92 CHAPTER 4 ELECTROSTATI

Figure 4-17: A dielectric medium polarized by 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 theelectron cloud [Fig. a-16(c)]. Each such dipole setsup a small electric field, pointing from the positivelycharged nucleus to the center of the equally but neg-atively charged electron cloud. This induced electricfield, called a polarizntionfield, is weaker than and op-posite in direction to Esx1. Consequently, the 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-ample 4-7. Within the dielectric material, the dipolesalign themselves in a linear arrangement, as shown inFig. 4-17. Along the upper and lower edges of the ma-terial, the dipole arrangement exhibits a positive surfacecharge density on the upper surface and a negative den-sity on the lower surface.

The relatively simple picture described in Figs. 4-16and 4-17 pertains to nonpolnr materials in which themolecules do not have permanent dipole moments.Nonpolar molecules become polarized only when anextemal electric field is applied, and when the field is

terminated, the molecules return to their originalpolarized state. In some materials, such as watcqmolccrrlar stnrcftrc is such that thc molcculcsbuilt-in permanent dipole moments that areoriented in the absence of an applied electric field-terials composed of permanent dipoles are callodmaterials. Owing to their random orientations,poles of polar materials produce no net dipolemacroscopically (at the macroscopic scale, eachthe material represents a small volume containingsands of molecules). Under the influence of anfield, the permanent dipoles tend to alignto some extent along the direction of the electicin an arrangement somewhat similar to thatFig.4-17 for nonpolar materials.

Whereas D and E are related by e6 in freepresence of 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 is said to be linear If tlrcnitude of the induced polarization field is directlyportional to the magnitude of E, and it is saidisotropic if the polarization field and E are in thedirection. In some crystals, the periodic structurematerial allows more polarization to take placecertain directions, such as the crystal axes, thanothers. In such anisotropic dielectrics, E and Dhave different directions. A medium is said tomogeneous if its constitutive parameters (e, pa,are constant throughout the medium. Our presentment will be limited to media that are linear, iand homogeneous. For such media theis directly proportional to E and is expressedrelationship

P: aoXeE,

r.8 DIELECTRICS 93

iginalwater,cs

field.dled,ns, thelech poiciningan

:ctrict shown

) spacgr d irial to

t ,al. Theheld E

if theirectlysaid to

in thecture dplacethan

and Dd t o b e, tt,resentlr. 1

lzanonssed by

Petoleum oil

Polystyrene

Glass

QuartzBakelite

Mica

2.12.6

4.5-103.8-5

55.44

aJ

1220

25403020200

Table 4-2: Relative permittivity (dielectric constant) and dielectric strength of common materials.

I : er6o and es : 8.854 x 10-12 F/m.

-E-te xe is called the electric susceptibility of the ma-:rl. Inserting Eq. (4.8a) into Eq. (4.83), we have

D: eoE * e6X"E

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

--',-h defines the permittivity e of the material as

e : 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, are

in Table L2 for a few common materials. androger list is given in Appendix B. In free space= l, and for most conductors sr : l. The dielectric:rant of air is approximately 1.0006 at sea level,rt decreases toward unity with increasing altitude.

::pt in some special circumstances, such as when:rlating electromagnetic wave refraction (bending)

the atmosphere over long distances, air is:-',d the same as free space.ae dielectric polarization model presented thus far:laced no restriction on the upper end of the strength

te applied electric field E. In reality, if E exceeds a

certain critical value, known as the dielectric strength ofthe material, it will free the electrons completely fromthe molecules and cause them to accelerate through thematerial in the form of a conduction current. When thishappens, sparking can occur, and the dielectric materialcan sustain permanent damage due to electron collisionwith the molecular structure. This abrupt change in be-havior is called a dielectric breakdown. The dielectricstrength 86, is the highest magnitude of E that the ma-terial can sustain without breakdown. Dielectric break-down can occur in gas, liquid, and solid dielectrics.The associated field strength depends on the materialcomposition, as well as other factors such as temper-ature and humidity. The dielectric strength for air is 3(MV/m); for glass, it is 25 to 40 (MV/m); and for mica,it is 200 (MV/m) [see Table 4-2].

A charged thundercloud with an electric potential V,relative to the ground, induces an 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 the dielectric strength of air, ionization occursand discharge (lightning) follows. The breakdownvoltage V61 of a parallel-plate capacitor is discussed inExample 4-11.

94 CHAPTER 4 ELECTROSTAN

Figure 4-18: Interface between two dielectric media.

REVIEW OUESTI(}NS

Q4.20 What is a polar material? A nonpolar material?

Q4.21 Do D and E always point in the same direction?If not, when do they not?

Q4.22 What happens when dielecfic breakdown oc-curs?

4-g Electric Boundary GonditionsAn electric field is said to be spatially continuous if itdoes not exhibit abrupt changes in 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 between them if surface charge existsalong that boundary. Boundary conditions specify howthe tangential and normal components of the field inone medium are related to the components of the fieldacross the boundary in another medium. We will derivea general set of boundary conditions, applicable at theinterface between any two dissimilar media, be theytwo different dielectrics or a conductor and a dielectric.

Also, any of the dielectrics may be free space. Erthough these boundary conditions will be derived 1electrostatic 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 general case, Iboundary may have a surface charge density p..

