5 Chapter Capacitance and Dielectrics

Transcription

1 5 Chaptr Capacitanc and Dilctrics 5.1 Introduction Calculation of Capacitanc Exampl 5.1: Paralll-Plat Capacitor Exampl 5.: Cylindrical Capacitor Exampl 5.3: Sphrical Capacitor Storing Enrgy in a Capacitor Enrgy Dnsity of th Elctric Fild Exampl 5.4: Elctric Enrgy Dnsity of Dry Air Exampl 5.5: Enrgy Stord in a Sphrical Shll Dilctrics Polarization Dilctrics without Battry Dilctrics with Battry Gauss s Law for Dilctrics Exampl 5.6: Capacitanc with Dilctrics Crating Elctric Filds Crating an Elctric Dipol Movi Crating and Dstroying Elctric Enrgy Movi Summary Appndix: Elctric Filds Hold Atoms Togthr Ionic and van dr Waals Forcs Problm-Solving Stratgy: Calculating Capacitanc Solvd Problms Capacitor Filld with Two Diffrnt Dilctrics Capacitor with Dilctrics Capacitor Connctd to a Spring Concptual Qustions Additional Problms Capacitors and Dilctrics Gauss s Law in th Prsnc of a Dilctric

2 Gauss s Law and Dilctrics A Capacitor with a Dilctric Forc on th Plats of a Capacitor Enrgy Dnsity in a Capacitor with a Dilctric

3 Capacitanc and Dilctrics 5.1 Introduction A capacitor is a dvic that stors lctric charg. Capacitors vary in shap and siz, but th basic configuration is two conductors carrying qual but opposit chargs (Figur 5.1.1). Capacitors hav many important applications in lctronics. Som xampls includ storing lctric potntial nrgy, dlaying voltag changs whn coupld with rsistors, filtring out unwantd frquncy signals, forming rsonant circuits and making frquncy-dpndnt and indpndnt voltag dividrs whn combind with rsistors. Som of ths applications will b discussd in lattr chaptrs. Figur Basic configuration of a capacitor. In th unchargd stat, th charg on ithr on of th conductors in th capacitor is zro. During th charging procss, a charg Q is movd from on conductor to th othr on, giving on conductor a charg, and th othr on a charg Q. A potntial diffrnc ΔV is cratd, with th positivly chargd conductor at a highr potntial than th ngativly chargd conductor. Not that whthr chargd or unchargd, th nt charg on th capacitor as a whol is zro. Th simplst xampl of a capacitor consists of two conducting plats of ara A, which ar paralll to ach othr, and sparatd by a distanc d, as shown in Figur Figur 5.1. A paralll-plat capacitor Exprimnts show that th amount of charg Q stord in a capacitor is linarly proportional to ΔV, th lctric potntial diffrnc btwn th plats. Thus, w may writ 5-3

4 Q = C ΔV. (5.1.1) whr C is a positiv proportionality constant calld capacitanc. Physically, capacitanc is a masur of th capacity of storing lctric charg for a givn potntial diffrnc ΔV. Th SI unit of capacitanc is th farad [F] : 1 F = 1 farad = 1 coulomb volt = 1 C V. A typical capacitanc that on finds in a laboratory is in th picofarad (1 pf = 1 1 F ) to millifarad rang, (1 mf = 1 3 F=1µF; 1µF = 1 6 F ). Figur 5.1.3(a) shows th symbol that is usd to rprsnt capacitors in circuits. For a polarizd fixd capacitor that has a dfinit polarity, Figur 5.1.3(b) is somtims usd. (a) (b) 5. Calculation of Capacitanc Figur Capacitor symbols. Lt s s how capacitanc can b computd in systms with simpl gomtry. Exampl 5.1: Paralll-Plat Capacitor Considr two mtallic plats of qual ara A sparatd by a distanc d, as shown in Figur 5..1 blow. Th top plat carris a charg +Q whil th bottom plat carris a charg Q. Th charging of th plats can b accomplishd by mans of a battry, which producs a potntial diffrnc. Find th capacitanc of th systm. Figur 5..1 Th lctric fild btwn th plats of a paralll-plat capacitor Solution: To find th capacitanc C, w first nd to know th lctric fild btwn th 5-4

5 plats. A ral capacitor is finit in siz. Thus, th lctric fild lins at th dg of th plats ar not straight lins, and th fild is not containd ntirly btwn th plats. This is known as dg ffcts, and th non-uniform filds nar th dg ar calld th fringing filds. In Figur 5..1, th fild lins ar drawn incorporating dg ffcts. Howvr, in what follows, w shall ignor such ffcts and assum an idalizd situation, whr fild lins btwn th plats ar straight lins., and zro outsid. In th limit whr th plats ar infinitly larg, th systm has planar symmtry and w can calculat th lctric fild vrywhr using Gauss s law givn in Eq. (3..5): S E d A = q nc ε. By choosing a Gaussian pillbox with cap ara A to nclos th charg on th positiv plat (s Figur 5..), th lctric fild in th rgion btwn th plats is EA' = q nc ε = σ A' ε E = σ ε. (5..1) Th sam rsult has also bn obtaind in Sction using th suprposition principl. Figur 5.. Gaussian surfac for calculating th lctric fild btwn th plats. Th potntial diffrnc btwn th plats is ΔV = V V + = E d s = Ed, (5..) whr w hav takn th path of intgration to b a straight lin from th positiv plat to th ngativ plat following th fild lins (Figur 5..). Bcaus th lctric fild lins ar always dirctd from highr potntial to lowr potntial, V < V +. Howvr, in computing th capacitanc C, th rlvant quantity is th magnitud of th potntial diffrnc: ΔV = Ed, (5..3) + 5-5

