Capacitance and Dielectrics. Q26.3 The parallel-connected capacitors store more energy, since they have higher equivalent capacitance.

Transcription

1 6 pcitnce n Dielectrics HAPTER OUTLNE 6 Deinition o pcitnce 6 lculting pcitnce 6 omintions o pcitors 64 Energy Store in hrge pcitor 65 pcitors with Dielectrics 66 Electric Dipole in n Electric iel 67 An Atomic Description o Dielectrics ANSWERS TO UESTONS 6 Nothing hppens to the chrge i the wires re isconnecte the wires re connecte to ech other, chrges in the single conuctor which now exists move etween the wires n the pltes until the entire conuctor is t single potentil n the cpcitor is ischrge 6 6 km The plte re woul nee to e m 6 The prllel-connecte cpcitors store more energy, since they hve higher equivlent cpcitnce 64 Seventeen comintions: niviul,, Prllel + +, +, +, + Series-Prllel Series , KJ + KJ, KJ, , KJ + KJ, KJ, + + KJ KJ, KJ + 65 This rrngement woul ecrese the potentil ierence etween the pltes o ny iniviul cpcitor y ctor o, thus ecresing the possiility o ielectric rekown Depening on the ppliction, this coul e the ierence etween the lie or eth o some other (most likely more expensive) electricl component connecte to the cpcitors 66 No not just using rules out cpcitors in series or in prllel See Prolem 7 or n exmple connections cn e me to comintion o cpcitors t more thn two points, the comintion my e irreucile KJ 77

2 78 pcitnce n Dielectrics 67 A cpcitor stores energy in the electric iel etween the pltes This is most esily seen when using issectle cpcitor the cpcitor is chrge, creully pull it prt into its component pieces One will in tht very little resiul chrge remins on ech plte When ressemle, the cpcitor is suenly rechrge y inuction ue to the electric iel set up n store in the ielectric This proves to e n instructive clssroom emonstrtion, especilly when you sk stuent to reconstruct the cpcitor without supplying him/her with ny ruer gloves or other insulting mteril (O course, this is ter they sign liility wiver) 68 The work you o to pull the pltes prt ecomes itionl electric potentil energy store in the cpcitor The chrge is constnt n the cpcitnce ecreses ut the potentil ierence increses to rive up the potentil energy V The electric iel etween the pltes is constnt in strength ut ills more volume s you pull the pltes prt 69 A cpcitor stores energy in the electric iel insie the ielectric Once the externl voltge source is remove provie tht there is no externl resistnce through which the cpcitor cn ischrge the cpcitor cn hol onto this energy or very long time To mke the cpcitor se to hnle, you cn ischrge the cpcitor through conuctor, such s screwriver, provie tht you only touch the insulting hnle the cpcitor is lrge one, it is est to use n externl resistor to ischrge the cpcitor more slowly to prevent mge to the ielectric, or weling o the screwriver to the terminls o the cpcitor 6 The work one, W V, is the work one y n externl gent, like ttery, to move chrge through potentil ierence, To etermine the energy in chrge cpcitor, we must the work one to move its o chrge rom one plte to the other nitilly, there is no potentil ierence etween the pltes o n unchrge cpcitor As more chrge is trnserre rom one plte to the other, the potentil ierence increses s shown in igure 6, mening tht more work is neee to trnser ech itionl it o chrge The totl work is the re uner the curve o igure 6, n thus W V 6 Energy is proportionl to voltge squre t gets our times lrger 6 Let the cpcitnce o n iniviul cpcitor, n s represent the equivlent cpcitnce o the group in series While eing chrge in prllel, ech cpcitor receives chrge 4 e j chrge 5 8 V 4 While eing ischrge in series, (or times the originl voltge) ischrge 4 8 kv Put mteril with higher ielectric strength etween the pltes, or evcute the spce etween the pltes At very high voltges, you my wnt to cool o the pltes or choose to mke them o ierent chemiclly stle mteril, ecuse toms in the pltes themselves cn ionize, showing thermionic emission uner high electric iels 64 The potentil ierence must ecrese Since there is no externl power supply, the chrge on the cpcitor,, will remin constnt tht is ssuming tht the resistnce o the meter is suiciently lrge Aing ielectric increses the cpcitnce, which must thereore ecrese the potentil ierence etween the pltes 65 Ech polr molecule cts like n electric compss neele, ligning itsel with the externl electric iel set up y the chrge pltes The contriution o these electric ipoles pointing in the sme irection reuces the net electric iel As ech ipole lls into conigurtion o lower potentil energy it cn contriute to incresing the internl energy o the mteril s

