CHAPTER 5 CAPACITORS


 Kelley Long
 2 years ago
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1 CHAPTER 5 CAPACITORS 5. Inroducion A capacior consiss of wo meal plaes separaed by a nonconducing medium (known as he dielecric medium or simply he dielecric, or by a vacuum. I is represened by he elecrical symbol. Capaciors of one sor or anoher are included in almos any elecronic device. Physically, here is a vas variey of shapes, sizes and consrucion, depending upon heir paricular applicaion. This chaper, however, is no primarily concerned wih real, pracical capaciors and how hey are made and wha hey are used for, alhough a brief secion a he end of he chaper will discuss his. In addiion o heir pracical uses in elecronic circuis, capaciors are very useful o professors for oruring sudens during exams, and, more imporanly, for helping sudens o undersand he conceps of and he relaionships beween elecric fields E and D, poenial difference, permiiviy, energy, and so on. The capaciors in his chaper are, for he mos par, imaginary academic conceps useful largely for pedagogical purposes. Need he elecronics echnician or elecronics engineer spend ime on hese academic capaciors, apparenly so far removed from he real devices o be found in elecronic equipmen? The answer is surely and decidedly yes more han anyone else, he pracical echnician or engineer mus horoughly undersand he basic conceps of elecriciy before even saring wih real elecronic devices. If a poenial difference is mainained across he wo plaes of a capacior (for example, by connecing he plaes across he poles of a baery) a charge +Q will be sored on one plae and Q on he oher. The raio of he charge sored on he plaes o he poenial difference V across hem is called he capaciance C of he capacior. Thus: Q = CV. 5.. If, when he poenial difference is one vol, he charge sored is one coulomb, he capaciance is one farad, F. Thus, a farad is a coulomb per vol. I should be menioned here ha, in pracical erms, a farad is a very large uni of capaciance, and mos capaciors have capaciances of he order of microfarads, µf. Q The dimensions of capaciance are = M L T Q. ML T Q
2 I migh be remarked ha, in older books, a capacior was called a condenser, and is capaciance was called is capaciy. Thus wha we would now call he capaciance of a capacior was formerly called he capaciy of a condenser. In he highly idealized capaciors of his chaper, he linear dimensions of he plaes (lengh and breadh, or diameer) are supposed o be very much larger han he separaion beween hem. This in fac is nearly always he case in real capaciors, oo, hough perhaps no necessarily for he same reason. In real capaciors, he disance beween he plaes is small so ha he capaciance is as large as possible. In he imaginary capaciors of his chaper, I wan he separaion o be small so ha he elecric field beween he plaes is uniform. Thus he capaciors I shall be discussing are mosly like figure V., where I have indicaed, in blue, he elecric field beween he plaes: FIGURE V. However, I shall no always draw hem like his, because i is raher difficul o see wha is going on inside he capacior. I shall usually much exaggerae he scale in one direcion, so ha my drawings will look more like his: FIGURE V. If he separaion were really as large as his, he field would no be nearly as uniform as indicaed; he elecric field lines would grealy bulge ouwards near he edges of he plaes. In he nex few secions we are going o derive formulas for he capaciances of various capaciors of simple geomeric shapes.
3 3 5. Plane Parallel Capacior A FIGURE V.3 d V We have a capacior whose plaes are each of area A, separaion d, and he medium beween he plaes has permiiviy ε. I is conneced o a baery of EMF V, so he poenial difference across he plaes is V. The elecric field beween he plaes is E = V/d, and herefore D = εe/d. The oal Dflux arising from he posiive plae is DA, and, by Gauss s law, his mus equal Q, he charge on he plae. Thus Q = εav /d, and herefore he capaciance is C εa =. 5.. d Verify ha his is dimensionally correc, and noe how he capaciance depends upon ε, A and d. In Secion.5 we gave he SI unis of permiiviy as C N m. Equaion 5.. shows ha a more convenien SI uni for permiiviy is F m, or farads per mere. Quesion: If he separaion of he plaes is no small, so ha he elecric field is no uniform, and he field lines bulge ouwards a he edge, will he capaciance be less han or greaer han εa/d?
4 4 5.3 Coaxial Cylindrical Capacior b FIGURE V.4 a ε The radii of he inner and ouer cylinders are a and b, and he permiiviy beween hem is ε. Suppose ha he wo cylinders are conneced o a baery so ha he poenial difference beween hem is V, and he charge per uni lengh on he inner cylinder is +λ C m, and on he ouer cylinder is λ C m. We have seen (Subsecion..3) ha he λ poenial difference beween he cylinders under such circumsances is ln( b / a). Therefore he capaciance per uni lengh, C ', is πε C' πε = ln( b / a) This is by no means solely of academic ineres. The capaciance per uni lengh of coaxial cable ( coax ) is an imporan propery of he cable, and his is he formula used o calculae i. 5.4 Concenric Spherical Capacior Unlike he coaxial cylindrical capacior, I don know of any very obvious pracical applicaion, nor quie how you would consruc one and connec he wo spheres o a baery, bu le s go ahead all he same. Figure V.4 will do jus as well for his one. The wo spheres are of inner and ouer radii a and b, wih a poenial difference V beween hem, wih charges +Q and Q on he inner and ouer spheres respecively. The Q poenial difference beween he wo spheres is hen, and so he 4πε a b capaciance is C 4πε = a b
5 5 If b, we obain for he capaciance of an isolaed sphere of radius a: C = 4πε a Exercise: Calculae he capaciance of plane Earh, of radius km, suspended in free space. I make i 79 µf  which may be a bi smaller han you were expecing. 5.5 Capaciors in Parallel FIGURE V.5 C C C 3 The poenial difference is he same across each, and he oal charge is he sum of he charges on he individual capacior. Therefore: C = C + C C3 5.6 Capaciors in Series +Q Q +Q Q +Q Q C C C 3 FIGURE V.6 The charge is he same on each, and he poenial difference across he sysem is he sum of he poenial differences across he individual capaciances. Hence C = C C C3
6 6 5.7 DelaSar Transform X Y X Y C 3 c c C C c 3 Z Z FIGURE V.7 As we did wih resisors in Secion 4., we can make a delasar ransform wih capaciors. I leave i o he reader o show ha he capaciance beween any wo erminals in he lef hand box is he same as he capaciance beween he corresponding wo erminals in he righ hand box provided ha c c CC3 + C3C + CC =, 5.7. C C C + C C C 3 3 = C C CC3 + C3C + CC and c3 = C The converse relaions are 3 C C = cc3, c + c + c c c + c + c = c 3 cc and C = 3. c + c + c
7 7 For example, jus for fun, wha is he capaciance beween poins A and B in figure V.8, in which I have marked he individual capaciances in microfarads? 3 A 4 B 8 FIGURE V.8 The firs hree capaciors are conneced in dela. Replace hem by heir equivalen sar configuraion. Afer ha i should be sraighforward. I make he answer.4 µf. 5.8 Kirchhoff s Rules We can even adap Kirchhoff s rules o deal wih capaciors. baery across he circui of figure V.8 see figure V.9 4 V Thus, connec a 4 V +Q 3 Q +Q 3 Q 3 +Q 5 Q 5 4 +Q Q 8 FIGURE V.9 +Q 4 Q 4 Calculae he charge held in each capacior. We can proceed in a manner very similar o how we did i in Chaper 4, applying he capaciance equivalen of Kirchhoff s second rule o hree closed circuis, and hen making up he five necessary equaions by applying Kirchhoff s firs rule o wo poins. Thus: 4 Q Q 3 3 =, 5.8.