6 and its sign is immatrial. From th dfinition of capacitanc, w hav C = Q ΔV = ε A (paralllplat). (5..4) d Not that C dpnds only on th gomtric factors A and d. Th capacitanc C incrass linarly with th ara A sinc for a givn potntial diffrnc ΔV, a biggr plat can hold mor charg. On th othr hand, C is invrsly proportional to d, th distanc of sparation bcaus th smallr th valu of d, th smallr th potntial diffrnc ΔV for a fixd Q. Exampl 5.: Cylindrical Capacitor Considr nxt a solid cylindrical conductor of radius a surroundd by a coaxial cylindrical shll of innr radius b, as shown in Figur Th lngth of both cylindrs is L and w tak this lngth to b much largr than b a, th sparation of th cylindrs, so that dg ffcts can b nglctd. Th capacitor is chargd so that th innr cylindr has charg +Q whil th outr shll has a charg Q. What is th capacitanc? (a) (b) Figur 5..3 (a) A cylindrical capacitor. (b) End viw of th capacitor. Th lctric fild is non-vanishing only in th rgion a < r < b. Solution: To calculat th capacitanc, w first comput th lctric fild vrywhr. Du to th cylindrical symmtry of th systm, w choos our Gaussian surfac to b a coaxial cylindr with lngth < L and radius r whr a < r < b. Using Gauss s law, w hav S E d A = EA = E( πr ) = λ E = ε λ πε r, (5..5) whr λ = Q / L is th charg pr unit lngth. Notic that th lctric fild is nonvanishing only in th rgion a < r < b. For r < a, th nclosd charg is q nc = bcaus in lctrostatic quilibrium any charg in a conductor must rsid on its surfac. Similarly, for r > b, th nclosd charg is qnc = λ λ = sinc th Gaussian surfac ncloss qual but opposit chargs from both conductors. Th potntial diffrnc is givn by 5-6

7 b λ b dr λ b Δ V = Vb Va = Er dr = = ln a πε r πε a, (5..6) a whr w hav chosn th intgration path to b along th dirction of th lctric fild lins. As xpctd, th outr conductor with ngativ charg has a lowr potntial. Th capacitanc is thn Q L L C = λ πε ΔV = λ ln( b / a) / πε = ln( b / a). (5..7) Onc again, w s that th capacitanc C dpnds only on th lngth L, and th radii a and b. Exampl 5.3: Sphrical Capacitor As a third xampl, lt s considr a sphrical capacitor which consists of two concntric sphrical shlls of radii a and b, as shown in Figur Th innr shll has a charg +Q uniformly distributd ovr its surfac, and th outr shll an qual but opposit charg Q. What is th capacitanc of this configuration? (a) (b) Figur 5..4 (a) sphrical capacitor with two concntric sphrical shlls of radii a and b. (b) Gaussian surfac for calculating th lctric fild. Solution: Th lctric fild is non-vanishing only in th rgion a < r < b. Using Gauss s law, w obtain E da = E r A = E r (4π r ) = Q. (5..8) ε Th radial componnt of th lctric fild is thn S 1 Q E = r 4πε r. (5..9) o 5-7

8 Thrfor, th potntial diffrnc btwn th two conducting shlls is: b Q b dr Q 1 1 Q b a Δ V = Vb Va = E a rdr = = = 4πε r 4πε a b 4πε ab, (5..1) a which yilds for th capacitanc C Q ab πε ΔV b a. (5..11) = = 4 Th capacitanc C dpnds only on th radii a and b. An isolatd conductor (with th scond conductor placd at infinity) also has a capacitanc. In th limit whr b, th abov quation bcoms ab a lim C = lim 4πε = lim 4πε = 4πε a. (5..1) b b b a b a 1 b Thus, for a singl isolatd sphrical conductor of radius R, th capacitanc is C = 4πε R. (5..13) Th abov xprssion can also b obtaind by noting that a conducting sphr of radius R with a charg Q uniformly distributd ovr its surfac has V = Q / 4πε R, whr infinity is th rfrnc point at zro potntial, V ( ) =. Using our dfinition for capacitanc, Q Q C = 4πε R ΔV = Q / 4πε R =. (5..14) As xpctd, th capacitanc of an isolatd chargd sphr only dpnds on th radius R. 5.3 Storing Enrgy in a Capacitor A capacitor can b chargd by conncting th plats to th trminals of a battry, which ar maintaind at a potntial diffrnc Δ V calld th trminal voltag. 5-8

9 Figur Charging a capacitor. Th connction rsults in sharing th chargs btwn th trminals and th plats. For xampl, th plat that is connctd to th (positiv) ngativ trminal will acquir som (positiv) ngativ charg. Th sharing causs a momntary rduction of chargs on th trminals, and a dcras in th trminal voltag. Chmical ractions ar thn triggrd to transfr mor charg from on trminal to th othr to compnsat for th loss of charg to th capacitor plats, and maintain th trminal voltag at its initial lvl. Th battry could thus b thought of as a charg pump that brings a charg Q from on plat to th othr. As discussd in th introduction, capacitors can b usd to stord lctrical nrgy. Th amount of nrgy stord is qual to th work don to charg it. During th charging procss, th battry dos work to rmov chargs from on plat and dposit thm onto th othr. Figur Work is don by an xtrnal agnt in bringing +dq from th ngativ plat and dpositing th charg on th positiv plat. Lt th capacitor b initially unchargd. In ach plat of th capacitor, thr ar many ngativ and positiv chargs, but th numbr of ngativ chargs balancs th numbr of positiv chargs, so that thr is no nt charg, and thrfor no lctric fild btwn th plats. W hav a magic buckt and a st of stairs from th bottom plat to th top plat (Figur 5.3.1). W show a movi of what is ssntially this procss in Sction 5.5. blow. 5-9

10 W start out at th bottom plat, fill our magic buckt with a charg + dq, carry th buckt up th stairs and dump th contnts of th buckt on th top plat, charging it up positiv to charg + dq. Howvr, in doing so, th bottom plat is now chargd to dq. Having mptid th buckt of charg, w now dscnd th stairs, gt anothr bucktful of charg +dq, go back up th stairs and dump that charg on th top plat. W thn rpat this procss ovr and ovr. In this way w build up charg on th capacitor, and crat lctric fild whr thr was non initially. Suppos th amount of charg on th top plat at som instant is + q, and th potntial diffrnc btwn th two plats is Δ V = q / C. To dump anothr buckt of charg + dq on th top plat, th amount of work don to ovrcom lctrical rpulsion is dw = Δ V dq. If at th nd of th charging procss, th charg on th top plat is + Q, thn th total amount of work don in this procss is Q Q q 1 Q W = dq Δ V = dq = C C This is qual to th lctrical potntial nrgy. (5.3.1) U E of th systm: 1 Q 1 1 E = = Δ = Δ. (5.3.) U Q V C V C Enrgy Dnsity of th Elctric Fild On can think of th nrgy stord in th capacitor as bing stord in th lctric fild itslf. In th cas of a paralll-plat capacitor, with C = ε A/ d and Δ V = Ed, w hav U E = 1 C ΔV = 1 ε A d (Ed) = 1 ε E ( Ad). (5.3.3) Bcaus th quantity Ad rprsnts th volum btwn th plats, w can dfin th lctric nrgy dnsity as u E U 1 Volum E = = ε E. (5.3.4) Th nrgy dnsity ue is proportional to th squar of th lctric fild. Altrnativly, on may obtain th nrgy stord in th capacitor from th point of viw of xtrnal work. Bcaus th plats ar oppositly chargd, forc must b applid to maintain a constant sparation btwn thm. From Eq. (3.4.7), w s that a small patch of charg Δ q = σ ( Δ A) xprincs an attractiv forc Δ F = σ ( Δ A) / ε. If th total ara of th plat is A, thn an xtrnal agnt must xrt a forc F = σ A/ ε to pull th two plats xt 5-1