3 hpter The mteril o the ielectric my e le to support lrger electric iel thn ir, without reking own to pss sprk etween the cpcitor pltes 67 The ielectric strength is mesure o the potentil ierence per unit length tht ielectric cn withstn without hving iniviul molecules ionize, leving in its wke conucting pth rom plte to plte or exmple, ry ir hs ielectric strength o out MV/m The ielectric constnt in eect escries the contriution o the electric ipoles o the polr molecules in the ielectric to the electric iel once ligne 68 n wter, the oxygen tom n one hyrogen tom consiere lone hve n electric ipole moment tht points rom the hyrogen to the oxygen The other O-H pir hs its own ipole moment tht points gin towr the oxygen Due to the geometry o the molecule, these ipole moments to hve non-zero component long the xis o symmetry n pointing towr the oxygen A non-polrize molecule coul either hve no intrinsic ipole moments, or hve ipole moments tht to zero An exmple o the ltter cse is O The molecule is structure so tht ech O pir hs ipole moment, ut since oth ipole moments hve the sme mgnitue n opposite irection ue to the liner geometry o the molecule the entire molecule hs no ipole moment 69 Heting ielectric will ecrese its ielectric constnt, ecresing the cpcitnce o cpcitor When you het mteril, the verge kinetic energy per molecule increses you reer ck to the nswer to uestion 65, ech polr molecule will no longer e nicely ligne with the pplie electric iel, ut will egin to ither rock ck n orth eectively ecresing its contriution to the overll iel 6 The primry choice woul e the ielectric You woul wnt to chose ielectric tht hs lrge ielectric constnt n ielectric strength, such s strontium titnte, where κ (Tle 6) A convenient choice coul e thick plstic or mylr Seconly, geometry woul e ctor To mximize cpcitnce, one woul wnt the iniviul pltes s close s possile, since the cpcitnce is proportionl to the inverse o the plte seprtion hence the nee or ielectric with high ielectric strength Also, one woul wnt to uil, inste o single prllel plte cpcitor, severl cpcitors in prllel This coul e chieve through stcking the pltes o the cpcitor or exmple, you cn lterntely ly own sheets o conucting mteril, such s luminum oil, snwiche etween your sheets o insulting ielectric Mking sure tht none o the conucting sheets re in contct with their next neighors, connect every other plte together igure 6 illustrtes this ie Dielectric onuctor G 6 This technique is oten use when home-rewing signl cpcitors or rio pplictions, s they cn withstn huge potentil ierences without lshover (without either ischrge etween pltes roun the ielectric or ielectric rekown) One vrition on this technique is to snwich together lexile mterils such s luminum roo lshing n thick plstic, so the whole prouct cn e rolle up into cpcitor urrito n plce in n insulting tue, such s PV pipe, n then ille with motor oil (gin to prevent lshover)

4 8 pcitnce n Dielectrics SOLUTONS TO PROBLEMS Section 6 Deinition o pcitnce 5 e j µ P6 () 4 V e j µ () 4 5 V 6 6 P6 () V V µ () V Section 6 lculting pcitnce P6 E kq e r : q e49 N j m e N m 9 j 4 µ () q 4 σ A 4π µ m e j p () 4π r 4π 885 P64 () 4π R R k e 9 e8 99 N m je j mm π e je j 4π 8 85 m () 4π R N m e j (c) V V p P65 () R R R + + R 5 7 µ KJ µ 5 µ () V V 5 µ 9 e899 m j 5 m V 899 kv

5 P66 e je j κ A 8 85 m N m 8 m The potentil etween groun n clou is e j 6 9 E N 8m 4 V 9 9 e je j V 4 V 6 6 n hpter 6 8 P67 () E V E 8 m kv m () (c) E σ 4 e je j σ N 8 85 N m 98 nm A e885 N m je76 cm j m cmg 8 m 74 p () e j V p P68 A 6 κ 5 e je j κ A m nm A P69 V σ A V e885 N m j5v σ 9 4 cm cm m e je j 44 µ m