8 8 Q4 4 Q =, Q 3 Q 4 5 Q = +, Q = Q +, Q5 and Q = Q I make he soluions Q 4 Q5 = µ C, Q = + 9.µ C, Q3 = +.44µ C, Q4 = µ C, Q5 = +.9 µ C. 5.9 Problem for a Rainy Day Anoher problem o while away a rainy Sunday afernoon would be o replace each of he resisors in he cube of subsecion 4.4. wih capaciors each of capaciance c. Wha is he oal capaciance across opposie corners of he cube? I would sar by supposing ha he cube holds a ne charge of 6Q, and I would hen ask myself wha is he charge held in each of he individual capaciors. And I would hen follow he poenial drop from one corner of he cube o he opposie corner. I make he answer for he effecive capaciance of he enire cube.c. 5. Energy Sored in a Capacior +q +δq q FIGURE V.
9 9 Le us imagine (figure V.) ha we have a capacior of capaciance C which, a some ime, has a charge of +q on one plae and a charge of q on he oher plae. The poenial difference across he plaes is hen q/c. Le us now ake a charge of +δq from he boom plae (he negaive one) and move i up o he op plae. We evidenly have o do work o do his, in he amoun of q δ q. The oal work required, hen, saring wih he plaes C compleely uncharged unil we have ransferred a charge Q from one plae o he oher is Q q dq Q /(C). C = This is, hen, he energy E sored in he capacior, and, by applicaion of Q = CV i can also be wrien E = QV, or, more usually, E = CV. 5.. Verify ha his has he correc dimensions for energy. Also, hink abou how many expressions for energy you know ha are of he form ab. There are more o come. The symbol E is becoming raher overworked. A presen I am using he following: Sorry abou ha! E = magniude of he elecric field E = elecric field as a vecor E = elecromoive force E = energy 5. Energy Sored in an Elecric Field Recall ha we are assuming ha he separaion beween he plaes is small compared wih heir linear dimensions and ha herefore he elecric field is uniform beween he plaes. The capaciance is C = εa/ d, and he poenial differnece beween he plaes is Ed, where E is he elecric field and d is he disance beween he plaes. Thus he energy sored in he capacior is ε E Ad. The volume of he dielecric (insulaing) maerial beween he plaes is Ad, and herefore we find he following expression for he energy sored per uni volume in a dielecric maerial in which here is an elecric field: ε E Verify ha his has he correc dimensions for energy per uni volume.. If he space beween he plaes is a vacuum, we have he following expression for he enrgy sored ber uni volume in he elecric field
10 ε E  even hough here is absoluely nohing oher han energy in he space. Think abou ha! I menioned in Secion.7 ha in an anisoropic medium D and E are no parallel, he permiiviy hen being a ensor quaniy. In ha case he correc expression for he energy per uni volume in an elecric field is D E. 5. Force Beween he Plaes of a Plane Parallel Plae Capacior We imagine a capacior wih a charge +Q on one plae and Q on he oher, and iniially he plaes are almos, bu no quie, ouching. There is a force F beween he plaes. Now we gradually pull he plaes apar (bu he separaion remains small enough ha i is sill small compared wih he linear dimensions of he plaes and we can mainain our approximaion of a uniform field beween he plaes, and so he force remains F as we separae hem). his mus hen equal he energy sored in he capacior,. he plaes is E = V/d, so we find for he force beween he plaes The work done in separaing he plaes from near zero o d is Fd, and QV The elecric field beween F = QE. 5.. We can now do an ineresing imaginary experimen, jus o see ha we undersand he various conceps. Le us imagine ha we have a capacior in which he plaes are horizonal; he lower plae is fixed, while he upper plae is suspended above i from a spring of force consan k. We connec a baery across he plaes, so he plaes will arac each oher. The upper plae will move down, bu only so far, because he elecrical aracion beween he plaes is counered by he ension in he spring. Calculae he equilibrium separaion x beween he plaes as a funcion of he applied volage V. (Horrid word! We don say mereage for lengh, kilogrammage for mass or secondage for ime so why do we say volage for poenial difference and acreage for area? Ugh!) We should be able o use our invenion as a volmeer i even has an infinie resisance! Refer o figure V.. k a  x x FIGURE V.
11 We ll suppose ha he separaion when he poenial difference is zero is a, and he separaion when he poenial difference is V is x, a which ime he spring has been exended by a lengh a x. ε AV V The elecrical force beween he plaes is QE. Now Q = CV = and E =, x x ε AV so he force beween he plaes is. Here A is he area of each plae and i is x assumed ha he experimen is done in air, whose permiiviy is very close o ε. The ension in he sreched spring is k(a x), so equaing he wo forces gives us V kx ( a x) =. 5.. ε A Calculus shows [do i! jus differeniae x (  x)] ha V has a maximum value of V max of max 3 8ka = for a separaion x = 3 a. If we express he poenial difference in unis 7ε A V and he separaion in unis of a, equaion 5.. becomes V 7x ( ) = x In figure V. I have ploed he separaion as a funcion of he poenial difference. FIGURE V.
