Electric Charges

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PHYS202 SPRING2009Lecturenotes Electrostatics

WhatisElectricity

Staticeffectsknownsinceancienttimes. Staticchargescanbemadebyrubbingcertainmaterialstogether.

DescribedbyBenjaminFranklinasafluid. Excessofthefluidwasdescribedaspositive.Deficit of the fluid was described as negative Deficitofthefluidwasdescribedasnegative.

Likechargesrepel,unlikechargesattract.

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AQuantitativeApproach

CharlesAugustin deCoulombfirstdescribedrelationshipofforce andcharges.

Coulombinventedthetorsionbalancetomeasuretheforcebetweencharges.

BasicunitofchargeinSIsystemiscalledtheCoulomb

Oneofthebasicirreduciblemeasurements.

Likemass,lengthandtime.

FundamentalCharge

Duetoatomicstructure.

Positively charged heavy nuclei and negative charged electrons.Positivelychargedheavynucleiandnegativechargedelectrons.

Chargedobjectshaveexcessordeficitofelectrons.

Protonsandelectronshaveequalandoppositecharge:e =1.602x1019 C

Rubbingsomeobjectscauseselectronstomovefromoneobjecttoanother.

Negativelychargedobjects haveanexcessofelectrons.

Positivelychargedobjects haveadeficitofelectrons.

Netchangeinchargeinaclosedsystemmustbezero.

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FundamentalCharge

Problem18.6Waterhasamasspermoleculeof18g/molandeach molecule has 10 electrons How many electrons are ineachmoleculehas10electrons.Howmanyelectronsareinoneliterofwater?

h i h h f h l ?

23 3

26

10electrons 6.02210 molecules 1mol 1g 10 ml 1molecule 1mol 18g 1ml 1liter

electrons3.3 10liter

N

N

=

=

Whatisthenetchargeoftheelectrons?

( )( )26 197

3.3 10 electrons 1.602 10 C / electron

5.4 10 C

q Ne

q

= = =

ConservationofCharge

Netchangeinchargeinanyclosedsystemiszero.

Or, the change of charge in any volume equals the net charge entering andOr,thechangeofchargeinanyvolumeequalsthenetchargeenteringandleavingthesystem:

WedenotetheamountofchargebytheletterQ orq

IfChargeiscreated,itmustbeequalandopposite.

( ) ( ) ( ) ( )Charge Original Charge Charge in Charge out= +

0 in out= + -q q q q

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ConservationofCharge

Twoidenticalmetalspheresofcharge5.0Cand1.6Carebrought together and then separated What is the charge onbroughttogetherandthenseparated.Whatisthechargeoneachsphere?

Sincethechargeisdistributedequally,eachmusthaveanequal charge:

1 2 = 3.4Ctotalq q q= +

equalcharge:

1 2 = 1.7C2totalqq q= =

ConductorsandInsulators

Insulator Charge stays where it is depositedChargestayswhereitisdeposited. Wood,mica,Teflon,rubber.

Conductor Chargemovesfreelywithintheobject. Metalsexhibithighconductivity

Duetoatomicstructure. Istheoutermostelectron(s)tightlybound?

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ChargingObjects

Byfriction Rubbing different materials Rubbingdifferentmaterials Example:furandebonite

Bycontact Bringingachargedobjectintocontactwithanother(chargedoruncharged)object.

Byinductionh d b d ll Bringingachargedobjectnearaconductortemporally

causesoppositechargetoflowtowardsthechargedobject.

Groundingthatconductordrainstheexcesschargeleavingtheconductoroppositelycharged.

ForceBetweenCharges

Coulombfoundthattheforcebetween two charges isbetweentwochargesisproportionaltotheproductoftheirchargesandinverselyproportiontothesquareofthedistancebetweenthem:

1 22

q qF Whereq1 andq2 representthechargesandr12 thedistancebetweenthem.

212r

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CoulombsLaw

Byusingthetorsionbalance,Coulombdiscoveredtheconstantofproportionality:k = 8.99 Nm2/C2

Ifbothq1 andq2 arebothpositive,theforceispositive.

1 22

12

q qF kr

=+ +

+

Ifbothq1 andq2 arebothnegative,theforceispositive. Ifbothq1 ispositiveandq2 isnegative,theforceisnegative. Hence,thesignoftheforceispositiveforrepulsionandnegativefor

attraction.

