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

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

Some capacitors

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.

surface

~E · d ~A = Qenclosed

�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>VB

B.VA=VB

C.VA<VB

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

Thetwoconductorsarenowconnectedbyawire.Howdothepoten7alsattheconductorsurfacescomparenow?

A.VA>VB

B.VA=VB

C.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.

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

R1!kQ2

R2⇒ Q2 !

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

20 nCQ2 ! Q # Q1 !

60 nCQ1 !R1

R1 " R2Q !

6.0 cm8.0 cm

(80 nC) !

Q1 "R1

R2Q1 ! Q

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

450 kN>C!

E2 !kQ2

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

(0.020 m)2

150 kN>C!

E1 !kQ1

!(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 !kQ1

(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,

F I G U R E 2 3 - 2 6

CheckPoint: Charged Parallel Plates 1

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

CompareQ1andQ0.oQ1<Q0

oQ1=Q0

oQ1>Q0

THECAPACITORQ

UESTIONSWERETOU

THEPLAN:

We’llwork

throughtheexam

pleinthe

PreLectureandth

endothepreflig

htques7ons.

Capacitanceisdefinedforanypairofspa7allyseparatedconductors.

Howdoweunderstandthisdefini7on?

!Considertwoconductors,onewithexcesscharge=+Qandtheotherwithexcesscharge=−Q

!Thesechargescreateanelectricfieldinthespacebetweenthem

!Thispoten7aldifferenceshouldbepropor7onaltoQ!•Thera7oofQtothepoten7aldifferenceisthecapacitanceandonlydependsonthegeometryoftheconductors

!Wecanintegratetheelectricfieldbetweenthemtofindthepoten7aldifferencebetweentheconductor

Capacitance

First determine E field produced by charged conductors:

Second,integrateEtofindthepotenHaldifferenceV

Aspromised,VisproporHonaltoQ!

Cdeterminedbygeometry!

Example (done in Prelecture 7)

Whatisσ ?

A=areaofplate

−Qd E

V = �Z d

0~E · d~y

dIniHalchargeoncapacitor=Q0

PlatesnotconnectedtoanythingA)Q1<Q0

B)Q1=Q0

C)Q1>Q0

HowisQ1relatedtoQ0?

Clicker Question Related to CheckPoint

Insertunchargedconductor

Chargeoncapacitornow=Q1

CHARGECANNOTCHANGE!

B)−Q1

C) 0D)PosiHvebutthemagnitudeunknownE)NegaHvebutthemagnitudeunknown

WhatisthetotalchargeinducedontheboHomsurfaceoftheconductor?

Where to Start ?

WHATDOWEKNOW?

Emustbe= 0inconductor!

ChargesinsideconductormovetocancelEfieldfromtop&bo8omplates.

NowcalculateVasafuncHonofdistancefromtheboSomconductor.

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

Theintegral=areaunderthecurve

Calculate V

V( y) = �Z y

0~E · d~y

CheckPoint Results: Charged Parallel Plates 1

CompareQ1andQ0.A.Q1<Q0

B.Q1=Q0

C.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)

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

cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength

L(L>>ai).

WhatisthecapacitanceCofthiscapacitor?

!StrategicAnalysis:• Put+Qonoutershelland −Qoninnershell

• Cylindricalsymmetry:UseGauss’LawtocalculateEeverywhere• IntegrateEtogetV• Takera7oQ/V:shouldgetexpressiononlyusinggeometricparameters(ai,L)

Calculationcross-sec7on

a3a4 Acapacitorisconstructedfromtwoconduc7ng

cylindricalshellsofradiia1,a2,a3,anda4andlength

L(L>>ai).

Calculation

Whereis+Qonouterconductorlocated?

A)atr=a4B)atr=a3C)bothsurfacesD)throughoutshell

WeknowthatE = 0 inconductor(betweena3anda4)

L(L>>ai).

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

Gauss’law:

surface

Calculation

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

−−−−−

−−− −

−−−−

−−−

L(L>>ai).

Whereis−Qoninnerconductorlocated?

A)atr=a2B)atr= a1C)bothsurfacesD)throughoutshell

+Q mustbeonoutersurface(a2),

sothatQenclosed = 0

Gauss’law:

surface

Calculation

−−−−−

−−− −

−−−−

−−−

−Why?

L(L>>ai).

a2<r<a3:WhatisE(r)?

A) 0 B) C)D)E)

E(2⇡rL) =Q✏0

Direc7on:RadiallyIn

Gauss’law:

surface

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)

−−−−−

−−− −

−−−

L(L>>ai).

Q(2πε0a2L)

WhatisV≡Vouter−Vinner?

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

Calculation

V propor7onaltoQ,aspromised

−−−−−

−−− −

−−−

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 R

has total charge Q . Find the potential atdistance z on the axis of the ring.

The distance r

The potential is then the sum:

14⇤�0

4⇤�01⌃

�Q =1

4⇤�0Q⌃

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

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