To derive the boundary conditions for the tangeilcomponents of E and D, we begin by constructing Iclosed rectangular loop abcda shown in Fig. 4-18, rthen we apply the conservative property of the eloclfield given by Eq. (4.40), which states that the lincitegral of the electrostatic field around a closed pdalways equal to zero. By letting A4* 0, the coltions to the line integral by segmenis bc and dazero. Hence.

f *. or: l"u Ez. dt * f"o E1 .dt : o,

where E1 and &z are the electric fields in media Irespectively. In terms of the tangential and normCrections shown in Fig. 4-18,

Er : Er t * Ern , (

Ez:Ezt * Ezn. (

y9 ELECTRIC BOUNDARY CONDHONS 95

rce.erivedillvshowst ) ' e r. case,

Ps .

rctrng+ 1 8 ,E C

re lined pat

do go

r lormal

lJttr segment ab, E7 and dl have the same direction,lul over segment cd, Ey and dl are in opposite direc-u. Conscquently, Eq. (4.87) gives

Ey Al - E11A/ = 0, (4.8e)

Er: Ezt ff/m). (4.90)

t'rordingly, the tangential component of the electricr'r.i is continuous across the boundary between any tw,o-, -;:,i. Since Dt : e1811 and Dz, : t2E2t, the bound-r. condition on the tangential component of the electricftu densitv is

Dt

a 1

Dzt

€2(4.e1)

\ext we apply Gauss's law, as expressed by\. G.29), to determine the boundary conditions on therurmal components of 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 in 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 to have free orn:rmd volume charge densities, the only charge remain-n,_: in the collapsed cylinder is that distributed on theI:undary. Thus, Q : p, As, and

where ff1 and ff2 are the outward normal unit vectors ofthe bottom and top surfaces, respectively. It is importantto rcmember ttnt the normal unit vector at the surface ofany mcdium is always defned to be in the outward direc_tionawayfromthatmediuru. Since fir : -ff2, W.9.92)simplifies to

ff2.@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 normal component of D changes abruptlt,at ctcharged boundary befween two dffirent media, and theatnount of change is equal to the surface charge density.The corresponding boundary condition for E is

€rEn- €zEzn - ps. (4.95)

In summary, (1) the conservative property of E,

V x E-0 <==+

led to the result that E has a continuous tangential com-ponent across a boundary, and (2) the divergence prop-erty of D,

V.D : pv <+ $ r .0 " : g , g97 )J S

led to the result that the normal component of D changesby p, across the boundary. A summary of the conditionsat the boundary between different types of media isgiven in Thble 4-3.

f l0 E.dl : 0, (4.96)J C

( 4

(4

f r fQ D.as : I Dr . f f2 ds + I D2 . f f1 dsJs Jtop ./bouo.

: Ps A,s, (4.92)

CHAPTER 4

Table 4-3: Boundary conditions for the clectric fields.

0 '� Application of Boundary Conditions

The x-y plane is a charge-free boundary separatingtwo dielectric media with permittivities e1 and 12, asshown in Fig. 4-19. lf the electric field in medium I isEr : iEr, * I Ery * iEp, find (a) the electric field E2in medium 2 and (b) the angles il and 02.

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

r-y plane

Figurt 4-19: Application of boundary conditions at theinterface between two dielectric media (Example 4-10).

is to find the components of E2 in terms of thecomponents of E1. The normal to the boundaryHence, the x and y components of the fields aretial to the boundary and the z components arethe boundary. At a charge-free interface, thecomponents of E and the normal components ofcontinuous. 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 componenrs oi ", and l- l l l - _ _ _ - - l - t - - - ] - - - - ] -En: tlEi, a Ef, and E21: J EL + E y.Tllr;91 and 02 are then given by

^ E t ,tan Ar :' E u

tanl.t: E"

:- Ez,

r--7---------;_\f Ef, + Eiy

E u 'r.-:-_^

\\f E;r + E;y

Ezz$,+ no(e1/e)Ep

Tangential E

Tirngential D

Normal E

Normal D

En= Ex

D*/et: Dzt/ez

ff.(erEr -tzBz)* p"

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

E*= Ex

Dt/er: Dzr/ez

stEln - e2B2n- ps

Dy1 - D21t: 4

I thendaryare ta: non

t : 0

r : 0

l s o f D

(

d F o

lhe

H.ECTRIC BOUNDARY CONDMONS

6e two angles are related by

tan02 (4.ee)tan01

4.16 With reference to Fig. 4-19,find E1 if8z : 8eo.- n - i3 + i3 (V/m), dr = 2€0, and

the boundary to be charge free.

Nrrrrrrrrrrrrrrrr�i. Er : i2 * y3 + 212 (Ylm). (See G;

ffiOSE 4.17 Repeat Exercise 4.16 for a boundaryuddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddsurface charge density ps:3.54 x l0-ll (C/m2).

.r"ns- Er : *2 * i3 + iI4 (Ylm). (See 0;

+9.1 Dielectric-ConductorBoundary

f,:'nsider the case when medium 1 in Fig. 4-18 is a die-s-tric and medium 2 is a perfect conductor. In a perfect:L'nductor, E = D : 0 everywhere in the conductor.Fience, Ez = Dz: 0, which requires the tangential and

normal components of 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)

L ' L

- l Z

Ps= stEo

Conductingsla$---..+.

tllre &20: When a conducting slab is placed in an external electric field &, charges that accumulate on the conductor

These two boundary conditions can be combined into

Dr : erBr : ffp, (at conductor surface), (4.101)

where ff is a unit vector directed normally outward fromthe conducting surface. This means that the electricfieldlines point directly awayftom the conductor surface whenp, is positive and directly toward the conductor surfacewhen p, is negative.

Figure 4-20 shows an infinitely long conducting slabplaced in a uniform electric field &. The medium aboveand below the slab has a permittivity e1. Because E6points away from the upper surface, it induces a pos-itive charge density ps : rllllol on the upper surface

Figure 4-2lz Metal sphere placed in an external electricfield Fa.

of the slab. On the bottom surface, E6 points towardthe surface, and therefore the induced charge density is-pr. The presence of these surface charges induces anelectric field E1 in the conductor, resulting in a total fieldE : Fo * Ei. To satisfy the condition that E must beeverywhere zero in the conductor, Ei must equal -S.

If we place a metallic sphere in an electrostatic field,as shown in Fig. 4-2I,negatwe charges will accumulateon the lower hemisphere and positive charges will ac-cumulate on the upper hemisphere. The presence of thesphere causes the field lines to bend to satisfy the con-dition given by Eq. (a.101); that is, E is alwav-s normalto the surface at the conductor boundary.