12 FIGURE V x/a V/V max As expeced, he poenial difference is zero when he separaion is or (and herefore you would expec i o go hrough a maximum for some inermediae separaion). We see ha for V < Vmax here are wo equilibrium posiions. For example, if V =.8, show ha x = or The quesion also arises wha happens if you apply across he plaes a poenial difference ha is greaer han V max? Furher insigh can be obained from energy consideraions. The poenial energy of he sysem is he work done in moving he upper plae from x = a o x = x while he poenial difference is V: ε AV ε AV E = + k( a ). a x x 5..4 You may need o refer o Secion 5.5 o be sure ha we have go his righ. If we express V in unis of V max, x in unis of a and E in unis of ka, his becomes E = 4 V ( / x) + ( x ) In figure V.3 I have ploed he energy versus separaion for hree values of poenial difference, 9% of V max, V max and % of V max.
13 3. FIGURE V.3 . V =.9 Poenial energy x We see ha for V < V max, here are wo equilibrium posiions, of which he lower one (smaller x) is unsable, and we see exacly wha will happen if he upper plae is displaced slighly upwards (larger x) from he unsable equilibrium posiion or if i is displaced slighly downwards (smaller x). The upper equilibrium posiion is sable. If V > V max, here is no equilibrium posiion, and x goes down o zero i.e. he plaes clamp ogeher. 5.3 Sharing a Charge Beween Two Capaciors Q C V C FIGURE V.4 We have wo capaciors. C is iniially uncharged. Iniially,C bears a charge Q and he poenial difference across is plaes is V, such ha
14 4 and he energy of he sysem is Q = C, 5.3. V V E = C We now close he swiches, so ha he charge is shared beween he wo capaciors: Q C V Q C V FIGURE V.5 The capaciors C and C now bear charges Q and Q such ha Q = Q + Q and Q C C = and Q = Q a,b C + C Q C + C The poenial difference across he plaes of eiher capacior is, of course, he same, so we can call i V wihou a subscrip, and i is easily seen, by applying Q = CV o eiher capacior, ha V C = V C + C We can now apply E = CV o each capacior in urn o find he energy sored in each. We find for he energies sored in he wo capaciors: 3 CV CC V E = and E. = 5.3.5a,b ( C + C ) ( C + C ) The oal energy sored in he wo capaciors is he sum of hese, which is which can also be wrien C V E =, ( C + C )
15 5 E C = E C + C Surprise, surprise! The energy sored in he wo capaciors is less han he energy ha was originally sored in C. Wha has happened o he los energy? A perfecly reasonable and no incorrec answer is ha i has been dissipaed as hea in he connecing wires as curren flowed from one capacior o he oher. However, i has been found in low emperaure physics ha if you immerse cerain meals in liquid helium hey lose all elecrical resisance and hey become superconducive. So, le us connec he capaciors wih superconducing wires. Then here is no dissipaion of energy as hea in he wires so he quesion remains: where has he missing energy gone? Well, perhaps he dielecric medium in he capaciors is heaed? Again his seems like a perfecly reasonable and probably no enirely incorrec answer. However, my capaciors have a vacuum beween he plaes, and are conneced by superconducing wires, so ha no hea is generaed eiher in he dielecric or in he wires. Where has ha energy gone? This will have o remain a mysery for he ime being, and a opic for lunchime conversaion. In a laer chaper I shall sugges anoher explanaion. 5.4 Mixed Dielecrics This secion addresses he quesion: If here are wo or more dielecric media beween he plaes of a capacior, wih differen permiiviies, are he elecric fields in he wo media differen, or are hey he same? The answer depends on. Wheher by elecric field you mean E or D;. The disposiion of he media beween he plaes i.e. wheher he wo dielecrics are in series or in parallel. Le us firs suppose ha wo media are in series (figure V.6). V Area = A FIGURE V.6 V ε E d ε D E d
16 6 Our capacior has wo dielecrics in series, he firs one of hickness d and permiiviy ε and he second one of hickness d and permiiviy ε. As always, he hicknesses of he dielecrics are supposed o be small so ha he fields wihin hem are uniform. This is effecively wo capaciors in series, of capaciances ε A/ d and ε A / d. The oal capaciance is herefore C εε A = ε d + ε d Le us imagine ha he poenial difference across he plaes is V. Specifically, we ll suppose he poenial of he lower plae is zero and he poenial of he upper plae is V. The charge Q held by he capacior (posiive on one plae, negaive on he oher) is jus given by Q = CV, and hence he surface charge densiy σ is CV /A. Gauss s law is ha he oal Dflux arising from a charge is equal o he charge, so ha in his geomery D = σ, and his is no alered by he naure of he dielecric maerials beween he plaes. Thus, in his capacior, D = CV /A = Q/A in boh media. Thus D is coninuous across he boundary. Then by applicaion of D = εe o each of he media, we find ha he Efields in he wo media are E = Q /( εa) and E = Q /( ε A), he Efield (and hence he poenial gradien) being larger in he medium wih he smaller permiiviy. The poenial V a he media boundary is given by V / d = E. Combining his wih our expression for E, and Q = CV and equaion 5.4., we find for he boundary poenial: V εd = V ε d + ε d Le us now suppose ha wo media are in parallel (figure V.7). FIGURE V.7 V d ε A A ε E D E D
17 7 This ime, we have wo dielecrics, each of hickness d, bu one has area A and permiiviy ε while he oher has area A and permiiviy ε. This is jus wo capaciors in parallel, and he oal capaciance is C ε A ε A = d d The Efield is jus he poenial gradien, and his is independen of any medium beween he plaes, so ha E = V/d. in each of he wo dielecrics. Afer ha, we have simply ha D = εe and D = εe. The charge densiy on he plaes is given by Gauss s law as σ = D, so ha, if ε < ε, he charge densiy on he lef hand porion of each plae is less han on he righ hand porion alhough he poenial is he same hroughou each plae. (The surface of a meal is always an equipoenial surface.) The wo differen charge densiies on each plae is a resul of he differen polarizaions of he wo dielecrics somehing ha will be more readily undersood a lile laer in his chaper when we deal wih media polarizaion. We have esablished ha:. The componen of D perpendicular o a boundary is coninuous;. The componen of E parallel o a boundary is coninuous. In figure V.8 we are looking a he Dfield and a he Efield as i crosses a boundary in which ε < ε. Noe ha D y and E x are he same on eiher side of he boundary. This resuls in: an θ ε = an θ ε D x E x ε θ θ D y ε D x /ε E y E x ε θ θ ε E y /ε D y FIGURE V.8
