Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx of 3 Last saved: /7/8; 8:4 PM Homework: See webste. Table of ontents: h.6. Defnton of apactance, 6. alculatng apactance, 6.a Parallel Plate apactor, 6.b ylndrcal apactor, 3 6.c Sphercal apactor, 4 6.3 ombnatons of apactors n Parallel and Seres, 4 6.4 Energy Stored n a harged apactor, 5 6.5 apactors Wth Delectrcs, 6 6.6 Electrc Dpole n an Electrc Feld, 8 6.7 Detals about delectrcs,
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx of 3 Last saved: /7/8; 8:4 PM h.6 Defnton of apactance. The capactance s the factor of proportonalty between voltage and charge. The hgher the capactance the more charge can be put on a devce called the capactor. (6.) n the unts of oulomb olt Farad We have already calculated the electrc feld between two parallel plane conductors charged wth equal and opposte charges. As we are dealng wth a conductor, the electrons put on one plate wll push the electrons on the opposte plate away (they flow nto the battery.) As a result we have a constant electrc feld between the two plates of the capactor. 6.a Parallel Plate apactor. We can easly fnd the electrc feld by usng the now famlar Gauss law. hoose the Gaussan surface to be the dashed rectangle: dved EdA d Δ - + -=-σa +=+σa volume Gauss volume surface EA A A ; hence E= B Eds Ed (6.) We see that the capactance of a parallel plate capactor s proportonal to the area dvded by the dstance between the plates. The surfaces wth opposte charges are called plates. The whole arrangement s called a capactor, n our case a parallel plate capactor. It s used to store electrc charge and energy. A new quantty s beng used n ths context, whch s the rato between the postve charge and the postve voltage dfference. Ths rato (always postve) s called capactance, and s abbrevated wth the symbol (n talc letterng, n contrast to oulomb whch s a normal nontalc ) The capactance depends only on the geometry of the devce. Δ s proportonal to the charge, and therefore drops out of the rato /Δ. The capactance depends only on the electrc constant and the geometry of the capactor. d A A d
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 3 of 3 Last saved: /7/8; 8:4 PM (6.) (6.) oulomb n the unts of Farad olt Farad F ; 6 mcro Farad F F; 9 pco Farad pf F ylndrcal and sphercal capactors: Let us next calculate the capactance of a cylndrcal capactor and a sphercal capactor. In both cases we have an outer and nner charge. In the case of the cylnder we have an nner cylnder wth radus a and an outer cylnder wth radus b. The thckness of each cylndrcal surface s neglgble. Let us calculate the electrc feld usng Gauss law, and then obtan the potental dfference through ntegraton. The electrc feld s perpendcular to the cylndrcal surfaces. It s not constant. Therefore, the calculaton of Δ leads to an ntegral. al a EdA E rl; E Gaussan EA r (6.3) surface al -σ +σ So we need : b (6.4) b a a b al L L dr ln a r a b b b a ln ln ke ln a a a 6.b apactance of a ylndrcal apactor: L L (6.5) b b ke ln ln a a We could have also calculated the charge by usng =λl. Note: the relatonshp between lne-charge λ, surface charge σ, and volume charge ρ can be obtaned by realzng that: (6.6) l A A capactor lke ths can be found n a coaxal cable, whch conssts of two concentrc cylndrcal conductors separated by an nsulator.