4-9.2 Conductor-Gonductor Boundarylffe 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 1has permittivity e1 and conductivity o1, medium 2 hass2 and 02, nn.d the interface between them has a surfacecharge density pr. For the electric fields, Eqs. (4.90) and(4.95) give

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

CHAPTER 4 ELECTROST

Since we are dealing with conducting media, theric fields give rise to current densities Jr and Jz,Jr being proportional to El and Jz beingto E2. From J - oE, we have

Jrn Jzn, t l

01 62

The tangential components .Irt and ,I21 representflowing in the two media in i direction parallel bboundary, and hence no transfer of charge is ibetween them. This is not the case for the normalponents. lf Jh +,/2n, then a different amount ofarrives at the boundary than leaves it. Hence, p,remain constant with time. which violates thetion of electrostatics requiring all fields andremain constant. Consequently, the nonnalof J has to be continuous across the boundarytwo dffirent media under electrostatic conditions.setting Jn: Jzn in Eq.(4.103), we have

Jn Jz,

o1 o2

ng media-

ia, thend Jz,

€ntrallel toIS

rormalnt of

e. Psi the

I chargest . (

(4.rM)

(4.1

O CAPACITANCE

0uEsTr0ltsWhat ar€ the boundary conditions for the elec-

6cld at a conductor-dielectric boundary?

Under electrostatic conditions, we requirebetween two conductors.= Jzt at the boundary

r,

10 Capacitance

separated by an insulating (dielectric) medium,1 r*'o conducting bodies, regardless of their shapes

sizes, form a capacitor. If a d-c voltage sourcecmnected to the conductors, as shown in Fie. 4-23n*'o arbitrary conductors, charge of equal and oppo-polarity is transferred to the conductors' surfaces.surface of the conductor connected to the posi-

side of the source will accumulate charge *Q, and-Q will accumulate on the surface of the other

From our discussion in Section 4-7, when a

conductor has excess charge, it distributes the chnrge onits surface in such a manner as to maintain a zero elec-tric field everywhere within the conductor This ensur€sthat a conductor is an equipotential body, meaning thatthe electric potential is the same at every point in theconductor. Capacitance of a two-conductor capacitor isdefined as

C : g (CA/ or F), (4.105)v

where V is the potential (voltage) difference betweenthe conductor with charge +Q aurrd the conductor withcharge -Q. Capacitance is measured in farads (F),which is equivalent to coulombs per volt (CA/).

The presence of free charges on th} conductors'surfaces gives rise to an electric field E, as shownin Fig. 4-23; the field lines originate on the positivecharges and terminate on the negative charges, and sincethe tangential component of E is always equal to zeroat a conductor's surface, E is always perpendicular tothe conducting surfaces. The normal component of E atany point on the surface of either conductor is given by

E n : f f ' E = (at conductor surface), (4.106)

where p, is the surface charge density at that point, ff isthe outward normal unit vector at the same location. ande is the permiaivity of the dielectric medium separatingthe conductors. The charge Q is equal to the integralof p, over surface S [Fig. 4-23]:

(4.r07)

where use was made of Eq. (a.106). The voltage V isrelated to E by Eq. (4.39):

Ps

t

n: Irp"ds -*

Ir'o r-ds:

lrel'as,

v : vtz : - I : '

E.dt, (4.108)

Surface S

Frgure 4-23: A d-c voltage source connected to a capac-nx composed of two conducting bodies.

100 CHAPTER 4 ELECTROSTAM

where / is the integration path from conductor 2 to con-ductor 1. To avoid making sign errors when applyingEq. (4.109), it is important to remember that surface ,Sis the *0 surface and P1 is on S. Because E appears inboth the numerator and denominator of Eq. (4.109), thevalue of C obtainedforany specific capacitor configura-tion is always independent of E. In fact, C depends onlyon the capacitor geometry (sizes, shapes and relativepositions of the two conductors) and the permittivity ofthe insulating material.

If the material between the conductors is 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 resistance R.The general expression for R for a resistor of arbitraryshape is given by Eq. (4.71):

I en.asC : "s

^ (F), (4.109)-

J,n' at

- [ n . a tR: v t , ( ( .2 ) .

J ,oE 'ds

where points P1 and Pz are any two points on conduc-tors I and 2, respectively. Substituting Eqs. (4.107) and(4.108) into Eq. (4.105) gives

Crpacitance and Ereakdown Voltaet IParallel-Plate Gapacilor

Obtain an expression for the capacitance C of a pclel-plate capacitor comprised of rwo parallel plates aof surface are,a A and separated by a distance d.lcapacitor is filled with a dielectric material with pmittivity e. Also, determine the breakdown voltapd : 1 cm and the dielectric material is quartz.

Solution: In Fig. 4-24, we place the lower plate of Icapacitor in the .r-y plane and the upper plate in Iplane z : d. Because of the applied voltage ditence V, charge *O accumulates uniformly on thclplate and -Q accumulates uniformly on the lowerflIn the dielectric medium betweerlthe plates, the ch{induce a uniform electric field in the -2-direction (lpositive to negative charges). In addition, someingfield lines will exist near the edges, but theirmay be ignored if the dimensions of the plates arelarger than the separation d between them,that case the bulk of the electric field lines will exithe medium between the plates. The chargethe upper plate is p' : QlA.Hence,

g : - i E ,

and from Eq. (4.106), the magnitude of E at tlrcductor-dielectric boundary is E : ptle : QltA-Eq. (4.108), the voltage difference is

and the capacitance is

where use was made of the relation B : QleA.From V - Ed, as given by Eq. (4.112), V

when E : Eds, the dielectric strength of the

v - - lr '

" dr: - Ioo ,-rr, .idz : Ed,

For a medium with uniform o and e, the product ofEqs. (4.109) and (4.110) gives

R C : I6

(4 .111 )

(4.1 l0)

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

O O e A- E d - d

(4 .113)

r-10 CAPACITANCE l0r

Yottagc

I of aplatesnce d.rl withI

1lZ.

plate ofplate inItage di

r on thclo$'er

, the

ectlon

sometheirEs are. becu'il l exie den

I at the

Q t A .

