A : l0 cm2,d : I cm, ande. - 4voltage crosst is 50 V.4.49* Dielectric breakdownoccurs n a materialever the magnitudeof the field E exceeds he distrengthanywhere n that material. n the coaxialitor of Example4-I2,(a) At whatvalue f r is lEl maximum?(b) What is rhe breakdown oltage f cb :2 cm, and hedielectricmaterialse r : 6 ?4.50 AnelectronwithchargeQ" : -l.6xl1-remass ne : 9.1 x 10-31 g is injectedat apointto the negativelychargedplate n the regionplatesof an air-filled parallel-platecapacitorwithration of I cm andrectangularplateseach 0 cm2

Figure4-32:Dielectric labsn problem4.47.


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PROBLEMS 1 1 5

: -' In a dielectricmediumwith e, : 4, theelectric: r givenbyE= i(r2 *22) *gr, - 2(y+ z) (v/m)

the electrostaticenergy stored in the regionn 5 r < I f i i , 0 < y < 2 m , a n d 0 S z < 3 m .

V o = l 0 VFigure4-33:Electron etweenharged lates f problem1.50.

:. 4-33).If thevoltage crosshecapacitors l0 V: the ollowing:.r The force acting on the electron.n The accelerationofthe electron.,: Thetime it takes he electron o reach he positivelychargedplate, assuming hat it starts rom rest.

Figure4-34(a)depictsa capacitor onsisting f:'":allel, onducting lates eparatedyadistance .;:ce betweenhe platescontainswo adjacent i_- . onewith permittivity e1andsurfaceareaA I and: u'ithe2 andA2.The objectiveof this problem s to:rt thecapacitanceC of theconfigurationshown n- -'-tla) is equivalent o two capacitancesn parallel,rstrated n Fig.4-34(b), withC : C r * C z

where^ e tA t. r = T^ zAzvz: T

Tothisend,proceedas ollows:(a) Findtheelectric ieldsE1andE2 n the wodielectriclayers.(b) Calculateheenergy toredn eachsection ndusetheresult o calculateCr andCz.(c) Use he otalenergy toredn thecapacitorto btainanexpressionor C. Show hat(4.125)s indeedavalidresult.

TdI

(b)Figure4-34: a)Capacitor ithparallel ielectric ection,and b) equivalent ircuit.

(4.126)(4.r27)

(4.rzs)


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CHAPTER4O (a) Conductingplates le on top andbottomthe cctangular tructuren Fig.4-35(a).(b) Conducting latesane il front ad back accsstrrrcturen Fig. 4-35(a).

(c) Conductingplatesare on top and bottomthecylindricalstructuren Fig.4-35(b).

4.54 The capacitor hown n Fig.4-36 consists fparallel dielectric layers. Use energyshow that the equivalent capacitance of the overallpacitor,C, is equal o the series ombination fpacitancesf the ndividual ayers,C1andC2,r - CrCzC r * C z

whereA A

C t : t - T , C z : z ;Clt 42(a) Let Vr and Vz be the electric potentials

upper and lower dielectrics, respectively.

4.53* Use the result of Problem 4.52 to determinethecapacitance or each of the following configurations:

;.$4--3 cm-.--.+l

rt=2mm12=4mm

(b) 13= 8mm

Icml e t = 8 e O ; 2 = 4 e , i e 3 = 2 e O

Figure 4-35: Dielectric sections or Problems 4.53 and4.5s.


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'R.OBLEMS Il7thecorrespondinglectric ieldsE1andE2?By ap_plyingtheappropriate oundaryconditionatthe n_terfacebetweenhetwo dielectrics,obtainexplicitexpressionsor E1 andE2 n termsof e1,e2,V, andthe ndicateddimensions f thecapacitor.

trr Calculatehe energystoredn eachof thedielectriclayers nd henuse he sum o obtainanexpressionfor C.

l Show hatC is givenby Eq.(4.128).

Ir, Iz-Figure4-38:Currents bove conductingplaneproblem4.57\.Aj-i Use the expressionsgiven in problem 4.54 to=rmine the capacitance for the configurations ins .1-35(a) hen heconducting latesareplacedonight and eft facesof the structure.

on4-12:mage ethod-i,r With reference o Fig. 4-37, chargee is locatedr listance d above a groundedhalf-plane located in:-\' planeandat a distanced from anothergrounded-,;lane n thex-z plane.Use the image methodton Ertablish the magnitudes,polarities, and locationsrithe imagesof chargee with respect o eachof thes'o groundplanes(as f each s infinite in extent).L' Frnd the electric potential and electric field at anrrbitrarypoint P(0, y, z) .

(S C.SZ Conducting iresabove conducting lanecarrycurrents11and12 n thedirections hownn Fig. 4_3g.Keepingn mind hat hedirectionof a currents definedin termsof themovement f positivecharges, hatarethedirections f the magecurrents onespondingo 11and12?4.58 Use he magemethodo findthecapacitanceerunit lengthof an nfinitely long conducting ylindeiofradius situated tadistance fromaparallel onductingplane, sshownn Fig.4-39.4.594.64 AdditionalSolvedhoblems - completesolutions n(0.

Figure 4-39: Conducting ylinderabovea conductinsplane Problem .58).

Electr Magnetic for Eng 4_2

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