11 apart. Sinc th lctric fild strngth in th rgion btwn th plats is givn by E = σ / ε, th xtrnal forc can b rwrittn as F xt ε = E A. (5.3.5) Th xtrnal forc F xt is indpndnt of d. Th total amount of work don xtrnally to sparat th plats by a distanc d is thn ε E A Wxt = F xt d s = Fxtd = d, (5.3.6) consistnt with Eq. (5.3.3). Bcaus th potntial nrgy of th systm is qual to th work don by th xtrnal agnt, w hav that th nrgy dnsity ue = Wxt / Ad = ε E /. In addition, w not that th xprssion for u E is idntical to Eq. (3.4.8) in Chaptr 3. Thrfor, th lctric nrgy dnsity ue can also b intrprtd as lctrostatic prssur P. Exampl 5.4: Elctric Enrgy Dnsity of Dry Air Th brakdown fild strngth at which dry air loss its insulating ability and allows a 6 discharg to pass through is E b = 3 1 V/m. At this fild strngth, th lctric nrgy dnsity is: u E = 1 ε E = 1 ( C /N m )(3 1 6 V/m) = 4 J/m 3. (5.3.7) Exampl 5.5: Enrgy Stord in a Sphrical Shll Find th nrgy stord in a mtallic sphrical shll of radius a and charg Q. Solution: Th lctric fild associatd of a sphrical shll of radius a is (Exampl 3.3) Q E = 4πε r ˆr, r > a, r < a. (5.3.8) Th corrsponding nrgy dnsity is u E 1 Q = ε =, (5.3.9) 3 E 4 π εr 5-11

12 outsid th sphr, and zro insid. Sinc th lctric fild is non-vanishing outsid th sphrical shll, w must intgrat ovr th ntir rgion of spac from r = a to r =. In sphrical coordinats, with dv = 4π r dr, w hav Q Q dr Q 1 U E = 4 a 4 π r dr = = = QV 3π ε r 8πε a r 8πε a, (5.3.1) whr V = Q / 4πε a is th lctric potntial on th surfac of th shll, with V ( ) =. W can radily vrify that th nrgy of th systm is qual to th work don in charging th sphr. To show this, suppos at som instant th sphr has charg q and is at a potntial V = q / 4πε a. Th work rquird to add an additional charg dq to th systm is dw = Vdq. Thus, th total work is Q q Q W = dw = Vdq = dq = 4πε a 8πε a. (5.3.11) 5.4 Dilctrics In many capacitors thr is an insulating matrial such as papr or plastic btwn th plats. Such matrial, calld a dilctric, can b usd to maintain a physical sparation of th plats. Sinc dilctrics brak down lss radily than air, charg lakag can b minimizd, spcially whn high voltag is applid. Exprimntally it was found that capacitanc C incrass whn th spac btwn th conductors is filld with dilctrics. To s how this happns, suppos a capacitor has a capacitanc C whn thr is no matrial btwn th plats. Whn a dilctric matrial is insrtd to compltly fill th spac btwn th plats, th capacitanc incrass to C = κ C, (5.4.1) whr κ is calld th dilctric constant. In th Tabl blow, w show som dilctric matrials with thir dilctric constant. Exprimnts indicat that all dilctric matrials hav 1 κ >. Not that vry dilctric matrial has a charactristic dilctric strngth that is th maximum valu of lctric fild bfor brakdown occurs and chargs bgin to flow. 5-1

13 Matrial κ Dilctric strngth (1 6 V / m) Air Papr Glass Watr 8 Th incras of capacitanc in th prsnc of a dilctric can b xplaind from a molcular point of viw. W shall show that κ is a masur of th dilctric rspons to an xtrnal lctric fild. Thr ar two typs of dilctrics. Th first typ ar polar dilctrics, which ar dilctrics that hav prmannt lctric dipol momnts. An xampl of this typ of dilctric is watr. Figur Orintations of polar molculs whn (a) E = and (b) E. As dpictd in Figur 5.4.1, th orintation of polar molculs is random in th absnc of an xtrnal fild. Whn an xtrnal lctric fild E is prsnt, a torqu is st up that causs th molculs to align with E. Howvr, th alignmnt is not complt du to random thrmal motion. Th alignd molculs thn gnrat an lctric fild that is opposit to th applid fild but smallr in magnitud. Th scond typ ar non-polar dilctrics, which ar dilctrics that do not possss a prmannt lctric dipol momnt. Placing a non-polar dilctric matrial in an xtrnally applid lctric fild can induc lctric dipol momnts. Figur 5.4. Orintations of non-polar molculs whn (a) E = and (b) E. 5-13

14 Figur 5.4. illustrats th orintation of non-polar molculs with and without an xtrnal fild E. Whn E, (Figur 5.4.(b)), th inducd surfac chargs on th facs producs an lctric fild E P in th dirction opposit to E, lading to E = E + E P, with E < E. Blow w show how th inducd lctric fild E P is calculatd Polarization W hav shown that dilctric matrials consist of many prmannt or inducd lctric dipols. On of th concpts crucial to th undrstanding of dilctric matrials is th avrag lctric fild producd by many littl lctric dipols that ar all alignd. Suppos w hav a pic of matrial in th form of a cylindr with ara A and hight h, as shown in Figur 5.4.3, and that it consists of N lctric dipols, ach with lctric dipol momnt p sprad uniformly throughout th volum of th cylindr. Figur A cylindr with uniform dipol distribution. W furthrmor assum for th momnt that all of th lctric dipol momnts p ar alignd with th axis of th cylindr. Sinc ach lctric dipol has its own lctric fild associatd with it, in th absnc of any xtrnal lctric fild, if w avrag ovr all th individual filds producd by th dipol, what is th avrag lctric fild just du to th prsnc of th alignd dipols? To answr this qustion, lt us dfin th polarization vctor P to b th nt lctric dipol momnt vctor pr unit volum: N 1 P = pi. (5.4.) Volum i=