Ed. (4-

Q ' e A -? \ . r l -

Dielectric e

Conducting plate

Figure 4-2A: A d-c voltage source connected to a parallel-plate capacitor (Example 4-l l).

to Table L2, E6s = 30 (MV/m) for quartz.

iI

? i lII

the

, the breakdown voltage is

l'- : Easd: 30 x 106 x 10-2 = 3 x 105 V. I

Gapacitance ol Goarial Line

'lbtain an expression for the capacitance of the coax-me shown inFig.4-25.

Solution: For a given voltage V across the capacitor(Fig.4-25), charge *Q and -Q will accumulate on thesurfaces of the outer and inner conductors, respectively.We assume that these charges are uniformly distributedalong the length of the conductors with line charge den-sity p; : Q/ I on the outer conductor and - pt onthe in-ner conductor. Ignoring fringing fields near the ends ofthe coaxial line, we can construct a cylindrical Gaussiansurface in the dielectric, around the inner conductor.with radius r such that c < r < b. The inner conductoris a line charge similar to that of Example 4-6, except

-Pt

Outer conductor

Figure 4-25: Coaxial capacitor filled with insulating material of perminivity e @xample 4-12).

Capacitive SensorsTo sense is to respond to a stimulus. (SeeResistiw Sercors.) A iruf::ftflrehion aga sensor if ttle stinulus 4{d ,&WlgSgeometry - usually the, $Bgirigl:,$G$its€fr,lt$conductive elementrs - orffidigt fic properiles'of the insulating material **ated between them.Capacitive sensors aie uSd'' in" a mullitude ofapplications. A few examples follow.'' '

Fluid Gauge

The two metal electrodes in (A), usually rods orplates, form a capacitor whose capacitance is di-rectly proportionalto the permittivity of the materialbetween them. lf the fluid section is of height H y and

TECHNOLOGY BRIEF: CAPACITTVE

the height of the empty space above it is (I/ - flthen the overallcapacitance is equivalent to trvopacitors in parallel, or

i , , : _ ^ ( w H t ) , ^ ( H - H iC z = C f * C o : e f = J !

* t . T

uhere .w is the elec{rode plate width, d isspaoing between electrodes, and ey and s4 arsperrnittiWles of the fluid and air, respectively.ranging the expression as a linear equation

Cz: kHf a Co, l

where the constarit ft : (e1 - e)w/d and Cs iscapacitance of the tank when totally empty.the linear equation, the fluid height can bemined by measuring C2, which can be realizedusing a bridge circuit (B). The output voltage I{proportionalto the deviation between C1 and C2-setting Cr : Co (a fixed capacitor) and bying the tank electrodes to the bridge circuit toC2, Vo." becomes proportionalto the fluid heigtrt

Y BRIEF: CAPACITTVE SENSORS 103

L aretly. Rn vid

H _ HD ttro

tr) .

d i s

Co is dity Sensortybealizedge lLnd cz.

metal electrodes shaped in an interdigi-pattern (to enhance the ratio A/d) are Iabri-on a silicon substrate (C). The spacing be-digits is typically on the order of 0.2 pm. The

of the materialseparating the electrodes

varies with the relative humidity of the surroundingenvironment Hence, the capacitor becomes a hu-midity sensor.

Pressure SensorA flexible metal diaphragm separates an oil-filledchamber with reference pressure p6 from a secondchamber exposed to the gas or fluid whose pres-sure P is to be measured by the sensor (D1). Themembrane is sandwiched, but electrically istfiated,between two conductive parallel surfaces, formingtwo capacitors in series (D2). When p > ps, themembrane bends in the direction of the lower plate(D3). Consequently, d1 increases and d2 decreasesand, in turn, C1 decreases and C2 increases. Theconverse happens when P < po. With the use of acapacitance bridge circuit, such as the one in (B),the sensor can be calibrated to measure the pres-sure P with good precision.

it to fiqheight

104

that ttre line charge of the inner conductor is negative.With a minus sign added to the expression for E givenby Eq.(4.33), we have

(4.r14\

CHAPTER4 ELECTROST

4-11 ElectrostaticPotential Energyrilhen a s{xrrue is conncctcd to a caryitm, itenergy in charging up the caprcitor. If thcplates are made of a good conductor withzero resistance and if thc dielectric separating thcconductors has negligible conductivity, then no realrent can flow through the dielcctric, and no ohmicoccur anywhere in the capacitor. Where then doescharging-up energy go? The energy ends upstored in the dielectric medium in the fqm ofstntic potential energy. The amount of stored energ:tis related to Q, C, and V.

Under the influence of the electric field in thelectric medium between the two conductors.accumulates on one of the conductors, and an equalopposite charge accumulates on the other conductc,effect, charge q has been transfened from one ofconductors to the other. The voltage u across theitor is related to q by

From the basic definition of the electric potential V,amount of work dW" required to transfer anincremental amount of charge dg is

d W s : u d q : t O n .

n ^ P t ^ OE : - f - : - | . -

Zner - Zrerl

The potential difference V between the outer and innerconductors is

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

: * " ( : )

The capacitance C is then given by

C _ ov

2nelln(b/a)

(4 .116)

The capacitance per unit length of the coaxial

2treln(b/a)

(F/m). I

REVIEW OUESTIONS

Q4.25 How is the capacitance of a two-conductorstructure related to the resistance of the insulatins ma-terial between the conductors?

Q4.26 What are fringing fields and when may they beignored?

(4.r

If we start with an uncharged capacitor andup from zero charge until a final charge O hasreached, then the total amount of work performed ir

Using C = Q/V, where V is the final voltage,also be written as

wu: |cvz (J).

For the parallel-plate capacitor discussed inple 4-11, its capacitance is given by Eq. (4.113)

(4 .115)

line is

(4.rr7)

(4.rq

U : _C

*": Ion t on ::f (D (4.

wc

: IMAGE METHOD 105

i \t l

t ,

I t? _ .