18 8 5.5 Changing he Disance Beween he Plaes of a Capacior If you gradually increase he disance beween he plaes of a capacior (alhough always keeping i sufficienly small so ha he field is uniform) does he inensiy of he field change or does i say he same? If he former, does i increase or decrease? The answer o hese quesions depends. on wheher, by he field, you are referring o he Efield or he Dfield;. on wheher he plaes are isolaed or if hey are conneced o he poles of a baery. We shall sar by supposing ha he plaes are isolaed. In his case he charge on he plaes is consan, and so is he charge densiy. Gauss s law requires ha D = σ, so ha D remains consan. And, since he permiiviy hasn changed, E also remains consan. The poenial difference across he plaes is Ed, so, as you increase he plae separaion, so he poenial difference across he plaes in increased. The capaciance decreases from εa/d o εa/d and he energy sored in he capacior increases from Adσ Adσ o. This energy derives from he work done in separaing he plaes. ε ε Now le s suppose ha he plaes are conneced o a baery of EMF V, wih air or a vacuum beween he plaes. A firs, he separaion is d. The magniudes of E and D are, respecively, V/d and ε V/d. When we have increased he separaion o d, he poenial difference across he plaes has no changed; i is sill he EMF V of he baery. The elecric field, however, is now only E = V/d and D = ε V/d. Bu Gauss s law sill dicaes ha D = σ, and herefore he charge densiy, and he oal charge on he plaes, is less han i was before. I has gone ino he baery. In oher words, in doing work by separaing he plaes we have recharged he baery. The energy sored in he capacior ε AV ε AV was originally ; i is now only. Thus he energy held in he capacior has d d been reduced by ε AV. d d ε AV The charge originally held by he capacior was. d ε AV been increased o d he charge held is. d The difference, Afer he plae separaion has ε AV d, is he d
19 9 charge ha has gone ino he baery. The energy, or work, required o force his amoun of charge ino he baery agains is EMF V is. ε AV d d Half of his came from he loss in energy held by he capacior (see above). The oher half presumably came from he mechanical work you did in separaing he plaes. Le s see if we can verify his. When he plae separaion is x, he force beween he plaes is QE, which is ε AV V ε. AV or. The work required o increase x from d o d is x x x ε AV d dx, which is indeed. d ε x AV d d Thus his amoun of mechanical work, plus an equal amoun of energy from he capacior, has gone ino recharging he baery. Expressed oherwise, he work done in separaing he plaes equals he work required o charge he baery minus he decrease in energy sored by he capacior. Perhaps we have invened a baery charger (figure V.9)! I x& x FIGURE V.9 AV When he plae separaion is x, he charge sored in he capacior is Q = ε. If x is x AV x increased a a rae x&, Q will increase a a rae Q& ε & =. Tha is, he capacior will x discharge (because Q & ε AV x& is negaive), and a curren I = will flow counerclockwise x in he circui. (Verify ha his expression is dimensionally correc for curren.)
20 5.6 Insering a Dielecric ino a Capacior Suppose you sar wih wo plaes separaed by a vacuum or by air, wih a poenial difference across he plaes, and you hen inser a dielecric maerial of permiiviy ε beween he plaes. Does he inensiy of he field change or does i say he same? If he former, does i increase or decrease? The answer o hese quesions depends. on wheher, by he field, you are referring o he Efield or he Dfield;. on wheher he plaes are isolaed or if hey are conneced o he poles of a baery. We shall sar by supposing ha he plaes are isolaed. See figure V.. σ Q = σa σ Q = σa V = σd/ε V = σd/ε ε ε D = σ E = D/ε + + D = σ + + E = D/ε + FIGURE V. Le Q be he charge on he plaes, and σ he surface charge densiy. These are unalered by he inroducion of he dielecric. Gauss s law provides ha D = σ, so his, oo, is unalered by he inroducion of he dielecric. The elecric field was, iniially, E = D. Afer inroducion of he dielecric, i is a lile less, namely E = D /. / ε ε Le us ake he poenial of he lower plae o be zero. Before inroducion of he dielecric, he poenial of he upper plae was V = σd. Afer inroducion of he dielecric, i is a lile less, namely V = σd /. ε / ε
21 Why is he elecric field E less afer inroducion of he dielecric maerial? I is because he dielecric maerial becomes polarized. We saw in Secion 3.6 how maer may become polarized. Eiher molecules wih preexising dipole momens align hemselves wih he imposed elecric field, or, if hey have no permanen dipole momen or if hey canno roae, a dipole momen can be induced in he individual molecules. In any case, he effec of he alignmen of all hese molecular dipoles is ha here is a sligh surplus of posiive charge on he surface of he dielecric maerial nex o he negaive plae, and a sligh surplus of negaive charge on he surface of he dielecric maerial nex o he posiive plae. This produces an elecric field opposie o he direcion of he imposed field, and hus he oal elecric field is somewha reduced. Before inroducion of he dielecric maerial, he energy sored in he capacior was QV. Afer inroducion of he maerial, i is QV, which is a lile bi less. Thus i will require work o remove he maerial from beween he plaes. The empy capacior will end o suck he maerial in, jus as he charged rod in Chaper araced an uncharged pih ball. Now le us suppose ha he plaes are conneced o a baery. (Figure V.) V V ε E = V/d D = ε E ε E = V/d D = εe FIGURE V. This ime he poenial difference remains consan, and herefore so does he Efield, which is jus V/d. Bu he Dfield increases from ε E o εe, and so, herefore, does he surface charge densiy on he plaes. This exra charge comes from he baery. ε A εa The capaciance increases from o and he charge sored on he plaes increases d d ε AV εav from Q o. = Q = The energy sored in he capacior increases from d d Q V o Q V.