15 In th cas of our cylindr, whr all th dipols ar prfctly alignd, th magnitud of P is qual to Np P =. (5.4.3) Ah Th dirction of P is paralll to th alignd dipols. Now, what is th avrag lctric fild ths dipols produc? All th littl ± chargs associatd with th lctric dipols in th intrior of th cylindr in Figur 5.4.4(a) ar rplacd by two quivalnt chargs, ± QP, on th top and bottom of th cylindr, rspctivly in Figur 5.4.4(b). (a) (b) Figur (a) A cylindr with uniform dipol distribution. (b) Equivalnt charg distribution. Th quivalnc can b sn by noting that in th intrior of th cylindr, positiv charg at th top of any on of th lctric dipols is cancld on avrag by th ngativ charg of th dipol just abov it. Th only placs whr cancllations do not tak plac ar at th top and bottom of th cylindr, whr thr ar no additional adjacnt dipols. Thus th intrior of th cylindr appars unchargd in an avrag sns (avraging ovr many dipols). Th top surfac of th cylindr carris a positiv charg and th bottom surfac of th cylindr carris a ngativ charg. How do w find an xprssion for th quivalnt charg Q P in trms of quantitis w know? Th simplst way is to rquir that th lctric dipol momnt Q P producs, QPh, is qual to th total lctric dipol momnt of all th littl lctric dipols. This givs Q h = Np, hnc P Np QP =. (5.4.4) h 5-15

16 To comput th lctric fild producd by Q P, w not that th quivalnt charg distribution rsmbls that of a paralll-plat capacitor, with an quivalnt surfac charg dnsity σ that is qual to th magnitud of th polarization: P Q A Np Ah σ = P = = P P. (5.4.5) 3 Th SI units of polarization dnsity, P, ar (C m)/m, or C/m, which ar th sam units as surfac charg dnsity. In gnral if th polarization vctor maks an angl θ with ˆn, th outward normal vctor of th surfac, th surfac charg dnsity would b σ P = P nˆ = P cosθ. (5.4.6) Th quivalnt charg systm will produc an avrag lctric fild of magnitud EP = P / ε. Bcaus th dirction of this lctric fild is opposit to th dirction of P, in vctor notation, w hav E P = P / ε. (5.4.7) Th avrag lctric fild of all ths dipols is opposit to th dirction of th dipols thmslvs. It is important to raliz that this is just th avrag fild du to all th dipols. If w go clos to any individual dipol, w will s a vry diffrnt fild. W hav assumd hr that all our lctric dipols ar alignd. In gnral, if ths dipols ar randomly orintd, thn th polarization P givn in Eq. (5.4.) will b zro, and thr will b no avrag fild du to thir prsnc. If th dipols hav som tndncy toward a prfrrd orintation, thn P, lading to a non-vanishing avrag fild E P. Lt us now xamin th ffcts of introducing a dilctric matrial into a systm. W shall first assum that th atoms or molculs comprising th dilctric matrial hav a prmannt lctric dipol momnt. If lft to thmslvs, ths prmannt lctric dipols in a dilctric matrial nvr lin up spontanously, so that in th absnc of any applid xtrnal lctric fild, P = du to th random alignmnt of dipols, and th avrag lctric fild E is zro as wll. Howvr, whn w plac th dilctric matrial in an P xtrnal fild E, th dipols will xprinc a torqu τ = p E that tnds to align th dipol vctors p with E. Th ffct is a nt polarization P paralll to E, and thrfor th dipols produc an avrag lctric fild, to rduc th total lctric fild strngth blow E ths two filds: E P, anti-paralll to E, i.., that will tnd. Th lctric fild E is th sum of E = E + EP = E P / ε. (5.4.8) 5-16

17 In most cass, th polarization P is not only in th sam dirction as E, but also linarly proportional to E, and hnc to E as wll. This is rasonabl bcaus without th xtrnal fild E thr would b no alignmnt of dipols and no polarization P. W writ th linar rlation btwn P and E as P = ε χ E, (5.4.9) whr χ is calld th lctric suscptibility. Matrials that oby this rlation ar calld linar dilctrics. Combing Eqs. (5.4.8) and (5.4.7) yilds whr E = (1 + χ) E = κe, (5.4.1) κ = (1 + χ ) (5.4.11) is th dilctric constant. Th dilctric constant κ is always gratr than on sinc χ >. This implis that E E = < E. (5.4.1) κ Thus, w s that th ffct of dilctric matrials is always to dcras th lctric fild blow what it would othrwis b. In th cas of dilctric matrial whr thr ar no prmannt lctric dipols, a similar ffct is obsrvd bcaus th prsnc of an xtrnal fild E inducs lctric dipol momnts in th atoms or molculs. Ths inducd lctric dipols ar paralll to E, again lading to a polarization P paralll to E, and a rduction of th total lctric fild strngth Dilctrics without Battry As shown in Figur 5.4.5, a battry with a potntial diffrnc ΔV across its trminals is first connctd to a capacitor C, which holds a charg Q = C ΔV. W thn disconnct th battry, laving Q =const. 5-17

18 Figur Insrting a dilctric matrial btwn th capacitor plats whil kping th charg Q constant If w thn insrt a dilctric btwn th plats, whil kping th charg constant, xprimntally it is found that th potntial diffrnc dcrass by a factor of κ : This implis that th capacitanc is changd to ΔV = ΔV κ. (5.4.13) Q Q Q C = = = κ = κc. (5.4.14) ΔV ΔV / ΔV κ Th capacitanc has incrasd by a factor of κ. Th lctric fild within th dilctric is now ΔV ΔV / κ 1 ΔV E E = = = d d κ = d. (5.4.15) κ In th prsnc of a dilctric, th lctric fild dcrass by a factor of κ Dilctrics with Battry Considr a scond cas whr a battry supplying a potntial diffrnc ΔV rmains connctd as th dilctric is insrtd (Figur 5.4.6). Exprimntally, it is found (first by Faraday) that th charg on th plats is incrasd by a factor κ : Q = κ Q, (5.4.16) whr Q is th charg on th plats in th absnc of any dilctric. 5-18