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 4 of 3 Last saved: /7/8; 8:4 PM 6.c apactance of two conductng spheres: (The total charge on one surface s ) For the electrc feld between two conductng sphercal shells we get E ke r (6.7) b dr a b k k k r b a ab e e e a Takng the absolute value, we re-arrange to k The charge cancels out and we get =/Δ ab (6.8) k b a e e b a ab If the radus of the outer sphere goes to nfnty we get the capactance of a smple sphere. (a becomes neglgble n the denomnator, then b cancels out): ab a (6.9) lm 4 a b k b a k a (6.) 4 a k 6.3 ombnatons of apactors n Parallel and Seres. e e Two or more capactors n parallel: The left sdes of the capactors are all connected to the same conductng wre, whch means they are at the same voltage, the same s true for the rght sde of the capactors. Ths means that all capactors n parallel have the same voltage dfference but ther total charge s the charge on all capactors combned. (6.) The equvalent capactor s one sngle capactor whch has the same capactance as all the capactors combned n parallel. As the postve plates are connected by a conductng wre the total charge on the two plates s equal to the sum of the charges. = + The exteror voltage s provded by a battery. e ; + ; equvalent (6.) ; = + - + - Δ Δ
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 5 of 3 Last saved: /7/8; 8:4 PM All capactors are at the same voltage:... The total charge s the sum of the charges of the ndvdual capactors: = The equvalent capactance s defned through: = = eq eq eq The equvalent capactance of a number of capactors n parallel s the sum of all capactances. (6.3) eq When we put a number of capactors n seres nsde of a crcut, we can see mmedately that now the charge on every capactor plate s the same and the total voltage dfference across all the capactors s the sum of the ndvdual voltage dfferences of all capactors. - + - + - + - + Δ Δ Δ 3 - + Δ = - + Δ Δ The total voltage dfference s the sum of the ndvdual voltage dfferences. Each voltage dfference has the same charge. (6.4) (6.5) eq eq The nverse equvalent capactance of a number of capactors n seres s the sum of all nverse capactances. 6.4 Energy Stored n a harged apactor. Recall our defntons for work and capactance n the case of a voltage across a parallel plate capactor: Note: we use the followng conventon: (captalzed) s the charge on the capactor, q (small cap) s the charge we move aganst the potental dfference Δ, whch n a capactor s equal to /. For a sngle charge q to be added to the charge, already present on a capactor wth Δ between the plates, the battery, chargng the capactor must do work: (6.6) W q ; ; W q
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 6 of 3 Last saved: /7/8; 8:4 PM Let us change the sngle charge nto a small (nfntesmal) amount of charge dq. In order to add ths small amount of charge dq to the postve plate of a capactor wth the varable charge q on t we have to do nfntesmal work aganst the electrc feld (aganst the potental dfference Δ) nsde of the capactor. We ncrease the charge q from to. q dw dq dq (6.7) W across the capactor q dq Ths work done to the capactor appears as electrc potental energy U of the capactor: (6.8) U ; wth = (Note: the work done aganst the electrc feld s equal to the potental energy of the electrc feld.) For a parallel plate capactor we found that. The potental dfference s smply d Ed. ombnng ths we get for the potental energy of a parallel plate capactor: A (6.9) U Ed E Ad d Thus, we can defne the electrc energy densty u E nsde the space between the plates of capactors as: U (6.) ue E ; u E volume contanng E A volume Ths result s correct for the electrc energy densty n any electrc feld. 6.5 apactors Wth Delectrcs. By fllng the space between the plates of a capactor wth non-conductng, so-called delectrc materal we can ncrease the capactance. (Wood, glass, paper) As s defned as /Δ the effect of the delectrc s to reduce the potental dfference and the electrc feld strength and ncrease the capactance.. The charge remans unaffected, as we can see from the pcture below. If we call the capactance wthout the delectrc then the capactance wth the delectrc s gven by: (6.) ; ( kapa ) delectrc constant > The capactance for a parallel plate capactor wth a delectrc s: A A A E d d d
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 7 of 3 Last saved: /7/8; 8:4 PM + - -σ nduced +E nd +σ nduced The delectrc strength n olts/m of a delectrc gves the maxmum voltage per unt length (of the thckness of the delectrc materal), beyond whch the delectrc materal becomes conductve. The orgnal electrc feld from + to -, nduces an opposte electrc feld E nd across the delectrc materal, whch reduces the orgnal electrc feld E. Example: We use a parallel plate capactor wth dmensons cm by 3cm, separated by. mm layer of paper wth a delectrc constant of 3.7. Its delectrc strength s 6 6 /m. The capactance s 4 A 3.7 6 m (6.) pf; wth 3.7 3 d m Fnd the maxmum charge that can be put on such a capactor. max max As the thckness of the delectrc smm, the maxmum voltage whch we can put on ths capactor s gven by 6 3 (6.3) Delectrc strength=6 6 6, m mm equal to 6 3 olts. 3 9 (6.4) max pf 6 olts 3 oulombs.3 oulombs The ntroducton of a delectrc materal nto a capactor decreases the orgnal feld strength, both and E, but leaves the charge unaffected: (6.5) Ed E d (6.6) E and E. alculate the energy nsde of a capactor wth the delectrc substance nserted: We saw n (6.8) that the potental energy of a capactor s equal to ½ / U (6.7) U Ths means that the total energy of a capactor goes down by / Ths was to be expected because What happens when we nsert some molecules nto the nsde of a capactor wth an electrc feld? Provded that the molecules have an asymmetrcal charge dstrbuton, they form a natural dpole. The dpoles experence a torque due to the electrc feld and get lned up wth the electrc feld lnes. In order to understand better what happens n such stuatons we need to study the dpole nserted nto an electrc feld. (See ch5 electrc potental.docx page 3.
