ChapterChapter 26A26A -- CapacitanceCapacitance AA PowerPointPowerPoint PresentationPresentation byby PaulPaul E.E. Tippens,Tippens, ProfessorProfessor ofof PhysicsPhysics SouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity

© 2007 Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: •• DefineDefine capacitancecapacitance inin termsterms ofof chargecharge andand voltage,voltage, andand calculatecalculate thethe capacitancecapacitance forfor aa parallelparallel plateplate capacitorcapacitor givengiven separationseparation andand areaarea ofof thethe plates.plates. •• DefineDefine dielectricdielectric constantconstant andand applyapply toto calculationscalculations ofof voltage,voltage, electricelectric fieldfield intensity,intensity, andand capacitance.capacitance. •• FindFind thethe potentialpotential energyenergy storedstored inin capacitors.capacitors. MaximumMaximum ChargeCharge onon aa ConductorConductor

AA batterybattery establishesestablishes aa differencedifference ofof potentialpotential thatthat cancan pumppump electronselectrons ee- fromfrom aa groundground (earth)(earth) toto aa conductorconductor

Battery Conductor - - - Earth -- - - e- e------

ThereThere isis aa limitlimit toto thethe amountamount ofof chargecharge thatthat aa conductorconductor cancan holdhold withoutwithout leakingleaking toto thethe air.air. ThereThere isis aa certaincertain capacitycapacity forfor holdingholding charge.charge. CapacitanceCapacitance

TheThe capacitance capacitance C C of of a a conductor conductor is is defined defined as as thethe ratio ratio of of the the charge charge Q Q on on the the conductor conductor to to thethe potential potential V V produced. produced. Battery Conductor - - - Earth -- - - e- e- - Q, V ------Q Capacitance: C ; Units : Coulombs per volt V CapacitanceCapacitance inin FaradsFarads

OneOne faradfarad (F)(F) isis thethe capacitancecapacitance CC ofof aa conductorconductor thatthat holdsholds oneone coulombcoulomb ofof chargecharge forfor eacheach voltvolt ofof potential.potential. Q coulomb (C) C; farad (F) V volt (V)

Example:Example: WhenWhen 4040 CC ofof chargecharge areare placedplaced onon aa concon-- ductorductor,, thethe potentialpotential isis 88 V.V. WhatWhat isis thethe capacitance?capacitance?

Q 40 C C  CC == 55 FF V 8 V  CapacitanceCapacitance ofof SphericalSpherical ConductorConductor At surface of sphere: At surface of sphere: Capacitance, C kQ kQ EV; r rr2 +Q 1 Recall:Recall: k  E and V at surface. 40 kQ Q Q And:And: V  Capacitance:Capacitance: C  rr40 V QQ C  Cr 40 VQ40 r ExampleExample 1:1: WhatWhat isis thethe capacitancecapacitance ofof aa metalmetal spheresphere ofof radiusradius 88 cm?cm? Capacitance, C Capacitance:Capacitance: CC == 44rr r -12 C C  4 (8.85 x 102 )(0.08 m) +Q Nm

r = 0.08 m CC == 8.908.90 xx 1010-12-12 FF

Note:Note: TheThe capacitancecapacitance dependsdepends onlyonly onon physicalphysical para-para- metersmeters (the(the radiusradius r)r) andand isis notnot determineddetermined byby eithereither chargecharge oror potential.potential. ThisThis isis truetrue forfor allall capacitors.capacitors. ExampleExample 11 (Cont.):(Cont.): WhatWhat chargecharge QQ isis neededneeded toto givegive aa potentialpotential ofof 400400 V?V?