19 Figur Insrting a dilctric matrial btwn th capacitor plats whil maintaining a constant potntial diffrnc Δ V. Th capacitanc bcoms C Q κ Q = = = κ ΔV ΔV C (5.4.17) incrasing bcaus th charg has dcrasd. In ithr cas, th nw valu of th capacitanc dos not dpnd on whthr or not th battry is connctd whil th dilctric matrial is insrtd. Howvr, th lctric fild, and charg on th plats do dpnd on whthr or not th battry was connctd whil th dilctric was insrtd Gauss s Law for Dilctrics Considr again a paralll-plat capacitor shown in Figur 5.4.7: Figur Gaussian surfac in th absnc of a dilctric. Whn no dilctric is prsnt, th lctric fild E in th rgion btwn th plats can b found by using Gauss s law: E da = E A = Q, E ε = Q = σ. Aε ε S 5-19

20 With capacitanc C = Q ΔV = Q E d = Aε d. (5.4.18) W hav sn that whn a dilctric is insrtd (Figur 5.4.8), th capacitanc incrass by an amount C = κ C = κ Q ΔV = κ Q E d = κ Aε. (5.4.19) d Thr is now an inducd charg Q P of opposit sign on th surfac, and th nt charg nclosd by th Gaussian surfac is Q Q. P Figur Gaussian surfac in th prsnc of a dilctric. Gauss s law bcoms S E da = EA = Q Q P ε. (5.4.) Th magnitud of th lctric fild has dcrasd btwn th plats Q Q E = P. (5.4.1) ε A Howvr, w hav just sn that th ffct of th dilctric is to wakn th original fild E by a factor κ. Thrfor, E Q Q QP E = = =. (5.4.) κ κ ε A ε A from which th inducd charg Q P can b obtaind as Q P 1 = Q 1 κ. (5.4.3) 5-

21 In trms of th surfac charg dnsity (divid Eq. (5.4.3)) by th ara of th plat, w hav 1 σ P = σ 1. (5.4.4) κ Th limit as κ = 1, th inducd charg is zro, Q P =, which corrsponds to th cas of no dilctric matrial. Substituting Eq. (5.4.3) into Eq. (5.4.), Gauss s law with dilctric can b rwrittn as, S E da = Q κ ε = Q ε, (5.4.5) whr ε = κε is calld th dilctric prmittivity. Altrnativly, w may also writ whr D = εκ E D da = Q, (5.4.6) S is calld th lctric displacmnt vctor. Exampl 5.6: Capacitanc with Dilctrics A non-conducting slab of thicknss t, ara A and dilctric constant κ is insrtd into th spac btwn th plats of a paralll-plat capacitor with spacing d, charg Q and ara A, as shown in Figur 5.4.9(a). Th slab is not ncssarily halfway btwn th capacitor plats. What is th capacitanc of th systm? (a) (b) Figur (a) Capacitor with a dilctric. (b) Elctric fild btwn th plats. Solution: To find th capacitanc C, w first calculat th potntial diffrnc Δ V. W hav alrady sn that in th absnc of a dilctric, th lctric fild btwn th plats is givn by E = Q / ε A, and E / D = E κ whn a dilctric of dilctric constant κ is prsnt, as shown in Figur 5.4.9(b). Th potntial can b found by intgrating th lctric fild along a straight lin from th top to th bottom plats: 5-1

22 ΔV = Edl = ΔV ΔV D = E (d t) E D t = Q (d t) Aε + = Q d t 1 1 Aε κ, Q Aε κ t (5.4.7) whr Δ VD = EDt is th potntial diffrnc btwn th two facs of th dilctric. Th capacitanc is Q ε A C = =. (5.4.8) ΔV 1 d t 1 κ It is usful to chck th following limits: (i) As t, i.., th thicknss of th dilctric approachs zro, w hav C = ε A/ d = C, which is th xpctd rsult for no dilctric. (ii) As κ 1, w again hav C ε A/ d = C, and th situation also corrspond to th cas whr th dilctric is absnt. (iii) In th limit whr t d, th spac is filld with dilctric, w hav C κ ε A/ d = κ C. 5.5 Crating Elctric Filds Crating an Elctric Dipol Movi Elctric filds ar cratd by lctric charg. If thr is no lctric charg prsnt, and nvr had bn any lctric charg prsnt in th past, thn thr would b no lctric fild anywhr is spac. How is lctric fild cratd and how dos it com to fill up spac? To answr this, considr th following scnario in which w go from th lctric fild bing zro vrywhr in spac to an lctric fild xisting vrywhr in spac. 5-

23 Figur Crating an lctric dipol. (a) Bfor any charg sparation. (b) Just aftr th chargs ar sparatd. (c) A long tim aftr sparation. Link to Movi Suppos w hav a positiv point charg sitting right on top of a ngativ lctric charg, so that th total charg xactly cancls, and thr is no lctric fild anywhr in spac. Now lt us pull ths two chargs apart slightly, so that a small distanc sparats thm. If w allow thm to sit at that distanc for a long tim, thr will now b a charg imbalanc an lctric dipol. Th dipol will crat an lctric fild. Lt us s how this cration of lctric fild taks plac in dtail. Figur shows thr frams of a movi of th procss of sparating th chargs. In Figur 5.5.1(a), thr is no charg sparation, and th lctric fild is zro vrywhr in spac. Figur 5.5.1(b) shows what happns just aftr th chargs ar first sparatd. An xpanding sphr of lctric filds is obsrvd. Figur 5.5.1(c) shows a long tim aftr th chargs ar sparatd (that is, thy hav bn at a constant distanc from ach othr for a long tim). An lctric dipol has bn cratd. What dos this squnc tll us? Th following conclusions can b drawn: (1) Elctric charg gnrats lctric fild no charg, no fild. () Th lctric fild dos not appar instantanously in spac vrywhr as soon as thr is unbalancd charg th lctric fild propagats outward from its sourc at som finit spd. This spd will turn out to b th spd of light, as w shall s latr. (3) Aftr th charg distribution sttls down and bcoms stationary, so dos th fild configuration. Th initial fild pattrn associatd with th tim dpndnt sparation of th charg is actually a burst of lctric dipol radiation. W rturn to th subjct of radiation at th nd of this cours. Until thn, w will nglct radiation filds. Th fild configuration lft bhind aftr a long tim is just th lctric dipol pattrn discussd abov. W not that th xtrnal agnt dos work whn pulling th chargs apart, and thn must apply a forc to kp thm sparat, sinc thy attract ach othr as soon as thy start to sparat. In addition, th work also gos into providing th nrgy carrid off by radiation, as wll as th nrgy ndd to st up th final stationary lctric fild that w s in Figur 5.5.1(c). 5-3

