Download - PHY 208 - capacitance€¦ · After Michael Faraday + C +q-q A d gaussian surface We begin by considering out most basic parallel plate capacitor with a gaussian surface as shown

Fig.1

Capacitance

1. Chargeseparation1. Theidealcapacitor

2. Capacitance1. Chargingacapacitor2. Calculatingthecapacitance3. SphericalCapacitor4. Capacitorsinparallel

3. Equivalentcircuits1. Findingtheequivalentcircuits2. CapacitanceinSeries3. CapacitanceinSeries4. EquivCircuit5. EnergyStoredinanElectricField6. Dielectric7. Gauss'Lawfordielectrics

1. Charge separation

Herewehavesomechargesseparated.Clearlythistakessomeworkandclearlythereissomepotentialenergyassociatedwiththisconfiguration.

Quick Question 1

Whereisthepotentialenergylocated?

1.Atthecenterofthecharges.2.Intheentireuniverse.3.Intheelectricfieldproducedbythesystemofcharges.4.Itisimpossibletosaywithoutmoreinformation.

PHY 208 - capacitance

updated on 2018-04-15 J. Hedberg | © 2018

Area = Ad

0V 1V 2V 3V 4V 5V

V

Q

Hereisthepotentialmapbetweenapositivelychargedblobofmetalandanegativelychargedblob.

The ideal capacitor

Thegeometryistwoparallelplateswitharea andseparation .

Thisproducesauniformfieldinsidetheplates.

Thepotentialdifferencebetweentheplatesisproportionaltotheamountofthechargewhichhasbeenseparated:

Theproportionalityconstantforthissystemis ,thecapacitance:

A d

q ∝ V

C

Q = CV



IthastheSIunitsofcoulombspervoltandiscalledafarad

1farad=1F=1C/V

AfterMichaelFaraday

+

C

+q

-q

A

d

gaussiansurface Webeginbyconsideringoutmostbasic

parallelplatecapacitorwithagaussiansurfaceasshown

Since and areparallel,wecanwritefortheelectricfield:

2. Capacitance

isthecapacitanceofthissystemoftwoconductors.

We'llseethatitisonlydependentontheshapeandsizeandmaterialsofthesystem,notthechargeorvoltage.

Charging a capacitor

Calculating the capacitance

C

∮ = E ⋅ dA = qenc

ε0

E dA

E =q

Aε0



+q

-q

A

d

gaussiansurface Withourdefinitionofwork electric

potential:

We'llfollowapathfromthenegative(low)platetothepositive(high)plate.Then:

Andwecanwriteforthepotentialdifferencebetweentheplates:

+q

-q

A

d

gaussiansurface Now,withourcapacitorgeometry

defined,wecansolvetheaboveintegralforV:

But,

So,wecanwriteforthecapacitance ,ofaparallelplatecapacitorwithplateareaandseparation :

→

− = − E ⋅ dsVf Vi ∫ f

i

E ⋅ ds = −E ds

V = E ds∫ +

−

V = E ds = Ed = ( ) d∫ d

0

q

Aε0

V =q

C

C

A

d

C =Aε0

d



Thecapacitorandbatterycircuitcanberepresentedbyapotentiallandscapemap.

1. Thebatteryseparatescharges.Thiscreatesahighandlowpotential.

2. Thewiresareconductors,sothepotentialisconstantallalongthem.

3. Inbetweenthecapacitorplatesweexpectthenormalpotentialdropacrossthecapacitorgap.

a

r

b

Theparallelplateisjustthemostbasicgeometry.Therearemanyotherrathersimplecaseswecanconsider.Forexample,twoconcentriccylinders.

Let'scalltheradiusoftheinner andtheradiusoftheouter andthelengthofthecylinderswillbe .

Thesamestrategywillapplysolet'smakeaGaussiansurfaceinbetweenthetwoconductors,withradius .

Again,Gauss'Lawwillgiveustheelectricfieldinbetweenthetwocylinders:

Becauseofcylindricalsymmetry,Gauss'LawgiveaneasyexpressionforE:

a

b

L

r

∮ = E ⋅ dA = qenc

ε0

q = EA = Eε0 ε0 (2πrL) areaofg.s.