CoulombsLawPrincipleofSuperposition

Electricforceobeysthelawoflinearsuperposition thenetforceonachargeisthevector sumoftheforcesduetoeachcharge:

Addthevectors:

2i ji ij

i j ij

q qk

r= F rr

1

3

Thenetforceoncharge4isF14+F24+F34=Fnet

Red arepositivechargesBlue arenegativecharges

1

24

F14F24

F34

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Theforcesaddforasmanychargesasyouhave:

2i ji ij

i j ij

q qk

r= F rr

36


1

2

3

4

5

7

8


Electricforceobeysthelawoflinearsuperposition thenetforceonachargeisthevector sumoftheforcesduetoeachcharge:

2i ji ij

i j ij

q qk

r= F rr

36 Thenetforceoncharge5 isthe

vectorsumofalltheforceson

1

2

3

4

5

7

8charge5.


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CoulombsLaw18.14 Twotinyconductingspheresareidenticalandcarrychargesof20.0Cand

+50.0C.Theyareseparatedbyadistanceof2.50cm.(a)Whatisthemagnitudeoftheforcethateachsphereexperiences,andistheforceattractiveorrepulsive?

( )( )( )( )

9 2 2 6 61 2

22 2

8.99 10 N m /C 20.0 10 C 50.0 10 Cq qF k = =

Thenetforceisnegativesothattheforceisattractive.

(b)Thespheresarebroughtintocontactandthenseparatedtoadistanceof2.50cm.Determinethemagnitudeoftheforcethateachspherenowexperiences,andstatewhethertheforceisattractiveorrepulsive.Si h h id i l h h h f b i d i

( )22 2124

2.50 10 m

1.44 10 N

r

F

=

Sincethespheresareidentical,thechargeoneachafterbeingseparatedisonehalfthenetcharge,soq1 +q2 =+15C.

Theforceispositivesoitisrepulsive.

( )( )( )( )

9 2 2 6 631 2

2 2212

8.99 10 N m /C 15.0 10 C 15.0 10 C3.24 10 N

2.50 10 m

q qF k

r

= = =

CoulombsLaw18.21 Anelectricallyneutralmodelairplaneisflyinginahorizontalcircleona3.0m

guideline,whichisnearlyparalleltotheground.Thelinebreakswhenthekineticenergyoftheplaneis50.0J.Reconsiderthesamesituation,exceptthatnowthereisapointchargeof+q ontheplaneandapointchargeofq attheotherendofh id li I hi h li b k h h ki i f h l itheguideline.Inthiscase,thelinebreakswhenthekineticenergyoftheplaneis51.8J.Findthemagnitudeofthecharges.

Fromthedefinitionofkineticenergy,weseethatmv2 =2(KE),sothatEquation5.3forthecentripetalforcebecomes

( )2c

2 KEmFr r

= =So,

r r

( ) ( ) ( ) ( )2

charged unchargedmax max2

CentripetalCentripetal force

force

2 KE 2 KE1 2

k qT T

r rr+ = =1424314243

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CoulombsLaw18.21 (cont.)

( ) ( ) ( ) ( )2

charged unchargedmax max2

CentripetalCentripetal force

f

2 KE 2 KE1 2

k qT T

r rr+ = =1424314243

SubtractingEquation(2)fromEquation(1)eliminatesTmax andgives

( ) ( )2 charged uncharged2

2 KE KEk qrr

=

force

Solvingfor|q|gives

( ) ( ) ( ) ( )charged uncharged 59 2 2

2 KE KE 2 3.0 m 51.8 J 50.0 J3.5 10 C

8.99 10 N m / C

rq

k

= = =

TheElectricField Electricchargescreateandexertandexperiencetheelectricforce.

q q q

WedefinetheElectricField astheforceatestchargewouldfeelifplacedatthatpoint:

2 2 i j ji ij i i ij

i j i jij ij

q q qk q k

r r = = F r F rr r

Theelectricfieldexistsaroundallchargedobjects.

2j

i j

qk q

r= =E r F Er r r

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ElectricForceLines

MichaelFaradaysawtheforcecaused by one charge to becausedbyonechargetobefromlinesemanatingfromthecharges.

LinesbeginonPositivechargesandendonNegativecharges,flowingsmoothlythroughspace,nevercrossing.

ElectricForceLines

Faradayimaginedtheselinesd h fandtheforcewasgreater

wherethelinesaredenser,weakerwheretheyarespreadout.

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ElectricFieldLines

Gooutfrompositivechargesandintoti hnegativecharges.

PositiveNegativeNeutral

TheInverseSquareLaw

Manylawsbasedonthegeometryofthreedi i ldimensionalspace:

1 22

2 24 4

g

sound light

Gm mFr

P LI Ir r

=

= =1 2

2eq qF kr

=

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TheElectricField

18.28 Fourpointchargeshavethesamemagnitudeof2.4x1012 Candarefixedtothecornersofasquarethatis4.0cmonaside.Threeofthechargesarepositived i ti D t i th it d f th t l t i fi ld th t i tandoneisnegative.Determinethemagnitudeofthenetelectricfieldthatexists

atthecenterofthesquare.