6 8 pcitnce n Dielectrics P6 With θ π, the pltes re out o mesh n the overlp re is zero With θ, the overlp re is tht o semi-circle, π R By proportion, the π θr eective re o single sheet o chrge is When there re two pltes in ech com, the numer o joining sheets o positive n negtive chrge is, s shown in the sketch When there re N pltes on ech com, the numer o prllel cpcitors is N n the totl cpcitnce is N Aeective N π θ R N π θ R istnce G P6 5 P6 () ln 8 99 ln k e 9 77 ch e j c58 h () Metho : ke λ ln H G K J 68 n q 8 7 λ 6 m 5 m e8 99 je 6 jln 58 Metho : 8 kv 9 68 H G K J kv P6 Let the rii e n with Put chrge on the inner conuctor n on the outer k e Electric iel exists only in the volume etween them The potentil o the inner sphere is V ; k e tht o the outer is V Then V k e k e V 4π Here π 4 8π K J n V V 8π π π π π π The intervening volume is Volume H G K J H G K J 8 π 84π 7 e N mj 6 Volume m 84π e8 85 N m j The outer sphere is 6 km in imeter P6 () k e g e8 99 j 4 7 g 5 6 p 7 () V kv

7 hpter 6 8 P64 y : Tcosθ mg x : Tsinθ Eq Diviing, tnθ Eq mg so E mg tnθ q n E mg tnθ q 6 4 P65 4π R 4π 8 85 Nm 6 7 m 7 8 e je j Section 6 omintions o pcitors P66 () pcitors in prllel Thus, the equivlent cpcitor hs vlue o eq + 5 µ + µ 7 µ () The potentil ierence cross ech rnch is the sme n equl to the voltge o the ttery g (c) 5 V 5 µ 9 V 45 µ g n V µ 9 V 8 µ 9 V P67 () n series cpcitors s µ µ eq n eq 5 µ (c) The chrge on the equivlent cpcitor is 5 µ 9 V 8 µ Ech o the series cpcitors hs this sme chrge on it So 8 µ eq eq g () The potentil ierence cross ech is n 8 µ 5 µ 8 µ µ 65 V 65 V

8 84 pcitnce n Dielectrics P68 The circuit reuces irst ccoring to the rule or cpcitors in series, s shown in the igure, then ccoring to the rule or cpcitors in prllel, shown elow K J eq G P68 P69 p + Sustitute p + s + s p p + e j p Simpliying, + p p s p ± p 4ps p ± p ps 4 We choose ritrrily the + sign (This choice cn e ritrry, since with the cse o the minus sign, we woul get the sme two nswers with their nmes interchnge) g g g g p p ps p + 9 p 9 p p 6 p 4 p p p ps 9 p g 5 p p 4 P6 p + n + s Sustitute : p + s p p + e j p Simpliying, + p p s n p p p s ± p p p s where the positive sign ws ritrrily chosen (choosing the negtive sign gives the sme vlues or the cpcitnces, with the nmes reverse) Then, rom p p p ps 4

9 P6 () + s 5 hpter 6 85 s p eq 5 µ µ + 85 µ µ KJ 596 µ g on () 596 µ 5 V 895 µ 89 5 µ 447 V µ V g µ 6 µ 5 V 6 µ on 6 µ µ on 5 µ n µ G P6 *P6 () pcitors n re in prllel n present equivlent cpcitnce 6 This is in series L O with cpcitor, so the ttery sees cpcitnce + 6 () NM P they were initilly unchnge, stores the sme chrge s n together With greter cpcitnce, stores more chrge thn Then > > (c) The g equivlent cpcitor stores the sme chrge s Since it hs greter cpcitnce, implies tht it hs smller potentil ierence cross it thn n prllel with ech other, n hve equl voltges: > () is increse, the overll equivlent cpcitnce increses More chrge moves through the ttery n increses As increses, must ecrese so ecreses Then must increse even more: n increse; ecreses P6 so 6 V n µ µ n : or 6 g µ µ 4 µ 8 µ 9 G P6

