Upload
others
View
4
Download
0
Embed Size (px)
Citation preview
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 Page 1
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 2
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 3
+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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 4
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 5
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 6
+
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.
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 7
+
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 8
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 9
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 10
+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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 11
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 ?
1.Thepotentialdifferencedecreases.2.Thepotentialdifferenceincreases.3.Thepotentialdifferencestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
Quick Question 9
Now,thetwoplatesareconnectedtoavoltagesource(i.e.abattery)thatcontinuouslyprovides9V.Whathappenstotheelectricfieldbetweentheplatesasthedistanceseparatingtheplatesisreducedfrom to ?
1.Theelectricfielddecreases.2.Theelectricfieldincreases.3.Theelectricfieldstaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
d d/3
d d/3
d d/3
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 12
d
i.
9 V
d/2
ii.
9 V
d/2
metal
Quick Question 10
Now,thetwoplatesareconnectedtoavoltagesource(i.e.abattery)thatcontinuouslyprovides9V.Whathappenstotheamountofchargeoneachplateasthedistanceseparatingtheplatesisreducedfrom to ?
1.Theamountofchargeincreases.2.Theamountofchargedecreases.3.Theamountofchargestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
So,whatdidwefigureout:reducingtheplateseparationfortwoisolatedplatesdidnothingtothechargesontheplatesandtheelectricfieldbetweentheplates,butitdiddecreasethepotentialdifferencebetweenthetwoplates.
Reducingtheplateseparationfortwoconnectedandfixedpotentialplatesincreasedtheamountofchargesontheplatesandtheelectricfieldbetweentheplates,whilethepotentialdifferencestayedthesame.
So,weknowthatwhateverproportionalityconstantexistsbetweenvoltageandcharge,thenitwillhaveincreasedforthistransformation.
Indeed,wehaveincreasedthecapacitanceofthesystem.Inotherwords,itnowhasagreatercapacitytostorechargegivenavoltage.
Quick Question 11
Herewehavetwoplates,connectedbyabattery.Ametalslabisinsertedinthemiddleoftheemptyspacebetweenthetwoplates.Whatwillhappentotheelectricfieldintheemptyregionsaftertheslabisinserted?
1.Theelectricfielddecreases.2.Theelectricfieldincreases.3.Theelectricfieldstaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
d d/3
Q = constant × V
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 13
d
i.
9 V
d/2
ii.
9 V
d/2
metal
dielectric
i.
ii.
+q –q
Quick Question 12
Herewehavetwoplates,connectedbyabattery.Ametalslabisinsertedinthemiddleoftheemptyspacebetweenthetwoplates.Whatwillhappentotheamountofchargeontheoriginaltwometalplatesaftertheslabisinserted?
1.Theamountofchargeincreases.2.Theamountofchargedecreases.3.Theamountofchargestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
Quick Question 13
Herewehavetwoplates,withchargeq,and-q,andnobatteryisconnected.Insteadofametalslab,let'sinsertapolarizabledielectricinthemiddle.Whatwillhappentotheamountofchargeontheoriginaltwometalplatesafterthedielectricisinserted?
1.Theamountofchargeincreases.2.Theamountofchargedecreases.3.Theamountofchargestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 14
dielectric
i.
ii.
+q –q
dielectric
i.
ii.
+q –q
Quick Question 14
Samesituation:WhatwillhappentotheElectricFieldbetweenthetwometalplates?
1.Theelectricfielddecreases.2.Theelectricfieldincreases.3.Theelectricfieldstaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
Quick Question 15
Again:WhatwillhappentothePotentialDifferencebetweenthetwometalplates?
1.Thepotentialdifferencedecreases.2.Thepotentialdifferenceincreases.3.Thepotentialdifferencestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 15
dielectric
i.
9 V
ii.
9 V
dielectric
i.
9 V
ii.
9 V
Quick Question 16
Herewehavetwoplates,connectedbyabattery,thistime.Let'sinsertapolarizabledielectricinthemiddle.WhatwillhappentothePotentialDifferencebetweenthetwometalplates?
1.Thepotentialdifferencedecreases.2.Thepotentialdifferenceincreases.3.Thepotentialdifferencestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
Quick Question 17
Again,let'sinsertapolarizabledielectricinthemiddle.Whatwillhappentotheelectricfieldintheregionwherethedielectric?
1.Theelectricfielddecreases.2.Theelectricfieldincreases.3.Theelectricfieldstaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 16
dielectric
i.
9 V
ii.
9 V
dielectric
i.
ii.
+q –q
Quick Question 18
Whatwillhappentotheamountofchargeontheoriginaltwometalplatesafterthedielectricisinserted?
1.Theamountofchargeincreases.2.Theamountofchargedecreases.3.Theamountofchargestaysthesame.4.Cannotbedeterminedfromtheinformationgiven.
Now,wehaveshownthataddingadielectricmaterialinbetweenthetwoplatescausestheamountofchargeontheplatestoincreaseforagivenpotentialdifference.Or,ifthechargeisheldconstant,thenthepotentialdifferenceisseentodecrease.Thesamerelationshipholds.Wehavechangedtheconstantbetweenthe andthe .
Wecandefineadielectricconstant, ,bytheratioofthepotentialdifferencechangeexperiencedbytwoplatesuponinsertionofadielectricmaterial:
Thiscanbeextendedtoshowthechangeincapacitancebasedonthedielectricconstantusing
Q V
Q = constant × V
κ
κ =V0Vd
Q = CV
= κCd C0
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 17
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
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 18
Gauss' Law for dielectrics
1. Thefluxintegralnowinvolves ,notjust .Thevector issometimescalledtheelectricdisplacement
2. ThechargeqenclosedbytheGaussiansurfaceisnowtakentobethefreechargeonly.
∮ κ ⋅ d =ϵ0 E⃗ A⃗ qfree
κE⃗ E⃗ κE⃗
D⃗
PHY 208 - capacitance
updated on 2018-04-15 J. Hedberg | © 2018 Page 19