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 8 of 3 Last saved: /7/8; 8:4 PM 6.6 Electrc Dpole n an Electrc Feld. Earler, we defned the concept of the electrc dpole moment: (the word moment s a hstorcal use of a length. For example: the magntude of the moment of nerta s mass tmes dstance squared, the dpole moment s charge tmes dstance.). Thnk of a dpole as a stck of length d=a wth a postve charge on one end and a negatve charge on the other end. (6.8) p aqur qd u s the unt vector pontng from the negatve to the postve charge q. r and "a" s the half-dstance between the two charges. Note that ths drecton s opposte to that of an electrc feld whch s drected from + to -. If we nsert a dpole nto an electrc feld, the electrc feld wll exert a force qe on the postve charge and a force qe on the negatve charge. As the charges are bound to the dpole structure (molecule), the electrc feld wll produce a torque around the center of the dpole. F qe P qd +q -q E θ F qe You can see that the torque attempts to rotate the dpole untl t s lned up wth the electrc feld, whch s the most stable equlbrum poston, correspondng to the mnmum amount of potental energy of the dpole-electrc feld system. r F; r ponts from the center of rotaton to the charge. (6.9) a qe; a q E Both torques have the same drecton (nto the plane) and produce a resultant torque, whch tres to lne up the dpole wth the electrc feld, such that the vector orentaton of the dpole s parallel to the exteror electrc feld. (6.3) a qe p E If an outsde agent rotates the dpole nsde of the feld t wll have do work, whch s equal to the potental energy change of the dpole n the feld (Note the changed orentaton of the dpole n the fgure below, as compared to the prevous fgure!):
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 9 of 3 Last saved: /7/8; 8:4 PM W done on the dpole by the outsde agent U U U d pe sn d (6.3) pe cos pe cos cos -q P qd +q θ E It s convenent f we choose our reference angle n such a way that the potental energy equals when the drecton of the dpole s perpendcular to the electrc feld, whch means we choose our reference angle as π/. That choce allows us to defne the potental energy of the dpole nsde of an electrc feld as a scalar product, and ensures that the poston of mnmum energy corresponds to that, n whch the drecton of the dpole and the electrc feld are parallel. (6.3) Udpole Ths potental energy has a mnmum when the dpole and the electrc feld pont n the same drecton (θ=º) and s equal to p E. It has a maxmum when the vectors are ant-parallel, and U=+pE. It takes a maxmum effort of energy for an outsde agent to twst the dpole from ts parallel poston, to ts antparallel poston. The maxmum change of the dpole potental energy s pe, correspondng to the rotaton of the dpole from an algned poston (-pe) to an ant parallel poston +pe. U U U pe ( pe) pe f The mnmum value of the energy of the dpole-electrc feld system corresponds to -pe, (the two vectors p and E are parallel, whch s shown n the next fgure on the left. As the potental energy of the dpole-e system s U p E parallel vectors means mnmum potental energy. In a manner of speakng: The postve charge of the dpole s held n place by the negatve end of the electrc feld vector, and the negatve charge of the dpole s held n place by the postve end of the electrc feld vector. The mnmum energy poston corresponds to the natural algnment between dpole and electrc feld. p E
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx of 3 Last saved: /7/8; 8:4 PM p E U p E mnmum U p ant-parallel to E U p E maxmum U When the outsde agent releases the dpole, the electrc feld rotates the dpole back nto the algned poston wth mnmum potental energy -pe. ompare ths to mgy whch s the maxmum potental energy of a mass lfted up from to a heght y. When you release the mass t wll be forced back by the downward gravtatonal force towards ts poston wth mnmum potental energy. The relatonshp between a conservatve force and ts potental energy can be generalzed to the relatonshp between a torque and ts potental energy. It s ths generalzed concept whch dctates the defnton of the dpole s drecton from to +. U p E pe cos (6.33) U p E; = pe sn ; ' ' We defned U= wth reference to a poston n whch the dpole and the electrc feld are perpendcular to each other. A rotaton from ths poston nvolves the angle θ whch s complementary to the angle θ between p and E as defned n the torque. Dpoles occur naturally n molecules, when the negatve charges are not symmetrcally dstrbuted around a net central postve charge. Such asymmetrc molecules have permanent dpole moments due to an excess of electrons on one end of the molecule, and a correspondng defct on the other. Ths means that the molecule has a negatve charge on one end (surplus electrons) and a postve charge on the other (defct of electrons). The typcal example occurs n a water molecule, n whch the two hydrogen atoms are arranged around the oxygen atom at a 5 degree angle, thus producng what s called a polarzed molecule wth a permanent dpole moment of 6.3x -3 oulomb-meter. The electrons tend to concentrate on the oxygen atom. Usng ths example, let us calculate how much energy s requred to rotate all molecules n one mole of water from an algned poston nto a poston perpendcular to the electrc feld of magntude 5k/m.