10 86 pcitnce n Dielectrics P64 () n series, to reuce the eective cpcitnce: + µ 4 8 µ s 5 s 98 µ µ () n prllel, to increse the totl cpcitnce: 9 8 µ + µ p p µ P65 n n n n cpcitors so n n n n *P66 or connecte y itsel, V 8 µ where is the ttery voltge: 8 µ or n in series: + V KJ µ 8 µ µ µ sustituting, + or n in series: P67 + V KJ 5 µ 8 µ 5 µ 5 µ + or ll three: V 8 µ + + KJ This is the chrge on ech one o the three s p p eq + K J µ µ µ K J µ 9 8 µ G P67

11 P V 6 eq 4 eq e j hpter 6 87 P69 p eq, so eq 6 p 866 p 6 4 e j e j 4 8 V p 48 V 86 µ s p K J µ µ G P69 *P6 Accoring to the suggestion, the comintion o cpcitors shown is equivlent to Then ± 4 + 4e j 4 Only the positive root is physicl e j g G P6 Section 64 Energy Store in hrge pcitor g P6 () U µ V 6µ J g () U µ 6 V 54 µ J P6 U U J 447 V V

12 88 pcitnce n Dielectrics P6 U The circuit igrm is shown t the right () p + 5 µ + 5 µ µ () e j J U 6 5 s + + KJ 5 µ 5 µ U U 5 47 KJ 68 V 47 µ G P6 P64 Use U n A, Thereore, the store energy oules e je j *P65 () 5 V 5 () U U 5 5 e Jj 8 V P66 u U E V 7 e885 jg V L 5 5 H G V m e m j K J 5 L m P67 W U z x so U x x H G KJ KJ x x A A

13 P68 With switch close, istnce 5 n cpcitnce A A e j () V 4 µ hpter 6 89 () The orce stretching out one spring is 4 A A A g One spring stretches y istnce x, so 4 k 4 V x H G 8 8 K J e j V e8 mj 6 e j 5 kn m 8 9 P69 The energy trnserre is HET V 5 V 5 J n % o this (or E int 5 7 J) is sore y the tree m is the mount o wter oile wy, g e j 6 7 then Eint m 4 86 Jkg + m 6 Jkg 5 J giving m 979 kg *P64 () U + () The ltere cpcitor hs cpcitnce V + V V V (c) H G K J + H G U K J The totl chrge is the sme s eore: 4 () The extr energy comes rom work put into the system y the gent pulling the cpcitor pltes prt P64 U U where 4 R R k e R k e k e k R R H G K J H G K J e π n k e k e R R

14 9 pcitnce n Dielectrics q q q *P64 () The totl energy is U U + U + 4π R or minimum we set U : q g q q + 4π R 4π R R Rq R Rq q R + R g q + 4π R Then q q R R + R q () V V kq kr R R R + R k R + R e e e kq kr R R R + R g k R + R e e e n V V g Section 65 P64 () pcitors with Dielectrics 4 κ e je j A 8 85 m 75 m 5 4 m 6 5 e je j () mx Emx 6 Vm 4 m 4 kv P644 V, mx mx ut mx Emx Also, κ A Thus, κ A E κ AE g mx mx mx () With ir etween the pltes, κ n E mx 6 Vm 8 8 p Thereore, mx AEmx 4 κ e je je j () With polystyrene etween the pltes, κ 56 n E mx 4 6 Vm e je je j 4 6 κ mx AEmx m 5 m 4 Vm 7 n

15 P645 κ A e j g or m hpter 6 9 P646 onsier two sheets o luminum oil, ech 4 cm y cm, with one sheet o plstic etween them Suppose the plstic hs κ, E mx ~ 7 Vm n thickness mil 54 cm Then, κ e je j A ~ N m m 5 54 m 7 5 E ~ Vm 54 m ~ V mx mx e je j A P647 Originlly, ~ V i () i A V The chrge is the sme eore n ter immersion, with vlue 4 e885 N m je5 m j5 V 5 m e j 69 p () inlly, κ A 4 e je j N m 5 m e5 mj A i i 5 V κ A κ A κ 8 8 p V i A V (c) Originlly, Ui V i i κ A A V i inlly, U V κ κ A i So, U U Ui κ κ 4 e885 N m je5 m j5v 79 U 45 5 nj 5 m 8 e j