24 Link to Movi Figur 5.5. Crating th lctric filds of two point chargs by pulling apart two opposit chargs initially on top of on anothr. W artificially trminat th fild lins at a fixd distanc from th chargs to avoid visual confusion. Finally, w ignor radiation and complt th procss of sparating our opposit point chargs that w bgan in Figur Th link in Figur 5.5. shows th complt squnc. Whn w finish and hav movd th chargs far apart, w s th charactristic radial fild in th vicinity of a point charg Crating and Dstroying Elctric Enrgy Movi Lt us look at th procss of crating lctric nrgy in a diffrnt contxt. W ignor nrgy losss du to radiation in this discussion. Figur shows on fram of a movi that illustrats th following procss. This movi is mor or lss analogous to th procss w discussd in Sction 5.3 abov for charging a capacitor. Figur Link to movi W start out with fiv ngativ lctric chargs and fiv positiv chargs, all at th sam point in spac. Sin thr is no nt charg, thr is no lctric fild. Now w mov on of th positiv chargs at constant vlocity from its initial position to a distanc L away along th horizontal axis. Aftr doing that, w mov th scond positiv charg in th sam mannr to th position whr th first positiv charg sits. Aftr doing that, w 5-4

25 continu on with th rst of th positiv chargs in th sam mannr, until all th positiv chargs ar sitting a distanc L from thir initial position along th horizontal axis. Figur shows th fild configuration during this procss. W hav color codd th grass sds rprsntation to rprsnt th strngth of th lctric fild. Vry strong filds ar whit, vry wak filds ar black, and filds of intrmdiat strngth ar yllow. Ovr th cours of th crat movi associatd with Figur 5.5.3, th strngth of th lctric fild grows as ach positiv charg is movd into plac. Th lctric nrgy flows out from th path along which th chargs mov, and is bing providd by th agnt moving th charg against th lctric fild of th othr chargs. Th work that this agnt dos to sparat th chargs against thir lctric attraction appars as nrgy in th lctric fild. W also hav a movi of th opposit procss linkd to Figur That is, w rturn in squnc ach of th fiv positiv chargs to thir original positions. At th nd of this procss w no longr hav an lctric fild, bcaus w no longr hav an unbalancd lctric charg. On th othr hand, ovr th cours of th dstroy movi associatd with Figur 5.5.3, th strngth of th lctric fild dcrass as ach positiv charg is rturnd to its original position. Th nrgy flows from th fild back to th path along which th chargs mov, and is now bing providd to th agnt moving th charg at constant spd along th lctric fild of th othr chargs. Th nrgy providd to that agnt as w dstroy th lctric fild is xactly th amount of nrgy that th agnt put into crating th lctric fild in th first plac, nglcting radiativ losss (such losss ar small if w mov th chargs at spds small compard to th spd of light). This is a rvrsibl procss if w nglct such losss. That is, th amount of nrgy th agnt puts into crating th lctric fild is xactly rturnd to that agnt as th fild is dstroyd. Thr is on final point to b mad. Whnvr lctromagntic nrgy is bing cratd, an lctric charg is moving (or bing movd) against an lctric fild ( q v E < ). Whnvr lctromagntic nrgy is bing dstroyd, an lctric charg is moving (or bing movd) along an lctric fild ( q v E > ). Whn w rturn to th cration and dstruction of magntic nrgy, w will find this rul holds thr as wll. 5.6 Summary A capacitor is a dvic that stors lctric charg and potntial nrgy. Th capacitanc C of a capacitor is th ratio of th charg stord on th capacitor plats to th potntial diffrnc btwn thm: Q C =. Δ V 5-5

26 Systm Capacitanc Isolatd chargd sphr of radius R C = 4πε R Paralll-plat capacitor of plat ara A and plat sparation d A C = ε d Cylindrical capacitor of lngth L, innr radius a and outr radius b πε L C = ln( b / a) Sphrical capacitor with innr radius a and outr radius b C = 4πε ab ( b a) Th work don in charging a capacitor to a charg Q is Q 1 1 U Q V C V C = = Δ = Δ. This is qual to th amount of nrgy stord in th capacitor. Th lctric nrgy can also b thought of as stord in th lctric fild E. Th nrgy dnsity (nrgy pr unit volum) is ue 1 = ε E. Th nrgy dnsity ue is qual to th lctrostatic prssur on a surfac. Whn a dilctric matrial with dilctric constant κ is insrtd into a capacitor, th capacitanc incrass by a factor κ : C = κ C. Th polarization vctor P is th magntic dipol momnt pr unit volum: P =. 1 N pi V i = 1 Th inducd lctric fild du to polarization is E P = P / ε. 5-6

27 In th prsnc of a dilctric with dilctric constant κ, th lctric fild bcoms E = E + E = E κ, whr E / P is th lctric fild without dilctric. 5.7 Appndix: Elctric Filds Hold Atoms Togthr In this Appndix, w illustrat how lctric filds ar rsponsibl for holding atoms togthr. As our mntal y pntrats into smallr and smallr distancs and shortr and shortr tims, w find natur bhaving so ntirly diffrntly from what w obsrv in visibl and palpabl bodis of our surroundings that no modl shapd aftr our larg-scal xprincs can vr b "tru". A compltly satisfactory modl of this typ is not only practically inaccssibl, but not vn thinkabl. Or, to b prcis, w can, of cours, think of it, but howvr w think it, it is wrong Ionic and van dr Waals Forcs Erwin Schrodingr Elctromagntic forcs provid th glu that holds atoms togthr that is, that kp lctrons nar protons and bind atoms togthr in solids. W prsnt hr a brif and vry idalizd modl of how that happns from a smi-classical point of viw. (a) Link to movi (b) Link to movi Figur (a) A ngativ charg and (b) a positiv charg mov past a massiv positiv particl at th origin and is dflctd from its path by th strsss transmittd by th lctric filds surrounding th chargs. Figur 5.7.1(a) illustrats th xampls of th strsss transmittd by filds, as w hav sn bfor. In Figur 5.7.1(a) w hav a ngativ charg moving past a massiv positiv charg and bing dflctd toward that charg du to th attraction that th two chargs 5-7