22 The energy supplied by he baery = he energy dumped ino he capacior + he energy required o suck he dielecric maerial ino he capacior: ( Q Q ) V = ( Q Q ) V + ( Q Q ) V. You would have o do work o remove he maerial from he capacior; half of he work you do would be he mechanical work performed in pulling he maerial ou; he oher half would be used in charging he baery. In Secion 5.5 I invened one ype of baery charger. I am now going o make my forune by invening anoher ype of baery charger. Example. x& ε a x ε x d V FIGURE V. A capacior is formed of wo square plaes, each of dimensions a a, separaion d, conneced o a baery. There is a dielecric medium of permiiviy ε beween he plaes. I pull he dielecric medium ou a speed x&. Calculae he curren in he circui as he baery is recharged. Soluion. When I have moved a disance x, he capaciance is εa( a x) εax εa ( ε ε) ax + = d d d. The charge held by he capacior is hen Q = εa ( ε ε) ax V. d
23 3 If he dielecric is moved ou a speed x&, he charge held by he capacior will increase a a rae Q& = ( ε ε ) axv & d. (Tha s negaive, so Q decreases.) A curren of his magniude herefore flows clockwise around he circui, ino he baery. You should verify ha he expression has he correc dimensions for curren. Example. x& ε d x V ε x FIGURE V.3 A capacior consiss of wo plaes, each of area A, separaed by a disance x, conneced o a baery of EMF V. A cup ress on he lower plae. The cup is gradually filled wih a nonconducing liquid of permiiviy ε, he surface rising a a speed x&. Calculae he magniude and direcion of he curren in he circui. I is easy o calculae ha, when he liquid has a deph x, he capaciance of he capacior is εε A C = εd ( ε ε ) x
24 4 and he charge held by he capacior is hen Q = εd εε AV ( ε ε ) x. If x is increasing a a rae x&, he rae a which Q, he charge on he capacior, is increasing is Q& = εε( ε ε) AV x& [ εd ( ε ε ) x]. A curren of his magniude herefore flows in he circui counerclockwise, draining he ε ( ε ε) AV x& baery. This curren increases monoonically from zero o. ε d 5.7 Polarizaion and Suscepibiliy When an insulaing maerial is placed in an elecric field, i becomes polarized, eiher by roaion of molecules wih preexising dipole momens or by inducion of dipole momens in he individual molecules. Inside he maerial, D is hen greaer han ε E. Indeed, D = εe + P The excess, P, of D over ε E is called he polarizaion of he medium. I is dimensionally similar o, and expressed in he same unis as, D; ha is o say C m. Anoher way of looking a he polarizaion of a medium is ha i is he dipole momen per uni volume. In vecor form, he relaion is If he medium is isoropic, all hree vecors are parallel. D = εe + P Some media are more suscepible o becoming polarized in a polarizing field han ohers, and he raio of P o ε E is called he elecric suscepibiliy χ e of he medium: P = χ ε E e This implies ha P is linearly proporional o E bu only if χ e is independen of E, which is by no means always he case, bu is good for small polarizaions.
25 5 When we combine equaions 5.7. and wih D = εe and wih εr = ε / ε, he relaive permiiviy or dielecric consan, we obain χ = ε e r 5.8 Discharging a Capacior Through a Resisor C + I = Q& R FIGURE V.4 Wha you have o be sure of in his secion and he following secion is o ge he signs righ. For example, if he charge held in he capacior a some is Q, hen he symbol Q &, or dq / d, means he rae of increase of Q wih respec o ime. If he capacior is discharging, Q & is negaive. Expressed oherwise, he symbol o be used for he rae a which a capacior is losing charge is Q &. In figure V.4 a capacior is discharging hrough a resisor, and he curren as drawn is given by I = Q&. The poenial difference across he plaes of he capacior is Q/C, and he poenial difference across he resisor is IR = QR &. Thus: Q Q IR = + Q & R =. C C 5.8. On separaing he variables (Q and ) and inegraing we obain Q dq Q = RC Q d, 5.8. where Q is he charge in he capacior a =. /( RC ) Hence Q = Q e
26 6 Here RC is he ime consan. (Verify ha i has he dimensions of ime.) I is he ime for he charge o be reduced o /e = 36.8% of he iniial charge. The half life of he charge is RC ln =.693RC. 5.9 Charging a Capacior Through a Resisor E I = Q& + C FIGURE V.5 This ime, he charge on he capacior is increasing, so he curren, as drawn, is + Q &. Thus Q E Q & R = C Whence: Q dq = E C Q RC d Q dq [Noe: Don be emped o wrie his as = d. Remember ha, Q E C RC a any finie, Q is less han is asympoic value E C, and you wan o keep he denominaor of he lef hand inegral posiive.] Upon inegraing, we obain Q /(RC) ( e ). = E C Thus he charge on he capacior asympoically approaches is final value E C, reaching 63% ( e ) of he final value in ime RC and half of he final value in ime RC ln =.693 RC. The poenial difference across he plaes increases a he same rae. Poenial difference canno change insananeously in any circui conaining capaciance.