16 9 pcitnce n Dielectrics κ A P648 () κ 4 7e8 85 je m j m 5 n () (c) The ttery elivers the ree chrge 9 e j 5 V 84 n The surce ensity o ree chrge is σ 84 m 4 A m The surce ensity o polriztion chrge is σ p σ σ κ K J 7 K J 8 4 m () We hve E E κ n E V ; hence, E κ V 4 7 m e j 694 Vm P649 The given comintion o cpcitors is equivlent to the circuit igrm shown to the right Put chrge on point A Then, g g g 4 µ µ 4 µ V G P649 AB B D So, B 4AB 4 D, n the center cpcitor will rek own irst, t B 5 V When this occurs, AB D Bg 75 V 4 n VAD VAB + VB + VD 75 V + 5 V + 75 V 5 V Section 66 Electric Dipole in n Electric iel P65 () The isplcement rom negtive to positive chrge is e j e j e j i + j mm 4i j mm 6i + 4 j m The electric ipole moment is e je j e j 9 p q 5 6i + 4 j m 9 i j m e j e j () τ p E 9 i j m 7 8i 4 9 j N e j N m N m τ k 65 5k 9 k continue on next pge

17 e j e j J nj (c) U pe 9 i j m 7 8i 4 9 j N U + hpter 6 9 m m N N () p E U U mx mx pe 4 nj, U 4 nj U min 8 nj min P65 () Let x represent the coorinte o the negtive chrge Then x+ cosθ is the coorinte o the positive chrge The orce on the negtive chrge is i The orce on the positive chrge is i i + E qe x cosθ qe x q cosθ i x p - θ E G P65() The orce on the ipole is ltogether + + q E cosθ i p E cosθ i x x + () The lloon cretes iel long the x-xis o kq e i x Thus, E x kq e x 9 e je j At x 6 cm, E x MN m e je j 6 m 8 78 N m cos i 55 i mn Section 67 An Atomic Description o Dielectrics P65 π re q in so E λ π r E r λ mx E π z r r r r r 6 5 e H G Vm je mjln K J mx 579 V z mx inner λ λ r ln π r π r r KJ G P65

18 94 pcitnce n Dielectrics P65 () onsier gussin surce in the orm o cylinricl pillox with ens o re A << A prllel to the sheet The sie wll o the cyliner psses no lux o electric iel since this surce is everywhere prllel to the iel Guss s lw ecomes EA + EA A, so E A irecte wy rom the positive sheet A () (c) n the spce etween the sheets, ech cretes iel wy rom the positive n A towr the negtive sheet Together, they crete iel o E A Assume tht the iel is in the positive x-irection Then, the potentil o the positive plte reltive to the negtive plte is z + plte E s i ix A plte z + plte plte e j + A () A A pcitnce is eine y: V A κ Aitionl Prolems L NM O L O NM P g g P654 () P 4 µ (c) V µ 9V 8µ () c c c Thereore, 6 8 µ µ 9 V µ G P i g 8 µ µ 6 8 µ 6 µ 6 µ µ 4 µ 4 µ 4 6 V V 6 V V e j () UT eq 9 V 4 mj

19 hpter 6 95 *P655 () Ech ce o P crries chrge, so the three-plte system is equivlent to P P P P Ech cpcitor y itsel hs cpcitnce 4 κ A e8 85 j 7 5 m 558 p N m 9 m Then equivlent cpcitnce p () + V 4 p (c) Now P hs chrge on two surces n in eect three cpcitors re in prllel: g 5 58 p 6 7 p () Only one ce o P 4 crries chrge: 558 V 669 p *P656 *P657 rom the exmple out cylinricl cpcitor, V V V keλ ln 9 45 kv 8 99 Nm 4 m ln V e je j e j Jln5 564 V V 56 V 89 V m 4 m mgine the center plte is split long its miplne n pulle prt We hve two cpcitors in prllel, supporting the sme n A crrying totl chrge The upper hs cpcitnce n the A lower hrge lows rom groun onto ech o the outsie pltes so tht + Then + A A G P657 () On the lower plte the chrge is On the upper plte the chrge is () A

