electricity magnetism lecture 9: conductors and capacitance lecture 10 - conductors … · two...

Electricity & MagnetismLecture 9: Conductors and Capacitance

Today’sConcept:

A)Conductors B)Capacitance

(

Electricity&Magne7smLecture7,Slide1

Some of your comments

Thischaptermakesabsolute0sense.

Iwouldreallyloveanexplana7ononhowcapacitorsareusedastemporarybaHeriesforstoringenergyinthem.

Capacitors vs. Batteries

‣ AbaHeryusesanelectrochemicalreac7ontoseparatecharges.ΔVbetweenterminalsiscalled“EMF”,symbol:,E orE.

‣ Capacitorsstorechargeonitsplates.

Capacitance and Capacitors (30.5)

We have been using capacitors a lot without defining capacitanceor describing how to charge-up these devices.

A battery will create a potential difference across a capacitorwhich is equal to the potential difference in the battery.

Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 4 / 14

E

Some capacitors

! 0

WE BELIEVE THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO ALL OF HOMEWORK!

1. E = 0 withinthematerialofaconductor:(atsta7cequilibrium)

Chargesmoveinsideaconductorinordertocanceloutthefieldsthatwouldbethereintheabsenceoftheconductor.Thisprincipledeterminestheinducedchargedensi7esonthesurfacesofconductors.

CONCEPTSDETERMINETHECALCULATION!

Our Comments

2.Gauss’Law:Ifchargedistribu7onshavesufficientsymmetry(spherical,

cylindrical,planar),thenGauss’lawcanbeusedtodeterminetheelectricfieldeverywhere.


I

surface

~E · d ~A = Qenclosed✏

0

�Va⇤b ⇥�Ua⇤b

q= �Z b

a~E · d~l

Conductors

" Chargesfreetomove" E=0inaconductor" Surface=Equipoten7al" Eatsurfaceperpendiculartosurface

TheMainPoints


CheckPoint: Two Spherical Conductors 1


Twosphericalconductorsareseparatedbyalargedistance.Theyeachcarrythesameposi7vechargeQ.ConductorAhasalargerradiusthanconductorB.

Comparethepoten7alatthesurfaceofconductorAwiththepoten7alatthesurfaceofconductorB.

A.VA>VBB.VA=VBC.VA<VB

“Thepoten7alisthesameasfromapointchargeatthecenterofAorB.So,followingV=kQ/r,Awouldhaveasmallerpoten7al“




Thetwoconductorsarenowconnectedbyawire.Howdothepoten7alsattheconductorsurfacescomparenow?

A.VA>VBB.VA=VBC.VA<VB

“Thepoten7alswillbecomeequalsincethechargeswillwanttogotoplacesoflowerpoten7al,un7litbalancesout.“




WhathappenstothechargeonconductorAaberitisconnectedtoconductorBbythewire?

A.QAincreasesB.QAdecreasesC.QAdoesnotchange

“Since B initially has a higher potential, charges move from B to A. “

786 | C H A P T E R 2 3 Electric Potential

Example 23-15 Two Charged Spherical Conductors

Two uncharged spherical conductors of radius and(Figure 23-26) and separated by a distance much greater than

are connected by a long, very thin conducting wire. A total chargeis placed on one of the spheres and the system is allowed to

reach electrostatic equilibrium. (a) What is the charge on each sphere?(b) What is the electric field strength at the surface of each sphere?(c) What is the electric potential of each sphere? (Assume that the chargeon the connecting wire is negligible.)

PICTURE The total charge will be distributed with on sphere 1 and on sphere 2 so that the spheres will be at the same potential. We can use

for the potential of each sphere.

SOLVE

V ! kQ>R Q2Q1

Q ! "80 nC6.0 cmR2 ! 2.0 cm

R1 ! 6.0 cm

(a) 1. Conservation of charge gives us one relation between thecharges and Q2:Q1

Q1 " Q2 ! Q

2. Equating the potential of the spheres gives us a secondrelation for the charges and Q2:Q1

kQ1R1

!kQ2R2

⇒ Q2 !R2R1Q1

3. Combine the results from steps 1 and 2 and solve for andQ2:

Q1 so

20 nCQ2 ! Q # Q1 !

60 nCQ1 !R1

R1 " R2Q !

6.0 cm8.0 cm

(80 nC) !

Q1 "R1R2Q1 ! Q

(b) Use these results to calculate the electric field strengths at thesurface of the spheres:

450 kN>C!E2 !

kQ2R22

!(8.99 $ 109 N # m2>C2)(20 $ 10#9 C)

(0.020 m)2

150 kN>C!E1 !

kQ1R21

!(8.99 $ 109 N # m2>C2)(60 $ 10#9 C)

(0.060 m)2

(c) Calculate the common potential from for either sphere:kQ>R9.0 kV!