27 7 How does he curren change wih ime? This is found by differeniaing equaion /(RC) wih respec o ime, o give I = E e. This suggess ha he curren grows R insananeously from zero o E / R as soon as he swich is closed, and hen i decays exponenially, wih ime consan RC, o zero. Is his really possible? I is possible in principle if he inducance (see Chaper ) of he circui is zero. Bu he inducance of any closed circui canno be exacly zero, and he circui, as drawn wihou any inducance whaever, is no achievable in any real circui, and so, in a real circui, here will no be an insananeous change of curren. Chaper Secion.5 will deal wih he growh of curren in a circui ha conains boh capaciance and inducance as well as resisance. Energy consideraions When he capacior is fully charged, he curren has dropped o zero, he poenial difference across is plaes is E (he EMF of he baery), and he energy sored in he capacior (see Secion 5.) is E = QE. Bu he energy los by he baery is Q E. Le us hope ha he remaining C QE is hea generaed in and dissipaed by he resisor. The rae a which hea is generaed by curren in a resisor (see Chaper 4 Secion 4.6) is I R. In his case, according o he previous paragraph, he curren a ime is /(RC) I = E E /( RC) e, so he oal hea generaed in he resisor is e R R = so all is well. The energy los by he baery is shared equally beween R and C. CE, Neon lamp Here s a way of making a neon lamp flash periodically. In figure V. 5 (sorry abou he fracion I slipped he figure in as an aferhough!), he hings ha looks somehing like a happy face on he righ is a discharge ube; he do inside i indicaes ha i s no a complee vacuum inside, bu i has a lile bi of gas FIGURE V. 5
28 8 inside. I will discharge when he poenial difference across he elecrodes is higher han a cerain hreshold. When an elecric field is applied across he ube, elecrons and posiive ions accelerae, bu are soon slowed by collisions. Bu, if he field is sufficienly high, he elecrons and ions will have enough energy on collision o ionize he aoms hey collide wih, so a cascading discharge will occur. The poenial difference rises exponenially on an RC imescale unil i reaches he hreshold value, and he neon ube suddenly discharges. Then i sars all over again. Problem Here is a problem ha will give pracice in charging a capacior, applying Kirchhoff s rules, and solving differenial equaions. I R I 3 R E I R C In he above circui, while he swich is open, I = I = E /(R) and I 3 =. This will also be he siuaion long afer he swich is closed and he capacior is charged. Bu we wan o invesigae wha happens in he brief momens while he capacior is being charged. And wha will be he final charge in he capacior? We apply Kirchhoff s rules: E = R I R I + I R + Q C I R / = I = +, I I3 Here Q is he charge on he capacior a some ime. Eliminae I and I o ge a single equaion in I 3. Q E = + 3I3R C
29 9 Bu dq I 3 =, so we have differenial equaion in Q and he ime. d dq d E + Q = RC 3R dy This is of he form + ay = b, and hose experienced wih differenial equaions will dx have no difficuly in arriving a he soluion Q C + Ae 3RC = E Wih he iniial condiion ha Q = when =, his becomes Thus he final charge in he capacior is E C e = Q 3 RC E C. The curren I 3 is found by differeniaing equaion 5.9. wih respec o ime, and he oher currens are found from Kirchhoff s rules (equaions ). I make hem: E + R I = e 3 RC E R I = e 3 RC 3 I = E e 3RC R Thus I goes from iniially E o finally 3R E. R I goes from iniially E o finally 3R E. R E I 3 goes from iniially o finally. 3R E E Before he swich was opened, hese currens were, and zero respecively. R R Readers migh wonder wheher he currens can change insananeously as soon as he
30 3 swich is closed. The answer is yes, provided ha he circui has no inducance (see Chaper, especially Secions.5, which deal wih he growh of curren in a circui ha has inducance). In pracice no circui can be enirely free from inducance; apar from he inducance of any circui componens, any circui ha forms a closed loop (as all circuis mus) mus have a small inducance. The inducance may be very small, which means ha he change of curren a he insan when he swich is closed is very rapid. I is no, however, insananeous. Here are graphs of he currens and of Q as a funcion of ime. Currens are expressed in 3 unis of E /R, Q in unis of E C, and ime in unis of RC..7.6 I.5 I.4 Q.3. I There is a similar problem involving an inducor in Chaper, Secion.. Inegraing and differeniaing circuis. We look now a wha happens if we connec a resisor and a capacior in series across a volage source ha is varying wih ime, and we shall show ha, provided some condiions are saisfied, he poenial difference across he capacior is he ime inegral of he inpu volage, while he poenial difference across he resisor is he ime derivaive of he inpu volage.
31 3 We have seen ha, if we connec a resisor and a capacior in series wih a baery of EMF E, he charge in he capacior will increase according o Q = E C e RC, asympoically approaching Q = E C, and reaching e =. 63 of his value in ime RC. Noe ha, when << RC, he curren will be large, and he charge in he capacior will be small. Mos of he poenial drop in he circui will be across he resisor, and relaively lile across he capacior. Afer a long ime, however, he curren will be low, and he charge will be high, so ha mos of he poenial drop will be across he capacior, and relaively lile across he resisor. The poenial drops across R and C will be equal a a ime = RC ln =.693RC. Suppose ha, insead of connecing R and C o a baery of consan EMF, we connec i o a source whose volage varies wih ime, V (). How will he charge in C vary wih ime? V = V () R C V C The relevan equaion is V = IR + Q/ C, in which I, Q and V are all funcions of ime. Since I = Q&, he differenial equaion showing how Q varies wih ime is dq d V + Q = RC R The inegraion of his equaion is made easy if we muliply boh sides by e RC. (Those who are experienced in solving differenial equaions will readily hink of his sep. Those who are less experienced migh no immediaely hink of i, bu will soon see ha i is a useful sep.) We hen obain e RC dq RC d RC V + e Q = Qe e RC d RC d = R Thus he answer o our quesion is Q RC RC e = Ve d R
32 3 If V = E and is independen of ime, his reduces o he familiar Q = E C e RC. The poenial difference across C increases, of course, as V C RC e RC = Ve d RC While is very much shorer han he ime consan RC, by which I mean shor enough ha e RC is very close o, his becomes V C = Vd RC Tha is why his circui is called an inegraing circui. The oupu volage across C is /( RC ) imes he ime inegral of he inpu volage V. This is also rue if he inpu volage is a periodic funcion of ime wih a period ha is very much shorer han he ime consan. By way of example, suppose ha V = a. If we pu his in he righ hand side of equaion and inegrae, wih iniial condiion V C = when =, (do i!), we obain V = + RC C ar C e R C RC For example, suppose he inpu volage varies as V = 5 vols, where is in seconds. If R = 5 Ω and C = 4 µf, wha will be he poenial differnece across he capacior afer.3 s? We immediaely see ha RC =. s and /(RC) =.5. Subsiue SI numbers in equaion o obain V C =.6 V. If I wrie y = VC ar C and becomes x = equaion in dimensionless form RC y x = x x + e If you Taylor expand his as far as x 3 (do i!), you ge y = 3 x 3, which is jus wha you would ge by using equaion 5.9.8, he equaion which is an approximaion for a ime ha is shor compared wih RC. The approximaion is good as long as 4 RC is
33 33 negligible. I show equaion 5.9. and y = x in he graph below, in which V C is in unis of C ar and T is in unis of RC V C.5..5 approximae exac ime Equaion (or, for shor ime inervals, equaion 5.9.8) gives us he volage across C as a funcion of ime. Wha abou he volage across R? Tha is evidenly V R RC RC e = V Ve d RC Differeniae wih respec o ime: dv d R dv e RC = e RCVe RC Ve RC d d RC RC dv dv VR = ( V VC ) = d RC d RC 5.9.