4Charge Q above grounded plane

F;ue 1'26zBy image theory a charge Q above a grounded perfectly conducting plane is equivalent ro e andits image - er"c de ground plane removed.Efg'

tbefiargr - tA/d, where A is the surface area of each of itsqud and d is the separation between them. Also, thelrctfr- sc V across the capacitor is related to the magni-c o f --nf the electric field, E, in the dielectric bv V - Ed.c 6ese two expressions in Eq. (4.121) gives

REVIEW OUESTIONS

Q4.27 To bring a charge 4 from infinity to a givenpoint in space, a certain amount of work llz is expended.Where does the energy corresponding to $r go?

EXERCISE 4.18 The radii of the inner and outer conduc-tors of a coaxial cable are 2 cm and 5 cm, respectively,and the insulating material between them has a relativepermittivity of 4. The charge density on the outer con-ductor is pl - 10-4 (C/m). Use the expression for Ederived in Example 4-12 to calculate the total energystored in a 20-cm length of the cable.

Ans. We:4.1 J. (See {S1

4-12 lmage Method

Consider a point charge Q at a distance d above a per-fectly conducting plane, as shown in the left-hand sec-tion of Fig. 4-26. We want to determine V,E, and D atany point in the space above the grounded conductor, aswell as the distribution of surface charge on the conduct-ing plate. Three different methods have been introduced

(4 .1

(4.1

(4.1

(4.1

, e Ar= = i jeaY: lef i1eg - |eE2v, (4.122)

(ttlEe v" : Ad is the volume of the capacitor.Ib cbctrostatic energy density u.r" is defined as the

taFr c hNed is

ic potential energy ly" peJ unit volume:

though this expression was derived for a,el-plate capacitor, it is equally valid for any di-

medium in an electric field E. Furthermore. forrolume v containing a dielectric e, the total elec-

potential energy stored in y is

w e - (J). (4.124)) 1,, n' a,

(4.r23)

in this chapter for finding E. The first, based on Cou-lomb's law, requires knowledge of the magnitudes andlocations of all the charges contributing to E at a givenpoint in space. In the present case, the charge Q will in-duce an unknown and nonuniform distribution of chargeon the surface of the conductor. Hence, we cannot uti-lize Coulomb's method. The second method is based onthe application of Gauss's law, and it is equally difficultto use because it is not clear how one would constructa Gaussian surface 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's or Laplace's equa-tion for V, subject to the available boundary conditions;that is, V :0 at any point on the grounded conductingsurface and at infinity. Although such an approach isfeasible in principle, the solution is quite complicatedmathematically. Alternatively, the problem at hand canbe solved with great ease using image theory, whichstates that cuttt given charge configuration above an inf-nite, perfectly conducting plane is electrically equivalentto the combination of the given charge configuration andits image configuration, with the conducting plane re-moved. The image-method equivalent of the charge eabove a conducting plane is shown in the right-handsection of Fig. 4-26. It consists of the charge p itselfand an image charge -Q at a distance 2d from e,

CHAPTER 4 ELECTROSTA

with nothing else between them. The electric fieldto the two isolated charges can now be easily foundany point (x,y,z) by applying Coulomb's method,demonstrated by Example 4-13. The combination oftwo charges will always produce a potential V = 0every point in the plane where the conductinghad been. If the charge is in the presence of moreone grounded plane, it is necessary to establish iof the charge relative to each of the planes andto establish images of each of those images againsremaining planes. The process is continued untilcondition V : O is satisfied at all points on allgrounded planes. The image method applies notto point charges, but also to any distributions ofsuch as the line and volume distributions depictodFig. 4-77.

Erample 4-13 lmage Method for Ghargeabove Conducling Plane

Use image theory to determine 7 and E at obitrary point P(.r, ), z) in the region z > 0 duccharge Q in free space at a distance d aboveaconducting plane.

Solution: In Fig. 4-28, charge Q is at (0,0, d)image -O is at (0,0, -d) in Cartesian coordi

HIGHLIGHTS

: field Eq. (4.19), the electric field at P(x, y, z); 'foundnetho4tion of! t : O

ngmore 4neo L[x2 + y2 + (z - dyzltlzish i i r +yy + i (z+d)

[ x 2 1 y z * ( z * d ) 2 ] 3 / 2i and

igainstI untilon alls notof 4.19 Use the result of Example 4-13 to findbpicted nrface charge density p, on the surface of the con_

plane.

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

n 6e trvo charges is given by

ORr -ORz\

oI*-{)i r + iy +t (z- d)

OUESTIONS

What is the fundamental premise of the image

$ Given a certain charge distribution, what are,inous approaches described in this chapter for:uring the electric field E at a given point in space?

CHAPTER HIGHLIGHTS 'o Maxwell's equations are the fundamental tenets of

electrornagnetic theory.o Under static conditions, Maxwell's equations s€p

arate into two uncoupled pairs, with one pair per_taining to electrostatics and the other to magneto_statics.

o Coulomb's law provides an explicit expression forthe electric field due to a specified charge distribu_tion.

o Gauss's law states that the total electric field fluxthrough a closed surface is equal to the net chargeenclosed by the surface.

o The electrostatic field E at a point is related to theelectric potential y at rhat point by E - _Vy,with V being referenced to zero potential at infin_ity.

o Because most metals have conductivities on theorder of 106 (S/m), they are treated in practice asperfect conductors. By the same token, insulatorswith conductivities smaller than l0-r0 (S/m) aretreated as perfect dielectrics.

o Boundary conditions at the interface between twomaterials specify the relations between the normaland tangential components of D, E, and J in oneof the materials to the conesponding componen$in the other.

r The capacitance of a two-conductor body and re_sistance of the medium between them can be com-puted from knowledge of the electric field in thatmedium.

o The electrostatic energy density stored in a dielec_tric medium is Lc," : )eZz 1ltm3S.

o When a charge configuration exists above an in_finite, perfectly conducting .plane, the inducedfield E is the same as that due to the configura_tion itself and its image with the conducting planeremoved.

point

; (

o fA L

I

4Tes

o

i a t a n0 dueI

d) ard

108

Noncontact Sensors

Precision positioning b q"4ritilgt,qtg .qf:#mFconductor device tabrbation, as w€ll as tlp gpera-tion and controlof many mechanical syste'ms. Non-contact capacitive sensors are U$ed tO $erF€ theposition of silicon wafers during the deposifu, etch-ing, and cutting processes, without coming in directcontactwith the wabrs. Theyare also usedtosenseand control robot arms in equipment manufacturingand to position hard disc drives, photocopier rollers,printing presses, and other similar systems.