V1 !kQ1R1

!(8.99 $ 109 N # m2>C2)(60 $ 10#9 C)

0.060 m

CHECK If we use sphere 2 to calculate we obtain An additional check is available, be-

cause the electric field strength at the surface of each sphere is proportional to its chargedensity. The radius of sphere 1 is three times the radius of sphere 2, so its surface area isnine times the surface area of sphere 2. And because sphere 1 has three times the charge,its charge density is one-third the charge density of sphere 2. Therefore, the electric fieldstrength at the surface of sphere 1 should be one-third of the electric field strength at thesurface of sphere 2, which is what we found in Part (b).

TAKING IT FURTHER The presence of the long, very thin wire connecting the spheresmakes the result of this example only approximate because the potential function is valid for the region outside an isolated conducting sphere. With the wire in place thespheres cannot accurately be modeled as isolated spheres.

V ! kQ>r

109 N # m2>C2)(20 $ 10#9 C)>0.020 m ! 9.0 $ 103 V. V2 ! kQ2 >R2 ! (8.99 $V,

R1R2

F I G U R E 2 3 - 2 6

CheckPoint: Charged Parallel Plates 1


TwoparallelplatesofequalareacarryequalandoppositechargeQ0.Thepoten7aldifferencebetweenthetwoplatesismeasuredtobeV0.Anunchargedconduc7ngplate(thegreenthinginthepicturebelow)isslippedintothespacebetweentheplateswithouttouchingeitherone.ThechargeontheplatesisadjustedtoanewvalueQ1suchthatthepoten7aldifferencebetweentheplatesremainsthesame.

CompareQ1andQ0.oQ1<Q0oQ1=Q0oQ1>Q0

THECAPACITORQ

UESTIONSWERE

TOUGH!

THEPLAN:

We’llwo

rkthroughtheexa

mpleinthe

PreLectureandth

endothepreflig

htques7ons.

Capacitanceisdefinedforanypairofspa7allyseparatedconductors.

Howdoweunderstandthisdefini7on?

+Q

−Qd

!Considertwoconductors,onewithexcesscharge=+Qandtheotherwithexcesscharge=−Q

!Thesechargescreateanelectricfieldinthespacebetweenthem

E

!Thispoten7aldifferenceshouldbepropor7onaltoQ!•Thera7oofQtothepoten7aldifferenceisthecapacitanceandonlydependsonthegeometryoftheconductors

V

!Wecanintegratetheelectricfieldbetweenthemtofindthepoten7aldifferencebetweentheconductor

Capacitance


First determine E field produced by charged conductors:

Second,integrateEtofindthepotenHaldifferenceV

x

y

Aspromised,VisproporHonaltoQ!

Cdeterminedbygeometry!

Example (done in Prelecture 7)

Whatisσ ?

A=areaofplate

+Q

−Qd E


V = �Z d

0~E · d~y

+Q0

−Q0

dIniHalchargeoncapacitor=Q0

PlatesnotconnectedtoanythingA)Q1<Q0B)Q1=Q0C)Q1>Q0

HowisQ1relatedtoQ0?

Clicker Question Related to CheckPoint

td

+Q1

−Q1

Insertunchargedconductor

Chargeoncapacitornow=Q1

CHARGECANNOTCHANGE!


td

+Q1

−Q1

A)+Q1B)−Q1C) 0D)PosiHvebutthemagnitudeunknownE)NegaHvebutthemagnitudeunknown

WhatisthetotalchargeinducedontheboHomsurfaceoftheconductor?

Where to Start ?


E

+Q0

−Q0

E

WHATDOWEKNOW?

E = 0

Emustbe= 0inconductor!

Why ?

+Q0

−Q0

ChargesinsideconductormovetocancelEfieldfromtop&bo8omplates.


NowcalculateVasafuncHonofdistancefromtheboSomconductor.

+Q0

−Q0

E = 0

y

V

y

d t

WhatisΔV=V(d)?A)ΔV=E0dB)ΔV=E0(d−t)C)ΔV=E0(d+t)

d

tEy

-E0

Theintegral=areaunderthecurve

Calculate V


V( y) = �Z y

0~E · d~y

CheckPoint Results: Charged Parallel Plates 1



CompareQ1andQ0.A.Q1<Q0B.Q1=Q0C.Q1>Q0

“ThedistanceforQ1issmallerthereforethechargemustdecreasetocompensateforthechangeindistance““Sincethepoten?alremainsthesame,thereisnochangeinthechargeofQ.““thefieldthroughtheconductoriszero,soithasconstantpoten?al.BecauseofthisitmusthavegreaterchargesothetotalVisthatsame.“




Comparethecapacitanceofthetwoconfigura7onsintheaboveproblem.A.C1>C0B.C1=C0C.C1<C0

“ThedistanceforQ1issmallerthereforethechargemustdecreasetocompensateforthechangeindistance““Sincethepoten?alremainsthesame,thereisnochangeinthechargeofQ.““thefieldthroughtheconductoriszero,soithasconstantpoten?al.BecauseofthisitmusthavegreaterchargesothetotalVisthatsame.“




Comparethecapacitanceofthetwoconfigura7onsintheaboveproblem.A.C1>C0B.C1=C0C.C1<C0

WecandetermineCfromeithercase sameV(preflight) sameQ(lecture)Cdependsonlyongeometry! E0 = Q0 /ε0A

C0 = Q0 /E0d

C1 = Q0 /[E0(d − t)]

C0 = ε0A /d

C1 = ε0A /(d − t)

BANG

Energy in Capacitors


!ConceptualAnalysis:

ButwhatisQ andwhatisV ?Theyarenotgiven?