Capacitance, C CC == 8.908.90 xx 1010-12-12 FF r Q CQCV; +Q V r = 0.08 m Q  (8.90 pF)(400 V)

TotalTotal ChargeCharge onon Conductor:Conductor: QQ == 3.563.56 nCnC

Note:Note: TheThe faradfarad (F)(F) andand thethe coulombcoulomb (C)(C) areare extremelyextremely largelarge unitsunits forfor staticstatic electricity.electricity. TheThe SISI prefixesprefixes micromicro ,, nanonano n,n, andand picopico pp areare oftenoften used.used. DielectricDielectric StrengthStrength TheThe dielectricdielectric strengthstrength ofof aa materialmaterial isis thatthat electricelectric intensityintensity EEm forfor whichwhich thethe materialmaterial becomesbecomes aa conductor.conductor. (Charge(Charge leakage.)leakage.) EE variesvaries considerablyconsiderably withwith m r physicalphysical andand environmentalenvironmental Q conditionsconditions suchsuch asas pressure,pressure, humidity,humidity, andand surfaces.surfaces. Dielectric

For air: E = 3 x 1066 N/C for spherical surfaces For air: Emm = 3 x 10 N/C for spherical surfaces andand asas lowlow asas 0.80.8 xx 101066 N/C N/C forfor sharpsharp points.points. ExampleExample 2:2: WhatWhat isis thethe maximummaximum chargecharge thatthat cancan bebe placedplaced onon aa sphericalspherical surfacesurface oneone metermeter inin diameter?diameter? (R(R == 0.500.50 m)m) Maximum Q kQ Er2 EQ; m r m 2 Air rk Q 62N (3 x 10C )(0.50 m) 6 Q  9 2 Em = 3 x 10 N/C Nm 9 x 10 C2

MaximumMaximum chargecharge inin air:air: Q = 83.3 C Qmm = 83.3 C

ThisThis illustratesillustrates thethe largelarge sizesize ofof thethe coulombcoulomb asas aa unitunit ofof chargecharge inin electrostaticelectrostatic applications.applications. CapacitanceCapacitance andand ShapesShapes TheThe chargecharge densitydensity onon aa surfacesurface isis significantlysignificantly affectedaffected byby thethe curvaturecurvature.. TheThe densitydensity ofof chargecharge isis greatestgreatest wherewhere thethe curvaturecurvature isis greatest.greatest.

+ + + + + ++ kQ + + + + + + m + ++ +++ + + + Em  + + + + + + + r2

LeakageLeakage (called(called coronacorona discharge)discharge) oftenoften occursoccurs atat sharpsharp pointspoints wherewhere curvaturecurvature rr isis greatest.greatest. ParallelParallel PlatePlate CapacitanceCapacitance +Q Area A For these two parallel plates: d QV -Q CE and Vd

YouYou willwill recallrecall fromfrom GaussGauss’’ lawlaw thatthat EE isis also:also:  Q Q is charge on either E  Q is charge on either plate. A is area of plate. 00 A plate. A is area of plate.

VQ QA E  AndAnd C 0 dA0 Vd ExampleExample 3.3. TheThe platesplates ofof aa parallelparallel plateplate capacitorcapacitor havehave anan areaarea ofof 0.40.4 mm2 andand areare 33 mmmm apartapart inin air.air. WhatWhat isis thethe capacitance?capacitance? QA C  A Vd0 0.4 m2 (8.85 x 10-12 C2 )(0.4 m2 ) C  Nm2 (0.003 m) d 3 mm CC == 1.181.18 nFnF ApplicationsApplications ofof CapacitorsCapacitors AA microphonemicrophone convertsconverts soundsound waveswaves intointo anan electricalelectrical signalsignal (varying(varying voltage)voltage) byby changingchanging dd..

Changing d Changing A C   ++ Area 0 d ++ - ++ - + --- A Q -- V  Variable d C Microphone Capacitor

TheThe tunertuner inin aa radioradio isis aa variablevariable capacitorcapacitor.. TheThe changingchanging areaarea AA altersalters capacitancecapacitance untiluntil desireddesired signalsignal isis obtained.obtained. DielectricDielectric MaterialsMaterials MostMost capacitorscapacitors havehave aa dielectricdielectric materialmaterial betweenbetween theirtheir platesplates toto provideprovide greatergreater dielectricdielectric strengthstrength andand lessless probabilityprobability forfor electricalelectrical discharge.discharge.