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx of 3 Last saved: /7/8; 8:4 PM Permanent dpole moment of a water molecule for one molecule: W U ( pe cos9 pe cos ) 4 U(9 ) U( ) pe.6 J 4 3 for N molecules: W=.6 J 6..96 J Dong work on the dpole ncreases the energy of the dpole-electrc feld system. We have to do the maxmum work f we want to turn all the dpole moments around to an ant-parallel poston: (6.34) U U U ( pe cos8 pe cos ) ( pe pe) pe f + 6.7 Detals about delectrcs: Molecules that do not have a permanent dpole moment may acqure a temporary dpole moment when placed nto an electrc feld. The opposte electrc forces on the negatve and postve charges of such materal can dstort the molecules and produce a charge separaton. Such a charge separaton, whch lasts only as long as the materal s mmersed n the electrc feld, s called an nduced dpole moment. As we can see from the pcture, the nduced charges create an nduced electrc feld opposng the orgnal feld. Thus, n summary: + - -σ nduced +σ nduced - If we ntroduce delectrc materal nto a capactor, the dpole moments of the materal are lned up and form an nduced electrc feld nsde of the delectrc, whch opposes the orgnal feld E. Insde of the delectrc we therefore have a reduced resultant feld E: E E End (6.35) E nd E The nduced electrc feld orgnates from the surface charges of the delectrc materal: nd E nd, whereas the orgnal feld s gven by E The electrc feld between the plates of the capactor wth the delectrc nserted s accordng to E (6.6) gven by E Ths s the same as the resultant feld.
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx of 3 Last saved: /7/8; 8:4 PM (6.36) E E E res nd nd E nd (6.37) nd nd (6.38) nd a) 3 nd 3 (6.39) b) no delectrc ; nd c) the materal s conductve E= E E. nd The sgn of the charge densty on the nserted conductor s opposte that on th plates. nd The nduced surface charge, the nduced electrc feld, and the nduced potental dfference are gven by: (6.4) nd ; End E; nd If we nsert a metallc slap of thckness a centrally nto the space of a parallel plate capactor wth plate dstance d, we create n effect two parallel plate capactors connected n seres, each havng the capactance A (6.4), d a A (6.4) eq d a It does not matter where we nsert the conductng slap. We see that f we make the slap very thn n comparson to the space d, the orgnal capactance remans unchanged. A A (6.43) lm d a d a Ths leads us to the next constellaton of a capactor wth a delectrc slap of thckness f d, f beng a fracton between and, nserted nto the space d between the orgnal plates and adjacent to one plate. If we nsert a thn conductng slap below the delectrc materal, we, n
Dr. Frtz Wlhelm, Physcs 3 E:\Excel fles\3 lecture\ch6 capactance.docx 3 of 3 Last saved: /7/8; 8:4 PM effect, create a seres of two capactors, one wth the delectrc and thckness fd, the other wthout delectrc and thckness (-f)d. The equvalent capactance turns out to be: A A eq ; ; fd f d (6.44) A eq f f d f f f d (-f) d