Basic Principle

The concentric plate capacitor (A1) consists of twometal plates, sharing the same plane, but electri-cally isolated from each other by an insulating ma-terial. When connected to a voltage source, chargesof opposite polarity will form on the two plates, re-sulting in the creation of electric-field lines betweenthem. The same principle applies to the adjacent-plates capacitor in (A2). In both cases, the capaci-tance is determined by the shapes and sizes of theconductive elements and by the permittivity of thedielectric medium containing the electric field linesbetween them. Otten, the capacitor surface is cov-ered by a thin film of nonconductive material, thepurpose of which is to keep the plate surfaces cleanand dust free. The introduction of an elternalobjectinto the proximity of the capacitor (A3) will perturbthe electric field lines, modifying the charge distribu-tion on the plates, as well as modifying the value ofthe capacitance as would be measured by a capac-itance meter or bridge circuit. Hence, the capacitorbecomes a proximity sensor, and its sensitivity de-pends, in part, on how different the permittivity ofthe object is from that of the unperturbed mediumand on whether it is or is not made of a conductivematerial.

TECHNOI-OCIY BRIEF: NONCONTACT S

Fingerprint lmager.

An interesting extension of noncontactthe development of a fingerprint imagering of a two-dimensional array of capacitivecells. constructed to record an electricaltation of a fingerprint (81 and B2). Eachis composed of an adjacent-plates capacitornected to a capacitance measurement circuil

Y BRIEF: NONCONIACT SENSORS

entire surface of the imager is covered by aayer of nonconductive oxide. When the finger is

on the oxide surface, it perturbs the field linesindividual sensor cells to varying degrees, de-

on the distance between the ridges andof the finge/s surface from the sensor cells.

that the dimensions of an individual sensorfi the order of 65 pm on the side, the imager is

of recording a fingerprint imag€ at a reso-corresponding to 400 dots per inch or better.

9r cofltive serreprer

9ensoracitorircuit * Courtesy of Dr. M. Tartagni, University of Bologna, ltaly

PROBLEMSSeclions tl-Z: Charge and Current Dislribulions

4.1* 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 charge density is 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 o (mC/m3).

4.4 If the line charge density is given by pr : I2y2(mC/m), find the total charge distributed 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 z :0 i f :

(a) p, - pso sind (Clm2)

(b) p. - pr6 sin2 Q (Ctmz)

O (c) p, : pso€-r (C/m2)

(d) p, : pso€-'sin2 4 1C/m2;where ps6 is a constant.

4.6 If J : jLxz (Nm2;, find the current / flow-ing through a square with corners at (0, 0,0), (2, 0, 0),(2,0,2), and (0, 0, 2).

4.7* ffJ: frZS1n (A/m2), find 1 through the surfaceR : 5 m .

4.8 An electron beam shaped like a circular cylinderof radius r0 carries a charge density given by

CHAPTER 4

where p6 is a positive constant and the b€am'scoincident with the e-axis.(a) Delermhe th€ totd charge conAincd in

of the beam.

(b) If the electrons are moving in the *zdirectiouniform speed u, determine the magnituderection of the eurrent crossing the z-plane.

Section 4-3: Coulomb's Law

4.9* A square with sides of 2 m has a charge ofat each of its four corners. Determine the electrbat a point 5 m above the center of the square.

4.10 Three point charges, each with 4 - 3located at the corners of a triangle in the .r-ywith one corner at the origin, another at (2 cm,and the third at (0,2 cm,0). Find the forcethe charge located at the origin.

4.11* Charge il:4 pC is located at (1 cm, Iand charge q2 is located at (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 in air along the z-axisz :0 and z :5 cm. Find E at (0,10 cm,0).

4.13* Electric charge is distributed along ancated in the -r-y plane and defined by r :2O < 0 S n 14. lf p1 - 5 (pClm), find E at (0, 0,then evaluate it at:

(a) The origin.

(b) z :5 cm

( c ) z : - 5 c m

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

(D Solution available in CD-ROM.

PROBLEMS l 1 l

3 n C

4'l{ A line of charge with uniform density O extendsbcrween z : -L/2 and z : L/2 along the z-axis.Apply Coulomb's law to obtain an expression for the

length ic field at any point P(r, @,0) on the x-y plane.that your result reduces to the expression given

ction (4.33) as the length Z is extended to infinity.le andrc.

of radius a, but in the present case, assume the:ace charge density to vary with r as

ps: psor2 (Clm2\

pr6 is a constant.

'r Multiple charges at different locations are said to- equilibrium if the force acting on any one of them isrical in magnitude and direction to the force acting onrf the others. Suppose we have two negative charges,.rrcated at the origin and carrying charge -9e, and the

located on the positive x-axis at a distance d from

15' Repeat Example 4-5 for the circular disk of

Figure 4-29: Kite-shaped arrangment of line charges forProblem 4.17.

your result to the special case where d is infinite andcompare it with E4. @.25).

Seclion 4-4: Gauss's Law

4,20 Given the electric flux density

D : *2(x * y) + y(3x - 2y) (C/m2)

determine

(a) pu by applying Eq. @.26).(b) The total charge Q enclosed in a cube 2 m on a

side, located in the first octant with three of itssides coincident with the x-, 1,-, and z-axes andone of its corners at the origin.