34 34 If he ime consan is small so ha dv R VR <<, his becomes d RC dv V R = RC, 5..3 dr so ha he volage across R is RC imes he ime derivaive of he inpu volage V. Thus we have a differeniaing circui. Noe ha, in he inegraing circui, he circui mus have a large ime consan (large R and C) and ime variaions in V are rapid compared wih RC. The oupu volage across C is hen Vd. In he differeniaing circui, he circui mus have a small ime RC consan, and ime variaions in V are slow compared wih RC. The oupu volage across dv R is hen. dr 5. Real Capaciors Real capaciors can vary from huge meal plaes suspended in oil o he iny cylindrical componens seen inside a radio. A grea deal of informaion abou hem is available on he Web and from manufacurers caalogues, and I only make he briefes remarks here. A ypical inexpensive capacior seen inside a radio is nohing much more han wo srips of meal foil separaed by a srip of plasic or even paper, rolled up ino a cylinder much like a Swiss roll. Thus he separaion of he plaes is small, and he area of he plaes is as much as can be convenienly rolled ino a iny radio componen. In mos applicaions i doesn maer which way round he capacior is conneced. However, wih some capaciors i is inended ha he ouermos of he wo meal srips be grounded ( earhed in UK erminology), and he inner one is shielded by he ouer one from sray elecric fields. In ha case he symbol used o represen he capacior is The curved line is he ouer srip, and is he one ha is inended o be grounded. I should be noed, however, ha no everyone appears o be aware of his convenion or adheres o i, and some people will use his symbol o denoe any capacior. Therefore care mus be aken in reading he lieraure o be sure ha you know wha he wrier inended, and, if you are describing a circui yourself, you mus make very clear he inended meaning of your symbols. There is a ype of capacior known as an elecrolyic capacior. The wo plaes are srips of aluminium foil separaed by a conducing pase, or elecrolye. One of he foils is covered by an exremely hin layer of aluminium oxide, which has been elecrolyically
35 35 deposied, and i is his layer han forms he dielecric medium, no he pase ha separaes he wo foils. Because of he exreme hinness of he oxide layer, he capaciance is relaively high, alhough i may no be possible o conrol he acual hickness wih grea precision and consequenly he acual value of he capaciance may no be known wih grea precision. I is very imporan ha an elecrolyic capacior be correced he righ way round in a circui, oherwise elecrolysis will sar o remove he oxide layer from one foil and deposi i on he oher, hus grealy changing he capaciance. Also, when his happens, a curren may pass hrough he elecrolye and hea i up so much ha he capacior may burs open wih consequen danger o he eyes. The symbol used o indicae an elecrolyic capacior is: + The side indicaed wih he plus sign (which is ofen omied from he symbol) is o be conneced o he posiive side of he circui. When you une your radio, you will usually find ha, as you urn he knob ha changes he wavelengh ha you wan o receive, you are changing he capaciance of a variable airspaced capacior jus behind he knob. A variable capacior can be represened by he symbol Such a capacior ofen consiss of wo ses of inerleaved pariallyoverlapping plaes, one se of which can be roaed wih respec o he oher, hus changing he overlap area and hence he capaciance. Thinking abou his suggess o me a couple of small problems for you o amuse yourself wih. Problem. FIGURE V.6 d A capacior (figure V.6) is made from wo ses of four plaes. The area of each plae is A and he spacing beween he plaes in each se is d. The wo ses of plaes are inerleaved, so ha he disance beween he plaes of one se and he plaes of he oher is d. Wha is he capaciance of he sysem? Problem FIGURE V.7 d
36 36 This is jus like Problem, excep ha one se has four plaes and he oher has hree. Wha is he capaciance now? Soluions. The answer o he firs problem is 7ε A/d and he answer o he second problem is 6ε A/d bu i isn good enough jus o asser ha his is he case. We mus give some reasons. Le us suppose ha he poenial of he lefhand (blue) plaes is zero and he poenial of he righhand (blue) plaes is V. The elecric field in each space is V/d and D = ε V/d. The surface charge densiy on each plae, by Gauss s heorem, is herefore ε V/d excep for he wo end plaes, for which he charge densiy is jus ε V/d. The oal charge held in he capacior of Problem is herefore ε AV/d + 3 ε AV/d = 7ε AV/d, and he capaciance is herefore 7ε A/d. For Problem, he blue se has wo endplaes and wo middleplaes, so he charge held is ε AV/d + ε AV/d = 6ε AV/d. The red se has hree middle plaes and no endplaes, so he charge held is 3 ε AV/d = 6ε AV/d. The capaciance is herefore 6ε A/d. 5. More on E, D, P, ec. I ll review a few hings ha we have already covered before going on. The elecric field E beween he plaes of a plane parallel capacior is equal o he poenial gradien i.e. he poenial difference beween he plaes divided by he disance beween hem. The elecric field D beween he plaes of a plane parallel capacior is equal o he surface charge densiy on he plaes. Suppose a firs here is nohing beween he plaes. If you now hrus an isoropic* dielecric maerial of relaive permiiviy ε r beween he plaes, wha happens? Answer: If he plaes are isolaed D remains he same while E (and hence he poenial difference across he plaes) is reduced by a facor ε r. If on he oher hand he plaes are conneced o a baery, he poenial difference and hence E remains he same while D (and hence he charge densiy on he plaes) increases by a facor ε r. *You will have noiced he word isoropic here. Refer o Secion.7 for a brief menion of an anisoropic medium, and he concep of permiiviy as a ensor quaniy. I m no concerned wih his aspec here.