E =q

2π Lrε0



a

r

b

Again,

And,withtheelectricfielddefinitionwejustfound:

And,since

Spherical Capacitor

Showthatthecapacitanceoftwoconcentricspheres(smallradius: andlargeradius: )is

Quick Question 2

TheplatesofanisolatedparallelplatecapacitorwithacapacitanceCcarryachargeQ.Whatisthecapacitanceofthecapacitorifthechargeisincreasedto4Q?

1.C/22.C/43.4C4.2C5.C

V = Eds∫ +

−

V = − dr = ln( )q

2π Lε0∫ a

b

1r

q

2π Lε0

b

a

C = q/V

C =2π Lε0

ln b

a

Example Problem #1:

a b

C = 4πε0ab

b− a



+

C1

C2

C3

Wecanhaveacircuitwithmorethanonecapacitorinit.Onewaytoarrangethemiscalledinparallel.

Let'screateapotentiallandscapeforthiscircuit.

1. Thebatteryseparatescharges.Thiscreatesahighandlowpotential.

2. Thewiresareconductors,sothepotentialisconstantallalongthem.

3. Inbetweenthecapacitorplatesweexpectthenormalpotentialdropacrossthecapacitorgap.

Capacitors in parallel

3. Equivalent circuits

Tomakethingseasier,we'llseekanequivalentcircuit,withjust1capacitorthatwouldhavethesameeffectasthe3inparallel.



+

C1 C2 C3

Finding the equivalent circuits

Thechargeoneachcapacitorwillbegivenby:

Thus,thetotalchargeonthethreecapacitorsis:

Andthentheequivalentcapacitancewillbegivenby:

So,tofind ,wejustneedtosumall ofthecapacitances:

Capacitance in Series

Capacitance in Series

ShowthatinseriestheequivalentCapacitance, ,isgivenby:

= V , = V , = Vq1 C1 q2 C2 q3 C3

q = + + = ( + + )Vq1 q2 q3 C1 C2 C3

= = + +Ceqq

VC1 C2 C3

Ceq n

=Ceq ∑i=1

n

Ci

Example Problem #2:

Ceq

=1

Ceq∑j=1

n 1Cj



Equiv Circuit

Energy Stored in an Electric Field

Oncewe'vechargedupacapacitor,thenwe'vedonesomeworkseparatingallthosecharges.Ifwedisconnectthecapacitorfromthecircuit,thechargeswillstayseparated.Wecanthenconsidertheenergystoredinsuchaconfiguration.

Movingeachlittlecharge acrossthepotentialdifference willrequirework,

Togetallthechargeseparatedtoreachthefullychargedvalue willrequire:

Thus,thepotentialenergystoredinacapacitorwillbe:

Dielectric

Insidethecapacitorssofar,we'veonlyhadair(orvacuum).Whathappenswhenthereissomethingelseinbetweentheplates?

Ifthematerialinbetweenismadeofdipoles.

dq′ V ′ dW

dW = d = dV ′ q′q′

Cq′

q

W = ∫ dW = d =1C

∫ q

0q′ q′

q2

2C

U = ⇒q2

2CU = C

12

V 2



field from electrodes

field from electrodes

field from polarized material inside

+q

-q

A

d

gaussiansurface

Ifthemiddleofthecapacitorwasvacuum,theonlythingwehadtoworryaboutwasthefieldsfromtheelectrodes.

However,ifweputsomethinginsidethecapacitor,itsmaterialmightbecomepolarized,andcreateitsownelectricfield.Thisfieldwillpointoppositetothemainfield.

Thus,thecapacitancecanbeshowntoincreasedependingonthechoiceofinnerdielectricmaterial.

Material DelectricConstantAir 1.00054Paper 3.5Oil 1.00054Silicon 3.5Water 1.00054StrontiumTitonate 310

Quick Question 3

Aparallelplatecapacitorisconnectedtoabatterythatmaintainsaconstantpotentialdifferenceacrosstheplates.Initially,thespacebetweentheplatescontainsonlyair.Then,aTeflon( =2.1)sheetisinsertedbetween,butnottouching,theplates.HowdoesthestoredenergyofthecapacitorchangeasaresultofinsertingtheTeflonsheet?