(c) The total charge Q inthe cube, obtained by apply-ing Eq. (4.29).

r. Ibat

:-\'cm.0, :-rrt one and carrying charge *36e, Determine theactint con. polarity and magnitude of a third charge whose

:ment would bring the entire system into equilibrium.;-' Three infinite lines of charge, all parallel to the

. are located at the three corners of the kite-shapedgement shown in Fig. 4-29.lt the two right trian-ue symmetrical and of equal corresponding sides,hdrat the electric field is zero at the origin.

Three infinite lines of charge, p4 = 5 (nC/m),= -5 (nC/m), and p4 = 5 (nC/m), are all parallel

:-axis. Ifthey pass through the respective points-:,. (0, 0), and (0, D) in the .r-y plane, find the

: field at (a,0,0). Evaluate your 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. If the strip is in air and has a uniform charge

ps, use Coulomb's law to obtain an explicitfor the electric field at a point P located ath above the centerline of the strip. Extend

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 law to find E in the regionsR < a , a < R < D , a n d R > b .

4.23* The electric flux density inside a dielectricsphere of radius a centered at the origin is given by

D: fi.psR (Clm2)

where po is a constant. Find the total charse inside thesphere.

4.24 In a certain region of space, the charge densityis given in cylindrical coordinates by the function:

P, :20re-' (C/m3)

Apply Gauss's law to find D.

4.25* An infinitely long cylindrical shell extendingbetween r : I m and r : 3 m contains a uniformcharge density puo. Apply Gauss's law to find D in allregions.

4.26 If the charge density increases linearly with dis-tance from the origin such that pv : 0 at the originand pu: l0 C/m3 at R : 2 m, find the correspondingvariation of D.

Seclion 4-5: Electric Potential

4.27 A square in the -r-y plane in free space has apoint charge of *Q at corner (a/2,a/2), the same atcorner (a/2, -a/2), and a point charge of -Q at eachof the other two corners.

(a) Find the electric potential at any point P along the-r-axis.

(b ) Eva lua teV a tx :a /2 .

CHAPTER 4

4.28 The circular disk of radius c shown in Fighas uniform charge density ps across its surface.(a) Obtain an exprcssion for the electric

at a point P(0,0,2) on the z-axis.(b) Use your result to find E and then evaluatc i

z = h. Compare your final expression wittr (which was obtained on the basis of Coulomb'r

4.29* A circular ring of charge of radius a lies ir.r-y plane and is centered at the origin. Assumethe ring is in air and carries a uniform density p1.

(a) Show that the electrical potential at (0, 0, z) isby V - p1a/[2es(a2 + z2)t/2].

(b) Find the corresponding electric field E.

4.30 Show that the electric potential difference V12tween two points in air at radial distances 11 &nd 12an infinite line of charge with density p1 along the zis 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 acharge with charge density p1 and of 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 lEl : 2 (mV/m) at R - I mandgF i n d E a t R : 2 m a n d 0 : 9 0 .

4.33 For each of the distributions of the electrktential V shown in Fig. 4-30, sketch thedistribution ofE (in all cases, the vertical axis is inand the horizontal axis is in meters).

4.34 Given the electric field

s

u:R# tv/m)find the electric potential of point A withpoint B where A is at *2 m and B at -4 m, bothz-axis.

ttOBLEMS 1 1 3

tsilid

lrte rtf r (mb's

:e Vp

d r zthe:

ion ar t O a. The:nds

Fsure 4-30: Electric potential distributions of Problem] ! t - -

An infinitely long line of charge with uniform': pt :6 (nCVm) lies in the ;-y plane parallel to

r-eris at x : 2 m. Find the potential Van at pointn- 0.4 m) in Cartesian coordinates with respect to8,0, 0, 0) by applying the result of Problem 4.30.

4.36 The r-y plane contains a uniform sheet of chargewith pr, : 0.2 (nC/m2). A second sheet with ps2 : -9.,(nClm) occupies the plane e : 6 m. Fnd V Au : V ac, illdVrc for A(0,0,6 m), B(0,0,0), and C(0, -Z m,2 m).

Section tl-7: Conductorc

4.37* A cylindrical bar of silicon has a radius of 2 mmand alengthof 5 cm.If avoltageof 5 V is appliedbetweenthe ends of the bar Md p": 0.13 1m2lV.s), &n : 0.05(m2lV.s), N" : 1.5 x 1016 electrons/m3, and Nn : N",find the following:(a) The conductivity of silicon.(b) The current.l flowing in the bar.(c) The drift velocities u. and u6.(d) The resistance of the bar.(e) The power dissipated in 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 electrons or holes/m3.

4.39 A 100-m-longconductorofuniformcross-sectionhas a voltage drop of 2 V between its ends. If the density ofthe current flowing through it is 7 x ld (A/m2), identifythe material of the conductor.

4.40 A coaxial resistor of length / consists of two con-centric cylinders. The inner cylinder has radius a andis made of a material with conductivity o1, and theoutercylinder,extendingbetween r : aandr = S,is made of a material with conductivity o2. lf thetwo ends of the resistor are capped with conductingplates, show that the resistance between the two endsis R : I l[n(opz * o2(b2 - a\)].

4.41" Apply the result of Problem 4.40 to find the resis-tance of a l0-cm-long hollow cylinder (Fig.4-31) madeofcarbon with o : 3 x 104 (S/m).

4.42 A2 x 10-3-mm-thick square sheet of aluminumhas l0 cm x 10 cm faces. Find the followine:

Figure 4-31: Cross-section ofhollow cylinder ofproblem4.41.

(a) The resistance between opposite edges on a squareface.

(b) The resistance between fhe two square faces. (SeeAppendix B for the electrical constants of materi_als.)

Seclion 4-9: Boundary Conditions

4.43* 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 the z-axis?