20 96 pcitnce n Dielectrics P658 () We use Eqution 6 to in the potentil energy o the cpcitor As we will see, the potentil ierence chnges s the ielectric is withrwn The initil n inl energies re U i K J i n U K J But the initil cpcitnce (with the ielectric) is i κ Thereore, U κ i Since the work one y the externl orce in removing the ielectric equls the chnge in potentil energy, we hve W U Ui H G ikj H G i K J H G κ i K Jκ To express this reltion in terms o potentil ierence i, we sustitute iig, n 9 5 evlute: W iig κ e j V 5 4 J The positive result conirms tht the inl energy o the cpcitor is greter thn the initil energy The extr energy comes rom the work one on the system y the externl orce tht pulle out the ielectric H G KJ () The inl potentil ierence cross the cpcitor is Sustituting i n κ i ig gives i κ V 5 V 5V Even though the cpcitor is isolte n its chrge remins constnt, the potentil ierence cross the pltes oes increse in this cse Vmx P659 κ, Emx 8 Vm κ A or 5 6 V A 5 4 mx e j g κ κ E 8 85 mx 8 e je j 88 m *P66 The originl kinetic energy o the prticle is e je j 6 4 K mv kg m s 4 J The potentil ierence cross the cpcitor is µ V µ or the prticle to rech the negtive plte, the prticle-cpcitor system woul nee energy 6 4 e j U q V J Since its originl kinetic energy is greter thn this, the prticle will rech the negtive plte As the prticle moves, the system keeps constnt totl energy K + U K + U t + plte t plte : 4 4 J + e j+ V e jv + v 4 e Jj 6 ms kg

21 P66 () κ + A KJ ; + κ A ; A κ κ κ + κ κ A KJ hpter 6 97 A + + KJ κ κ κ + κ + κ G P66 () Using the given vlues we in: totl p KJ *P66 The initil chrge on the lrger cpcitor is µ 5 V 5 µ An itionl chrge q is pushe through the 5-V ttery, giving the smller cpcitor chrge q n the lrger chrge 5 µ + q q 5 µ + q Then 5 V + 5 µ µ 5 µ q+ 5 µ + q q 7 µ So cross the 5-µ cpcitor q 7 µ 5 µ Across the -µ cpcitor 5 µ + 7 µ µ V 6 7 V P66 () Put chrge on the sphere o rius n on the other sphere Reltive to V t ininity, the potentil t the surce o is n the potentil o is V V k e k e k e k e + k e k e k e k e The ierence in potentil is V V + n 4π V V + g g gkj () As, ecomes negligile compre to Then, 4π + n 4π + 4π s or two spheres in series

22 98 pcitnce n Dielectrics P664 () x + κ x + x κ () U U (c) i x V + xκ KJ H G K J κ to the let (out o the cpcitor) g e885 j5 g45 () 55 N P665 The portion o the cpcitor nerly ille y metl hs κ x cpcitnce n store energy The unille portion hs cpcitnce The chrge on this portion is e j x x () () The store energy is U x x x U x x x KJ + to the right (into the cpcitor) (c) Stress 4 σ () u E KJ KJ 4

23 gl m g g 786 m K J H G 67 kg K J J g s g 6 sg P666 Gsoline: 6 Btu gl 54 J Btu Bttery: 6 kg 7 5 Jkg 54 7 hpter 6 99 Jkg pcitor: V kg 7 Jkg Gsoline hs 94 times the speciic energy content o the ttery n 77 times tht o the cpcitor P667 ll the unknown cpcitnce u + u g g i i µ g V V V u i u ig i 49 µ *P668 She cn clip together series comintion o prllel comintions o two -µ cpcitors The equivlent cpcitnce is µ When 9 V is connecte cross the µ + µ g g comintion, only 45 V ppers cross ech iniviul cpcitor G P668 P669 () A V When the ielectric is inserte t constnt voltge, κ ; U g g e j U V κ V U n κ U The extr energy comes rom (prt o the) electricl work one y the ttery in seprting the extr chrge () n κ so κ