1.Theenergywilldecrease.2.Theenergywillnotbeaffected.3.Theenergywillincrease.4.Theenergywillbezerojoules.

Showthatthefieldinsidethiscapacitorisequalto

C = κCvaccum

κ

κ

Example Problem #3:

E = q

κ Aε0



+q –q

d

d / 3

i.

ii.

Quick Question 4

Canahumanbeingbeacapacitor?

1.Yes2.No3.Onlyiftheywearanaluminumfoilhat

Quick Question 5

Shownaretwopanels,i.andii.Equalamountsofcharge,butofoppositesignsareplacedontwoplates,asshownini.Theplatesarenotconnectedtoanything,andthemediumbetweentheplatesisvacuum.Whathappenstotheamountofchargeontheplatesifthedistanceseparatingtheplatesisreducedfromto ?

1.Theamountofchargeincreases.2.Theamountofchargedecreases.3.Theamountofchargestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.

Quick Question 6

Shownaretwopanels,i.andii.Equalamountsofcharge,butofoppositesignsareplacedontwoplates,asshownini.Theplatesarenotconnectedtoanything,andthemediumbetweentheplatesisvacuum.Whathappenstotheelectricfieldinbetweentheplatesifthedistanceseparatingtheplatesisreducedfrom to ?

1.Theelectricfielddecreases.2.Theelectricfieldincreases.3.Theelectricfieldstaysthesame.4.Cannotbedeterminedfromtheinformationgiven.

d

d/3

d d/3



d

d / 3

i.

ii.

9 V

9 V

Quick Question 7

Shownaretwopanels,i.andii.Equalamountsofcharge,butofoppositesignsareplacedontwoplates,asshownini.Theplatesarenotconnectedtoanything,andthemediumbetweentheplatesisvacuum.Whathappenstothepotentialdifferencebetweentheplatesifthedistanceseparatingtheplatesisreducedfrom to ?

1.Thepotentialdifferencedecreases.2.Thepotentialdifferenceincreases.3.Thepotentialdifferencestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.

Quick Question 8

Now,thetwoplatesareconnectedtoavoltagesource(i.e.abattery)thatcontinuouslyprovides9V.Whathappenstothepotentialdifferencebetweentheplatesasthedistanceseparatingtheplatesisreducedfrom to ?


Quick Question 9

Now,thetwoplatesareconnectedtoavoltagesource(i.e.abattery)thatcontinuouslyprovides9V.Whathappenstotheelectricfieldbetweentheplatesasthedistanceseparatingtheplatesisreducedfrom to ?


d d/3

d d/3

d d/3



d

i.

9 V

d/2

ii.

9 V

d/2

metal

Quick Question 10

Now,thetwoplatesareconnectedtoavoltagesource(i.e.abattery)thatcontinuouslyprovides9V.Whathappenstotheamountofchargeoneachplateasthedistanceseparatingtheplatesisreducedfrom to ?


So,whatdidwefigureout:reducingtheplateseparationfortwoisolatedplatesdidnothingtothechargesontheplatesandtheelectricfieldbetweentheplates,butitdiddecreasethepotentialdifferencebetweenthetwoplates.

Reducingtheplateseparationfortwoconnectedandfixedpotentialplatesincreasedtheamountofchargesontheplatesandtheelectricfieldbetweentheplates,whilethepotentialdifferencestayedthesame.

So,weknowthatwhateverproportionalityconstantexistsbetweenvoltageandcharge,thenitwillhaveincreasedforthistransformation.

Indeed,wehaveincreasedthecapacitanceofthesystem.Inotherwords,itnowhasagreatercapacitytostorechargegivenavoltage.

Quick Question 11

Herewehavetwoplates,connectedbyabattery.Ametalslabisinsertedinthemiddleoftheemptyspacebetweenthetwoplates.Whatwillhappentotheelectricfieldintheemptyregionsaftertheslabisinserted?


d d/3

Q = constant × V



d

i.