4.44 An infinitely long dielectric cylinder with e 1, : {and described by r = l0 cm is surrounded by a materialwith e2, - 8. If Er - ?r2 sinQ + 6Zr2cos @ f i3 (V/m)in the cylinder region, find E2 and D2 in the surround_ing region. Assume that no free charges exist along thecylinder's boundary.

4.45* A 2-cm dielectric sphere with e11 : 3 is embed_dedinamediumwithe2, : 9. I fEz : f r3cosg-63sing(V/m) in the surrounding region, find E1 and D1 in thesphere.

4.46 If E : ftSO rylm) ar the surface of a 5-cm con-ducting sphere centered at the origin, what is the totalcharge p on the sphere's surface?

CHAPTER 4

4.47* Figure 4-32 shows three planar dielectricequal thickness but with different dielectricE6 in airmakes an angle of 45. with respect to thefind the angle of E in each of the other lavers.

Seclions 4-10 and 4-11: Gapacitance and Electrical4.48 Determine the force of attraction in acapacitor with A : l0 cm2, d : I cm, and e. - 4voltage across it is 50 V.

4.49* Dielectric breakdown occurs in a materialever the magnitude of the field E exceeds the distrength anywhere in that material. In the coaxialitor of Example 4-I2,(a) At what value of r is lEl maximum?(b) What is rhe breakdown voltage if c

b :2 cm, and the dielectric material ise r : 6 ?

4.50 Anelectronwithcharge Q" : -l.6xl1-remass tn e : 9.1 x 10-31 kg is injected at a pointto the negatively charged plate in the regionplates of an air-filled parallel-plate capacitor withration of I cm and rectangular plates each l0 cm2 i

Figure 4-32: Dielectric slabs in problem 4.47.

PROBLEMS1 1 5

r)

3eo

ffiir)

: -' In a dielectric medium with e, : 4, the electric: r given by

:rialE = i(r2 *22) *gr, - 2(y + z) (v/m)

P the electrostatic energy stored in the regionxial n 5 r < I f i i , 0 < y < 2 m , a n d 0 S z < 3 m .

V o = l 0 V

Figure 4-33: Electron between charged plates of problem1.50.

:. 4-33).If the voltage across the capacitor is l0 V: the following:

.r The force acting on the electron.n The acceleration ofthe electron.,: The time it takes the electron to reach the positively

charged plate, assuming that it starts from rest.

Figure 4-34(a) depicts a capacitor consisting of:'":allel, conducting plates separated by a distance d.;:ce between the plates contains two adjacent di_- i. one with permittivity e 1 and surface area A I and: u'ith e2 and A2.The objective of this problem is to

:rt the capacitance C of the configuration shown in- -'-tl

a) is equivalent to two capacitances in parallel,rstrated in Fig.4-34(b), with

C : C r * C z

where

^ e tA t. r = T

^ €zAzvz: T

To this end, proceed as follows:(a) Find the electric fi elds E1 and E2 in the two dielectric

layers.(b) Calculate the energy stored in each section and use

the result to calculate Cr and Cz.(c) Use the total energy stored in the capacitorto obtain

an expression for C. Show that (4.125) is indeed avalid result.

Td

I

(b)

Figure 4-34: (a) Capacitor with parallel dielectric section,and (b) equivalent circuit.

(4.126)

(4.r27)

(4.rzs)

CHAPTER 4

O (a) Conducting plates ile on top and bottomthe rcctangular structure in Fig. 4-35(a).

(b) Conducting plates ane oil front ad back faccsstrrrcture in Fig. 4-35(a).

(c) Conducting plates are on top and bottomthe cylindrical structure in Fig. 4-35(b).

4.54 The capacitor shown in Fig.4-36 consists ofparallel dielectric layers. Use energyshow that the equivalent capacitance of the overallpacitor, C, is equal to the series combination ofpacitances of the individual layers, C1 and C2,

r - CrCz

C r * C z

where

A AC t : € t - T , C z : € z ;

Clt 42

(a) Let Vr and Vz be the electric potentialsupper and lower dielectrics, respectively.

4.53* Use the result of Problem 4.52 to determine thecapacitance for each of the following configurations:

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

r t=2mm

12=4mm

(b)13 = 8mm

I2cm

le t = 8 e O ; s 2 = 4 e , i e 3 = 2 e O

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

'R.OBLEMSIl7

r faces

res of

r faccr

sts ofratiosnerallof the

the corresponding electric fields E1 and E2? By ap_plying the appropriate boundary condition at the in_terface between the two dielectrics, obtain explicitexpressions for E1 and E2 in terms of e1, e2, V, andthe indicated dimensions of the capacitor.

tr r Calculate the energy stored in each of the dielectriclayers and then use the sum to obtain an expressionfor C.

l Show that C is given by Eq. (4.128).

Ir,Iz

-

Figure 4-38: Currents above aconductingplane (problem4.57\.

Ij

Aj-i Use the expressions given in problem 4.54 to=rmine the capacitance for the configurations ins .1-35(a) when the conducting plates are placed on

ight and left faces of the structure.

on 4-12: lmage Method

-i,r With reference to Fig. 4-37, charge e is locatedr listance d above a grounded half-plane located in:-\' plane and at a distance d from another grounded-,;lane in the x-z plane. Use the image method to

n Ertablish the magnitudes, polarities, and locationsrithe images of charge e with respect to each of thes'o ground planes (as if each is infinite in extent).

L' Frnd the electric potential and electric field at anrrbitrary point P(0, y, z).

(S C.SZ Conducting wires above a conducting plane carrycurrents 11 and 12 in the directions shown in Fig. 4_3g.Keeping in mind that the direction of a current is definedin terms of the movement of positive charges, what arethe directions of the image currents conesponding to 11and 12?

4.58 Use the image method to find the capacitance perunit length of an infinitely long conducting cylindeiofradius a situated at a distance d from a parallel conductingplane, as shown in Fig.4-39.

4.594.64 Additional Solved hoblems - completesolutions on (0.

Figure 4-39: Conducting cylinder above a conductinsplane (Problem 4.58).