!ImportantPoint:C isapropertyoftheobject!(concentriccylindershere)• AssumesomeQ (i.e.,+Q ononeconductorand−Q ontheother)• ThesechargescreateEfieldinregionbetweenconductors• This E fielddeterminesapoten7aldifferenceV betweentheconductors• V shouldbepropor7onaltoQ; thera7oQ/V isthecapacitance.

Calculation

metal

metal

a1a2

a3a4 cross-sec7on Acapacitorisconstructedfromtwoconduc7ng

cylindricalshellsofradiia1,a2,a3,anda4andlength

L(L>>ai).

WhatisthecapacitanceCofthiscapacitor?


!StrategicAnalysis:• Put+Qonoutershelland −Qoninnershell• Cylindricalsymmetry:UseGauss’LawtocalculateEeverywhere• IntegrateEtogetV• Takera7oQ/V:shouldgetexpressiononlyusinggeometricparameters(ai,L)

Calculationcross-sec7on

metal

metal

a1a2

a3a4 Acapacitorisconstructedfromtwoconduc7ng

cylindricalshellsofradiia1,a2,a3,anda4andlength

L(L>>ai).



Calculation

metal

metal

+Q

a1a2

a3a4

−Q

Why?

Whereis+Qonouterconductorlocated?A)atr=a4B)atr=a3C)bothsurfacesD)throughoutshell

WeknowthatE = 0 inconductor(betweena3anda4)

++

+

+

+++

+

+

++

+

+

+++

+

+

cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength

L(L>>ai).


+Q mustbeoninsidesurface(a3), sothat Qenclosed = + Q − Q = 0


Gauss’law:

I

surface


0

Calculation

metal

+Q

a1a2

a3a4

−Q

Why?

+ ++

+

+

++

+

+

+ ++

+

+

+++

+

+

We know that E = 0 in conductor (between a1 and a2)

−

metal

−−−−−

−−− −

−−−−

−

−−−


L(L>>ai).


Whereis−Qoninnerconductorlocated?A)atr=a2B)atr= a1C)bothsurfacesD)throughoutshell

+Q mustbeonoutersurface(a2),sothatQenclosed = 0


Gauss’law:

I

surface


0

Calculation

metal

+Q

a1a2

a3a4

−Q

+ ++

+

+

++

+

+

+ ++

+

+

++

+

+

−

metal

−−−−−

−−− −

−−−−

−

−−−

−Why?


L(L>>ai).


a2<r<a3:WhatisE(r)?

A) 0 B) C)D)E)

E(2⇡rL) =Q✏0

Direc7on:RadiallyIn


Gauss’law:

I

surface


0

Calculation

r < a2: E(r) = 0

since Qenclosed = 0

a2<r<a3:

WhatisV ?Thepoten7aldifferencebetweentheconductors.

WhatisthesignofV=Vouter−Vinner?A)Vouter − Vinner<0B)Vouter − Vinner= 0C)Vouter−Vinner>0

Q(2πε0a2L)

metal

+Q

a1a2

a3a4

−Q

+ ++

+

+

++

+

+

+ ++

+

+

++

+

+

−

metal

−−−−−

−−− −

−−−

−

−−−

−


L(L>>ai).



Q(2πε0a2L)

WhatisV≡Vouter−Vinner?

(A)(B)(C)(D)

Calculation

V propor7onaltoQ,aspromised

metal

+Q

a1a2

a3a4

−Q

+ ++

+

+

++

+

+

+ ++

+

+

++

+

+

−

metal

−−−−−

−−− −

−−−

−

−−−

−


L(L>>ai).


a2<r<a3:


C ⌘ QV=

2⇡✏0Lln(a3/a2)

The Potential of a Ring of Charge

The Potential of a Ring of Charge (Example 29.11)

A thin, uniformly charged ring of radius Rhas total charge Q . Find the potential atdistance z on the axis of the ring.

The distance ri

isr

i

=p

R

2 + z2

The potential is then the sum:

V =NX

i=1

14⇤�0

�Q

r

i

=1

4⇤�01⌃

R

2 + z2

NX

i=1

�Q =1

4⇤�0Q⌃

R

2 + z2Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 5 / 19

electricity magnetism lecture 9: conductors and capacitance lecture 10 - conductors … · two...

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