9 V

d/2

ii.

9 V

d/2

metal

dielectric

i.

ii.

+q –q

Quick Question 12

Herewehavetwoplates,connectedbyabattery.Ametalslabisinsertedinthemiddleoftheemptyspacebetweenthetwoplates.Whatwillhappentotheamountofchargeontheoriginaltwometalplatesaftertheslabisinserted?


Quick Question 13

Herewehavetwoplates,withchargeq,and-q,andnobatteryisconnected.Insteadofametalslab,let'sinsertapolarizabledielectricinthemiddle.Whatwillhappentotheamountofchargeontheoriginaltwometalplatesafterthedielectricisinserted?




dielectric

i.

ii.

+q –q

dielectric

i.

ii.

+q –q

Quick Question 14

Samesituation:WhatwillhappentotheElectricFieldbetweenthetwometalplates?


Quick Question 15

Again:WhatwillhappentothePotentialDifferencebetweenthetwometalplates?




dielectric

i.

9 V

ii.

9 V

dielectric

i.

9 V

ii.

9 V

Quick Question 16

Herewehavetwoplates,connectedbyabattery,thistime.Let'sinsertapolarizabledielectricinthemiddle.WhatwillhappentothePotentialDifferencebetweenthetwometalplates?


Quick Question 17

Again,let'sinsertapolarizabledielectricinthemiddle.Whatwillhappentotheelectricfieldintheregionwherethedielectric?




dielectric

i.

9 V

ii.

9 V

dielectric

i.

ii.

+q –q

Quick Question 18

Whatwillhappentotheamountofchargeontheoriginaltwometalplatesafterthedielectricisinserted?


Now,wehaveshownthataddingadielectricmaterialinbetweenthetwoplatescausestheamountofchargeontheplatestoincreaseforagivenpotentialdifference.Or,ifthechargeisheldconstant,thenthepotentialdifferenceisseentodecrease.Thesamerelationshipholds.Wehavechangedtheconstantbetweenthe andthe .

Wecandefineadielectricconstant, ,bytheratioofthepotentialdifferencechangeexperiencedbytwoplatesuponinsertionofadielectricmaterial:

Thiscanbeextendedtoshowthechangeincapacitancebasedonthedielectricconstantusing

Q V

Q = constant × V

κ

κ =V0Vd

Q = CV

= κCd C0



free charges free charges

bound charges bound charges Lookingatthemicroscopicviewofthechargesinadielectricfilledcapacitor,wecanseetherearetwotypes:freeandbound.

Thefreechargesrefertotheonesinthemetalplates,sincetheyarefreetomoveabout.Theboundchargesarelocatedinthedielectricandarenotfreetomove.

Theelectricfieldduetothefreechargeswillbegivenby:

whiletheelectricfieldduetotheboundchargesis:

Thus,thetotalfieldinsidethecapacitor, ,willbethedifferencebetweenthesetwofields:

Sincethedielectricconstantisgivenby: ,wecanwrite:

Or,

whichgivesusanexpressionfor :

Showthatthefieldinsidethiscapacitorisequalto

= =Efreeσfreeϵ0

qfree

Aϵ0

= =Eboundσbound

ϵ0

qbound

Aϵ0

E

E = − = =Efree Ebound−σfree σboundϵ0

−qfree qbound

Aϵ0

κ = /V0 Vd

κ =Efree

E

=qfree

κ Aϵ0

−qfree qbound

Aϵ0

qbound

=qboundκ− 1

κqfree

Example Problem #4:

E = q

κ Aε0



Gauss' Law for dielectrics

1. Thefluxintegralnowinvolves ,notjust .Thevector issometimescalledtheelectricdisplacement

2. ThechargeqenclosedbytheGaussiansurfaceisnowtakentobethefreechargeonly.

∮ κ ⋅ d =ϵ0 E⃗ A⃗ qfree

κE⃗ E⃗ κE⃗

D⃗



Download - PHY 208 - capacitance€¦ · After Michael Faraday + C +q-q A d gaussian surface We begin by considering out most basic parallel plate capacitor with a gaussian surface as shown

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