28 fl. This attraction is mdiatd by th strsss transmittd by th lctromagntic fild, and th simpl intrprtation of th intraction shown in Figur 5.7.1(b) is that th attraction is primarily du to a tnsion transmittd by th lctric filds surrounding th chargs. In Figur 5.7.1(b) w hav a positiv charg moving past a massiv positiv charg and bing dflctd away from that charg du to th rpulsion that th two chargs fl. This rpulsion is mdiatd by th strsss transmittd by th lctromagntic fild, as w hav discussd abov, and th simpl intrprtation of th intraction shown in Figur 5.7.1(b) is that th rpulsion is primarily du to a prssur transmittd by th lctric filds surrounding th chargs. Considr th intraction of four chargs of qual mass shown in Figur Two of th chargs ar positivly chargd and two of th chargs ar ngativly chargd, and all hav th sam magnitud of charg. Th particls intract via th Coulomb forc. W also introduc a quantum-mchanical Pauli forc, which is always rpulsiv and bcoms vry important at small distancs, but is ngligibl at larg distancs. Th critical distanc at which this rpulsiv forc bgins to dominat is about th radius of th sphrs shown in Figur This Pauli forc is quantum mchanical in origin, and kps th chargs from collapsing into a point (i.., it kps a ngativ particl and a positiv particl from sitting xactly on top of on anothr). Additionally, th motion of th particls is dampd by a trm proportional to thir vlocity, allowing thm to "sttl down" into stabl (or mta-stabl) stats. Link to Movi Figur 5.7. Four chargs intracting via th Coulomb forc, a rpulsiv Pauli forc at clos distancs, with damping. Whn ths chargs ar allowd to volv from th initial stat, th first thing that happns (vry quickly) is that th chargs pair off into dipols. This is a rapid procss bcaus th Coulomb attraction btwn unbalancd chargs is vry larg. This procss is calld "ionic binding", and is rsponsibl for th intr-atomic forcs in ordinary tabl salt, NaCl. Aftr th dipols form, thr is still an intraction btwn nighboring dipols, but this is a much wakr intraction bcaus th lctric fild of th dipols falls off much 5-8

29 fastr than that of a singl charg. This is bcaus th nt charg of th dipol is zro. Whn two opposit chargs ar clos to on anothr, thir lctric filds almost cancl ach othr out. Although in principl th dipol-dipol intraction can b ithr rpulsiv or attractiv, in practic thr is a torqu that rotats th dipols so that th dipol-dipol forc is attractiv. Aftr a long tim, this dipol-dipol attraction brings th two dipols togthr in a bound stat. Th forc of attraction btwn two dipols is trmd a van dr Waals forc, and it is rsponsibl for intrmolcular forcs that bind som substancs togthr into a solid. 5.8 Problm-Solving Stratgy: Calculating Capacitanc In this chaptr, w hav sn how capacitanc C can b calculatd for various systms. Th procdur is summarizd blow: (1) Idntify th dirction of th lctric fild using symmtry. () Calculat th lctric fild vrywhr. (3) Comput th lctric potntial diffrnc ΔV. (4) Calculat th capacitanc C using C = Q / Δ V. In th Tabl blow, w illustrat how th abov stps ar usd to calculat th capacitanc of a paralll-plat capacitor, cylindrical capacitor and a sphrical capacitor. 5-9

30 Capacitors Paralll-plat Cylindrical Sphrical Figur (1) Idntify th dirction of th lctric fild using symmtry () Calculat lctric fild vrywhr E d A = EA = Q ε S E = Q Aε = σ ε S E d A = E( πrl ) = Q ε E = λ πε r S E da = E r 4π r E r = 1 4πε o ( ) = Q ε Q r (3) Comput th lctric potntial diffrnc ΔV Δ V = V V+ = E d s + = Ed Δ V = V V = E dr b a a r λ b = ln πε a b Δ V = V V = E dr b a a r Q b a = 4πε ab b (4) Calculat C using C = Q / Δ V ε C = d A C πε l ln( b / a) = C 4 ab = πε b a 5-3

31 5.9 Solvd Problms Capacitor Filld with Two Diffrnt Dilctrics Two dilctrics with dilctric constants κ 1 and κ ach fill half th spac btwn th plats of a paralll-plat capacitor as shown in Figur Each plat has an ara A and th plats ar sparatd by a distanc d. Comput th capacitanc of th systm. Figur Capacitor filld with two diffrnt dilctrics. Solution: Bcaus th potntial diffrnc ΔV on ach half of th capacitor is th sam, w may trat th systm as bing composd of two capacitors, C 1 and C, with chargs ±Q 1 and ±Q on ach half. Th magnitud of th lctric fild is th sam on ach sid bcaus E = ΔV d. W can apply Eq. (5.4.5) to dtrmin th charg on ach plat in trms of th lctric fild btwn th plats: Q i = κ i ε E( A / ). Thrfor using our rsult for lctric fild, th charg is givn by Q i = κ ε ( A / ) ΔV i d. Th capacitanc of th systm is thn whr C = Q 1 + Q ΔV = ε A d (κ 1 + κ ) = C 1 + C, κ ( / ) iε A Ci =, i = 1,. d 5-31

32 5.9. Capacitor with Dilctrics Considr a conducting sphrical shll with an innr radius a and outr radius c. Lt th spac btwn two surfacs b fild with two diffrnt dilctric matrials so that th dilctric constant is κ 1 btwn a and b, and κ btwn b and c, as shown in Figur Dtrmin th capacitanc of this systm. Figur Sphrical capacitor filld with dilctrics. Solution: Th systm can b tratd as two capacitors connctd in sris, sinc th total potntial diffrnc across th capacitors is th sum of potntial diffrncs across individual capacitors, ΔV = ΔV 1 + ΔV. Each shll has th sam magnitud charg Q. Th charg on ach capacitor is rlatd to th potntial diffrnc by Each individual capacitor, satisfis ΔV i = Q C i. C i = κ i C i, whr C is th capacitanc for a vacuum sphrical capacitor of innr radius r 1 and outr radius r, which w calculatd in Exampl 5.3, Thrfor th capacitancs ar r C i, = 4πε 1 r r r 1. C 1 = κ 1 4πε C = κ 4πε Th capacitanc for th systm thrfor is ab b a bc c b. 5-3