EEo reducedreduced EE EE << EEo + - + - + - + - + -- ++ -- ++ - + -- ++ - + AirAir - + -- ++ -- ++ - + - + - + - + - + - + - + - + -- ++ -- ++ - + -- ++ - + - + - + - C Co DielectricDielectric CC >> CCo TheThe separationseparation ofof dielectricdielectric chargecharge allowsallows moremore chargecharge toto bebe placedplaced onon thethe platesplates——greatergreater capacitancecapacitance CC >> CCo. . AdvantagesAdvantages ofof DielectricsDielectrics

••• SmallerSmaller Smaller plateplateplate separationseparationseparation withoutwithoutwithout contact.contact.contact. ••• IncreasesIncreases Increases capacitancecapacitancecapacitance ofofof aaa capacitor.capacitor.capacitor. ••• HigherHigher Higher voltagesvoltagesvoltages cancancan bebebe usedusedused withoutwithoutwithout breakdown.breakdown.breakdown. ••• OftenOften Often ititit allowsallowsallows forforfor greatergreatergreater mechanicalmechanicalmechanical strength.strength.strength. InsertionInsertion ofof DielectricDielectric Field decreases. Air +Q Dielectric E < Eo Co Vo Eo  + + ++ + Voltage decreases. -Q + V < Vo Insertion of a dielectric +Q Capacitance increases. C > C C V E  o + Same Q + -Q Permittivity increases. Q = Qo > o DielectricDielectric Constant,Constant, KK TheThe dielectricdielectric constantconstant KK forfor aa materialmaterial isis thethe ratioratio ofof thethe capacitancecapacitance CC withwith thisthis materialmaterial asas comparedcompared withwith thethe capacitancecapacitance CCo inin aa vacuum.vacuum. C K  DielectricDielectric constant:constant: C0 KK == 11 forfor AirAir KK cancan alsoalso bebe givengiven inin termsterms ofof voltagevoltage VV,, electricelectric fieldfield intensityintensity EE,, oror permittivitypermittivity :: VE K  00 VE0 TheThe PermittivityPermittivity ofof aa MediumMedium TheThe capacitancecapacitance ofof aa parallelparallel plateplate capacitorcapacitor withwith aa dielectricdielectric cancan bebe foundfound from:from: A A CKC or CK or C  00dd

TheThe constantconstant  isis thethe permittivitypermittivity ofof thethe mediummedium whichwhich relatesrelates toto thethe densitydensity ofof fieldfield lines.lines. K ;  8.85 x 10-12 C2 00 Nm2 ExampleExample 4:4: FindFind thethe capacitancecapacitance CC andand thethe chargecharge QQ ifif connectedconnected toto 200200--VV battery.battery. AssumeAssume thethe dielectricdielectric constantconstant isis KK == 5.05.0..

-12 2 5(8.85 x 10 C/Nm )  -12 2  44.2544.25 xx 1010 C/NmC/Nm A -12 C2 2 0.5 m2 A (44.25 x 102 )(0.5 m ) C  Nm d 0.002 m

CC== 11.111.1 nFnF d 2 mm QQ ifif connectedconnected toto VV == 200200 V?V?

QQ == CVCV == (11.1(11.1 nF)(200nF)(200 V)V) QQ == 2.222.22 CC ExampleExample 44 (Cont.):(Cont.): FindFind thethe fieldfield EE betweenbetween thethe plates.plates. RecallRecall QQ == 2.222.22 CC;; VV == 200200 VV..  Q Gauss' law: E     A 44.2544.25 xx 1010-12 C/NmC/Nm2 A 2.22 x 10-6 C 0.5 m2 E  (44.25 x 10-12C2 )(0.5 m 2 ) Nm2 200 V