37 37 In eiher case, he block of dielecric maerial becomes polarized. I develops a charge densiy on he surfaces ha adjoin he plaes. The block of maerial develops a dipole momen, and he dipole momen divided by he volume of he maerial i.e. he dipole momen per uni volume is he polarizaion P of he maerial. P is also equal o D εe and, of course, o ε ε E. The raio of he resuling polarizaion P o he E polarizing field ε E is called he elecric suscepibiliy χ of he medium. I will be worh spending a few momens convincing yourself from hese definiions and conceps ha ε = ε ( + χ) and χ = ε r, where ε r is he dimensionless relaive permiiviy (or dielecric consan) ε / ε. Wha is happening physically inside he medium when i becomes polarized? One possibiliy is ha he individual molecules, if hey are asymmeric molecules, may already possess a permanen dipole momen. The molecule carbon dioxide, which, in is ground sae, is linear and symmeric, O=C=O, does no have a permanen dipole momen. Symmeric molecules such as CH 4, and single aoms such as He, do no have a permanen dipole momen. The waer molecule has some elemens of symmery, bu i is no linear, and i does have a permanen dipole momen, of abou 6 % 3 C m, direced along he bisecor of he HOH angle and away from he O aom. If he molecules have a permanen dipole momen and are free o roae (as, for example, in a gas) hey will end o roae in he direcion of he applied field. (I ll discuss ha phrase end o in a momen.) Thus he maerial becomes polarized. A molecule such as CH 4 is symmeric and has no permanen dipole momen, bu, if i is placed in an exernal elecric field, he molecule may become disored from is perfec erahedral shape wih nea 9º angles, because each pair of CH aoms has a dipole momen. Thus he molecule acquires an induced dipole momen, and he maerial as a whole becomes polarized. The raio of he induced dipole momen p o he polarizing field E polarizabiliy α of he molecule. Review Secion 3.6 for more on his. How abou a single aom, such as Kr? Even ha can acquire a dipole momen. Alhough here are no bonds o bend, under he influence of an elecric field a preponderance of elecrons will migrae o one side of he aom, and so he aom acquires a dipole momen. The same phenomenon applies, of course, o a molecule such as CH 4 in addiion o he bond bending already menioned. Le us consider he siuaion of a dielecric maerial in which he molecules have a permanen dipole momen and are free (as in a gas, for example) o roae. We ll suppose ha, a leas in a weak polarizing field, he permanen dipole momen is significanly larger han any induced dipole momen, so we ll neglec he laer. We have said ha, under he influence of a polarizing field, he permanen dipole will end o align hemselves wih he field. Bu hey also have o conend wih he consan josling and collisions beween molecules, which knock heir dipole momens haywire, so hey can immediaely all align exacly wih he field. We migh imagine ha he maerial may become fairly srongly polarized if he emperaure is fairly low, bu only relaively
38 38 weakly polarized a higher emperaures. Dare we even hope ha we migh be able o predic he variaion of polarizaion P wih emperaure T? Le s have a go! We recall (Secion 3.4) ha he poenial energy U of a dipole, when i makes an angle θ wih he elecric field, is U = pe cos θ = p E. The energy of a dipole whose direcion makes an angle of beween θ and θ + d θ wih he field will be beween U and U + du, where du = pe sin θdθ. Wha happens nex requires familiariy wih Bolzmann s equaion for disribuion of energies in a saisical mechanics. See for example my Sellar Amospheres noes, Chaper 8, Secion 8.4. The fracion of molecules having energies beween U and U + du will be, following Bolzmann s equaion, e U /( kt ) du + pe U /( kt ) e pe du 5.. (Cauion: Remember ha here I m using U for poenial energy, and E for elecric field.) Tha is, he fracion of molecules making angles of beween θ and is θ + d θ wih he field pee π pee pe cos θ /( kt ) pe cos θ /( kt ) sin θdθ sin θdθ = e π pe cos θ /( kt ) e pe cos θ /( kt ) sin θdθ sin θdθ. 5.. The componen in he direcion of E of he dipole momen of his fracion of he molecules is pe pe cos θ /( kt ) π e pe cos θ /( kt ) sin θcos θdθ sin θdθ, 5..3 so he componen in he direcion of E of he dipole momen all of he molecules is π p e pe cos θ /( kt ) π e pe cosθ /( kt ) sin θcos θdθ sin θdθ, 5..4 and his expression represens he induced dipole momen in he direcion of he field of he enire sample, which I ll call p s. The polarizaion of he sample would be his divided by is volume. Le pe x = cos θ = a cos θ kt
39 39 Then he expression for he dipole momen of he enire sample becomes (some care is needed): + a xe dx a a a e + e p = = s p p a x a a a e dx e e a a x The expression in parenheses is called he Langevin funcion, and i was firs derived in connecion wih he heory of paramagneism. If your calculaor or compuer suppors he hyperbolic coh funcion, i is mos easily calculaed as coh a /a. If i does no + suppor coh, calculae i as b a, where b = e. In any case i is a raher b a ineresing, even challenging, funcion. Le us call he expression in parenheses f(a). Wha would he funcion look like i you were o plo f(a) versus a? The derivaive wih 4b respec o a is. I is easy o see ha, as a, he funcion approaches a ( b) and is derivaive, or slope, approaches zero. Bu wha are he funcion and is derivaive (slope) a a =? You may find ha a bi of a challenge. The answer is ha, as a, he funcion approaches zero and is derivaive approaches /3. (In fac, for small a, he a Langevin funcion is approximaely, and for very small a, i is a.) Thus, for 3 3( a) pe small a (i.e. ho emperaures) p s approaches p and no higher. The Langevin 3kT funcion looks like his:
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