24 pcitnce n Dielectrics P67 The verticl orienttion sets up two cpcitors in prllel, with equivlent cpcitnce g g A A p κ + κ + K J A where A is the re o either plte n is the seprtion o the pltes The horizontl orienttion prouces two cpcitors in series is the rction o the horizontl cpcitor ille with ielectric, the equivlent cpcitnce is g L N M g O P L A, or s O NM κ + P κ g p s gives κ + κ + κ g, or κ + + κ g κ + κ + s κ A A M κ P A Requiring tht or κ, this yiels 4, with the solution P67 nitilly (cpcitors chrge in prllel), g g q 6 µ 5 V 5 µ q V µ 5V 5µ Ater reconnection (positive plte to negtive plte), qtotl qtotl q q µ n V µ 5 V 8 µ Thereore, g g totl q 6 µ 5V 75µ q V µ 5V 5µ P67 Assume potentil ierence cross n, n notice tht the potentil ierence cross the 8 µ cpcitor must e zero y symmetry Then the equivlent cpcitnce cn e etermine rom the ollowing circuit: G P67 µ

25 hpter 6 P67 E mx occurs t the inner conuctor s surce ke Emx λ rom Eqution 47 ke H G λ ln K J rom Exmple 6 Emx ln g Vmx Emxln H G ln K J 6 H G e8 Vm je 8 m j K J 9 kv 8 P674 E κλ ; mx Emxln V L mx EmxMln H G K J H G κλ ln K J H G K J + H G NM K J ln K J or e so e K J O P P P675 By symmetry, the potentil ierence cross is zero, so the circuit reuces to eq + 8 K J G P675

26 pcitnce n Dielectrics P676 The electric iel ue to the chrge on the positive wire is perpeniculr to the wire, ril, n o mgnitue λ E+ π r The potentil ierence etween wires ue to the presence o this chrge is z + wire λ r λ E r π r π wire z D K J D ln The presence o the liner chrge ensity λ on the negtive wire mkes n ienticl contriution to the potentil ierence etween the wires Thereore, the totl potentil ierence is g λ K J D ln π n the cpcitnce o this system o two wires, ech o length, is λ λ π λπ ln D ln D g The cpcitnce per unit length is: π ln D *P677 The conition tht we re testing is tht the cpcitnce increses y less thn %, or, < Sustituting the expressions or n rom Exmple 6, we hve, k e ln ln < ln c h k ln e c ch h ch This ecomes, ln ln ln ln ln ln K J < H G K J H G K J + H G K J H G K J We cn rewrite this s, H G ln K J < ln ln K J > ln ln where we hve reverse the irection o the inequlity ecuse we multiplie the whole expression y to remove the negtive signs ompring the rguments o the logrithms on oth sies o the inequlity, we see tht, > 85 Thus, i > 85, the increse in cpcitnce is less thn % n it is more eective to increse

27 ANSWERS TO EVEN PROBLEMS hpter 6 P6 () µ ; () V ; () 4 ; P64 () ; (c) 4 P64 () 899 mm; () p ; (c) p () Positive work is one on the system y the gent pulling the pltes prt P66 n; 6 6 P68 nm P6 N π θ R P64 () q R R + R () see the solution n q P644 () n ; () 7 n R R + R ; P6 6 m P64 mg tnθ q P66 () 7 µ ; () 9 V; (c) 45 µ n 8 µ P68 8 P6 p p + p s n p p p 4 4 P6 () ; () > > ; (c) > ; () n increse n ecreses P64 () 98 µ in series; () µ in prllel P µ P68 86 µ P6 e j P6 447 kv P64 energy oules P66 5 m 5 L P68 () 4 µ ; () 5 kn m s P646 ~ n ~ V or two 4 cm y cm sheets o luminum oil snwiching thin sheet o plstic P648 () 5 n; () 8 4 n ; (c) 84 µ m ree; 8 µ m inuce; () 694 V m e j p m ; P65 () 9 i + 84 j () 9 nn mk ; (c) nj ; () 8 nj P V P654 () µ ; () 6 V ; 6 V ; 6 V ; 4 V ; P kv (c) 6 8 µ ; 4 µ ; () 4 mj P658 () 4 µ J; () 5 V P66 yes; Mm s P66 V ; 6 7 V ; P664 () + x κ () + xκ ; (c) κ to the let ; () 55 mn let

28 4 pcitnce n Dielectrics P666 Gsoline hs 94 times the speciic energy content o the ttery, n 77 times tht o the cpcitor P67 µ P674 see the solution P668 see the solution; 45 V P676 see the solution P67