EE== 100100 N/CN/C d 2 mm

SinceSince VV == 200200 VV,, thethe samesame resultresult isis foundfound ifif EE == V/dV/d isis usedused toto findfind thethe field.field. ExampleExample 5:5: AA capacitorcapacitor hashas aa capacitancecapacitance ofof 66FF withwith airair asas thethe dielectric.dielectric. AA batterybattery chargescharges thethe capacitorcapacitor toto 400400 VV andand isis thenthen disconnected.disconnected. WhatWhat isis thethe newnew voltagevoltage ifif aa sheetsheet ofof micamica ((KK == 55)) isis inserted?inserted? WhatWhat isis newnew capacitancecapacitance CC ?? C VV AirAir dielectricdielectric KV00;  CV K 0 V = 400 V 400 V Vo = 400 V V  ; VV== 80.080.0 VV 5 MicaMica dielectricdielectric CC == KcKco == 5(65(6 F)F) Mica, K = 5 CC == 3030 FF ExampleExample 55 (Cont.):(Cont.): IfIf thethe 400400--VV batterybattery isis reconnectedreconnected afterafter insertioninsertion ofof thethe mica,mica, whatwhat additionaladditional chargecharge willwill bebe addedadded toto thethe platesplates duedue toto thethe increasedincreased CC?? AirAir CCo == 66 F QQ0 == CC0V V0 == (6(6 F)(400F)(400 V)V)

VVo == 400400 VV QQ0 == 24002400 CC QQ == CVCV == (30(30 F)(400F)(400 V)V) MicaMica CC == 3030 F QQ == 12,00012,000 CC Mica, K = 5 QQ == 12,00012,000 CC –– 24002400 CC

QQ == 96009600 CC QQ == 9.609.60 mCmC EnergyEnergy ofof ChargedCharged CapacitorCapacitor

TheThe potentialpotential energyenergy UU ofof aa chargedcharged capacitorcapacitor isis equalequal toto thethe workwork ((qVqV)) requiredrequired toto chargecharge thethe capacitor.capacitor. IfIf wewe considerconsider thethe averageaverage potentialpotential differencedifference fromfrom 00 toto VVf toto bebe V/2V/2::

WorkWork == Q(V/2)Q(V/2) == ½QV½QV

Q2 UQVUCVU11; 2 ;  222C ExampleExample 6:6: InIn ExEx--4,4, wewe foundfound capacitancecapacitance toto bebe 11.111.1 nFnF,, thethe voltagevoltage 200200 VV,, andand thethe chargecharge 2.222.22 CC.. FindFind thethe potentialpotential energyenergy UU..

1 2 Capacitor of UCV 2 Capacitor of ExampleExample 55.. U  1 (11.1 nF)(200 V)2 2 C = 11.1 nF

UU== 222222 JJ 200 V VerifyVerify youryour answeranswer fromfrom thethe U = ? otherother formulasformulas forfor P.E.P.E. Q2 UQVU1 ; Q = 2.22 C 2 2C EnergyEnergy DensityDensity forfor CapacitorCapacitor EnergyEnergy densitydensity uu isis thethe energyenergy perper unitunit volumevolume ((J/mJ/m3).). ForFor aa capacitorcapacitor ofof areaarea AA andand separationseparation dd,, thethe energyenergy densitydensity uu isis foundfound asas follows:follows:

EnergyEnergy DensityDensity UU AA u  uu forfor anan EE--field:field: d Vol. Ad  A Recall CVEd0 and : EnergyU Density1  A dEu: 2 d u 2 0 1 2 11220 A uEAd  Ad UCV22() Ed 2 0 d SummarySummary ofof FormulasFormulas

Q coulomb (C) Cfarad; (F) Vvolt(V) QA CK Cr 4 Vd0 0 C VE K 00 1 2 uE 2 0 CVE00

Q2 UQVUCVU11; 2 ;  222C CONCLUSION:CONCLUSION: ChapterChapter 2525 CapacitanceCapacitance