33 C = Q ΔV 1 + ΔV = Q Q / C 1 + Q / C = C 1 C C 1 + C. Using our rsults abov w hav that th capacitanc of this systm is givn by C = C C 1 C 1 + C ab κ 1 4πε b a κ 4πε bc c b = ab κ 1 4πε b a + κ 4πε bc c b ab κ 1 4πε b a κ 4πε bc c b (b a)(c b) = κ 1 4πε ab(c b) + κ 4πε bc(b a) Thus aftr som simplification w hav that 4πε κ1κ abc C =. κ c( b a) + κ a( c b) 1 It is instructiv to chck th limit whr κ1, κ 1. In this cas, th abov xprssion rducs to C 4πε abc 4πε abc 4πε ac = = = c ( b a ) + a ( c b ) b ( c a ) ( c a ) which agrs with Eq. (5..11) for a sphrical capacitor of innr radius a and outr radius c Capacitor Connctd to a Spring Considr an air-filld paralll-plat capacitor with on plat connctd to a spring having a forc constant k, and anothr plat hld fixd. Th systm rsts on a tabl top as shown in Figur Figur Capacitor connctd to a spring. 5-33

34 If th chargs placd on plats a and b ar + Q and Q, rspctivly, how much dos th spring xpand? Solution: Th spring forc F s acting on plat a is givn by Fs = kx ˆi. Similarly, th lctrostatic forc F du to th lctric fild cratd by plat b is ˆ σ ˆ Q F ˆ = QE i = Q i = i, ε Aε whr A is th ara of th plat. Th chargs on plat a cannot xrt a forc on itslf, as rquird by Nwton s third law. Thus, only th lctric fild du to plat b is considrd. At quilibrium th two forcs cancl and w hav which givs kx Q Q =, Aε Q x =. kaε 5.1 Concptual Qustions 1. Th chargs on th plats of a paralll-plat capacitor ar of opposit sign, and thy attract ach othr. To incras th plat sparation, is th xtrnal work don positiv or ngativ? What happns to th xtrnal work don in this procss?. How dos th stord nrgy chang if th potntial diffrnc across a capacitor is tripld? 3. Dos th prsnc of a dilctric incras or dcras th maximum oprating voltag of a capacitor? Explain. 4. If a dilctric-filld capacitor is coold down, what happns to its capacitanc? 5-34

35 5.11 Additional Problms Capacitors and Dilctrics (a) A paralll-plat capacitor of ara A and spacing d is filld with thr dilctrics as shown in Figur Each occupis 1/3 of th volum. What is th capacitanc of this systm? [Hint: Considr an quivalnt systm to b thr paralll capacitors, and justify this assumption.] Show that you obtain th propr limits as th dilctric constants approach unity, κ i 1.] Figur (b) This capacitor is now filld as shown in Figur What is its capacitanc? Us Gauss's law to find th fild in ach dilctric, and thn calculat ΔV across th ntir capacitor. Again, chck your answr as th dilctric constants approach unity, κ i 1. Could you hav assumd that this systm is quivalnt to thr capacitors in sris? Figur Gauss s Law in th Prsnc of a Dilctric A solid conducting sphr with a radius R 1 carris a fr charg Q and is surroundd by a concntric dilctric sphrical shll with an outr radius R and a dilctric constant κ. This systm is isolatd from othr conductors and rsids in air (κ 1), as shown in Figur (a) Dtrmin th displacmnt vctor D vrywhr, i.. its magnitud and dirction in th rgions r < R1, R1 < r < R and r > R. (b) Dtrmin th lctric fild E vrywhr. 5-35

36 Figur Gauss s Law and Dilctrics A cylindrical shll of dilctric matrial has innr radius a and outr radius b, as shown in Figur Figur Th matrial has a dilctric constant κ = 1. At th cntr of th shll thr is a lin charg running paralll to th axis of th cylindrical shll, with fr charg pr unit lngth λ. (a) Find th lctric fild for: r < a, a < r < b and r > b. (b) What is th inducd surfac charg pr unit lngth on th innr surfac of th sphrical shll? [Ans. 9 λ /1.] (c) What is th inducd surfac charg pr unit lngth on th outr surfac of th sphrical shll? [Ans. + 9 λ /1.] 5-36

37 A Capacitor with a Dilctric A paralll plat capacitor has a capacitanc of 11 pf, a plat ara of 96.5 cm, and a mica dilctric (κ = 5.4 ). At a 55 V potntial diffrnc, calculat (a) th lctric fild strngth in th mica; [Ans kv/m.] (b) th magnitud of th fr charg on th plats; [Ans nc.] (c) th magnitud of th inducd surfac charg; [Ans. 5. nc.] (d) th magnitud of th polarization P [Ans. 5 nc/m.] Forc on th Plats of a Capacitor Th plats of a paralll-plat capacitor hav ara A and carry total charg ±Q (s Figur 5.1.6). W would lik to show that ths plats attract ach othr with a forc givn by F = Q /(ε o A). Figur (a) Calculat th total forc on th lft plat du to th lctric fild of th right plat, using Coulomb's Law. Ignor fringing filds. (b) If you pull th plats apart, against thir attraction, you ar doing work and that work gos dirctly into crating additional lctrostatic nrgy. Calculat th forc ncssary to incras th plat sparation from x to x + dx by quating th work you do, F dx, to th incras in lctrostatic nrgy, assuming that th lctric nrgy dnsity is ε E /, and that th charg Q rmains constant. 5-37

38 (c) Using this xprssion for th forc, show that th forc pr unit ara (th lctrostatic strss) acting on ithr capacitor plat is givn by ε E /. This rsult is tru for a conductor of any shap with an lctric fild E at its surfac. (d) Atmosphric prssur is 14.7 lb/in, or 11,341 N/m. How larg would E hav to b to produc this forc pr unit ara? [Ans. 151 MV/m. Not that Van d Graff acclrators can rach filds of 1 MV/m maximum bfor brakdown, so that lctrostatic strsss ar on th sam ordr as atmosphric prssurs in this xtrm situation, but not much gratr] Enrgy Dnsity in a Capacitor with a Dilctric Considr th cas in which a dilctric matrial with dilctric constant κ compltly fills th spac btwn th plats of a paralll-plat capacitor. Show that th nrgy dnsity of th fild btwn th plats is u = E D / by th following procdur: (a) Writ th xprssion u = E D / E E as a function of E and κ (i.. liminat D ). (b) Givn th lctric fild and potntial of such a capacitor with fr charg q on it (problm 4-1a abov), calculat th work don to charg up th capacitor from q = to q = Q, th final charg. (c) Find th nrgy dnsity u E. 5-38