A

B C

D-

++

+

EA

ECEB

ED

Thedrawingattherightshowseachofthefieldcontributionsatthecenterofthesquare(seeblackdot).Eachisdirectedalongadiagonalofthesquare.NotethatED andEB pointinoppositedirectionsand,therefore,cancel,sincetheyhavethesamemagnitude.IncontrastEAandEC pointinthesamedirectiontowardcorner A and therefore combine to give a net field thatcornerAand,therefore,combinetogiveanetfieldthatistwicethemagnitudeofEA orEC.Inotherwords,thenetfieldatthecenterofthesquareisgivenbythefollowingvectorequation:

2

= + + += + + = +=

A B C D

A B C B

A C

A

E E E E EE E E EE EE

TheElectricField

18.28 (cont.)

B C++

ED

Themagnitudeofthenetfieldis A 22 2k q

E Er

= =

A D- +

EA

ECEB

Inthisresultr isthedistancefromacornertothecenterofthesquare,whichisonehalfofthediagonaldistanced.UsingL forthelengthofasideofthesquareandtakingadvantageofthePythagoreantheorem,wehave.Withthissubstitutionforr,themagnitudeofthenetfieldbecomes

r

/ 2r L=

4k k

( )( ) ( )

( )

2 2

9 2 2 12

2

42

/ 2

4 8.99 10 N m / C 2.4 10 C

0.040 m

54 N/C

k q k qE

LL

E

E

= =

=

=

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ElectricFieldLines

18.34 ReviewConceptualExample12beforeattemptingtoworkthisproblem.ThemagnitudeofeachofthechargesinFigure1821is8.6x1012 C.Thelengthsofth id f th t l 3 00 d 5 00 Fi d th it d f ththesidesoftherectanglesare3.00cmand5.00cm.FindthemagnitudeoftheelectricfieldatthecenteroftherectangleinFigures1821aandb.

1 2

34

1

4

+ q

+ q + q

- qThefigureattherightshowstheconfigurationgivenintextFigure18.21a.Theelectricfieldatthecenteroftherectangleistheresultantoftheelectricfieldsatthecenterduetoeachofthefourcharges.AsdiscussedinConceptualExample11,themagnitudesoftheelectricfieldatthecenterduetoeachofthefourchargesareequal.However,thefieldsproducedbythechargesincorners1and3areinoppositedirections.Sincetheyhavethesamemagnitudes,theycombinetogivedirections. Since they have the same magnitudes, they combine to givezeroresultant.Thefieldsproducedbythechargesincorners2and4pointinthesamedirection(towardcorner2).Thus,EC =EC2 +EC4,whereEC isthemagnitudeoftheelectricfieldatthecenteroftherectangle,andEC2 andEC4 arethemagnitudesoftheelectricfieldatthecenterduetothechargesincorners2and4respectively.SincebothEC2 andEC4 havethesamemagnitude,wehaveEC =2EC2.

ElectricFieldLines18.34 (cont.)

Thedistancer,fromanyofthechargestothecenteroftherectangle,canbefoundusingthePythagoreantheorem:

1 21

d5.00 cm

2 2(3 00 ) +(5 00 ) 5 83d

ThefigureattherightshowstheconfigurationgivenintextFi 18 21b All f h t ib t t

344

3.00 cm

2 2(3.00 cm) +(5.00 cm) 5.83 cmd = =

2Therefore, 2.92 cm 2.92 10 m2dr = = =

9 2 2 1222

2 2 2 22 2(8.99 10 N m /C )( 8.60 10 C)2 1.81 10 N/C

(2.92 10 m)C Ckq

E Er

= = = = - q - q

Figure18.21b.Allfourchargescontributeanonzerocomponenttotheelectricfieldatthecenteroftherectangle.AsdiscussedinConceptualExample11,thecontributionfromthechargesincorners2and4pointtowardcorner2andthecontributionfromthechargesincorners1and3pointtowardcorner1.

1 2

34

1

4+ q + q

C

E 13 E 24

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ElectricFieldLines18.34 (cont.)

Noticealso,themagnitudesofE24andE13areequal,and,fromthefirstpartofthisproblem,weknowthat

E24 =E13 =1.81 102 N/C1 21- q - q

E 13 E 24

TheelectricfieldatthecenteroftherectangleisthevectorsumofE24 andE13.ThexcomponentsofE24 andE13 areequalinmagnitudeandoppositeindirection;hence

(E13)x (E24)x =0Therefore,

F Fi 2 h th t

344+ q + q

C

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E = + = =

FromFigure2,wehavethat

5.00 cm 5.00 cmsin 0.8585.83 cmd

= = =

( ) ( )( )2 2C 132 sin 2 1.81 10 N/C 0.858 3.11 10 N/CE E = = =

TheElectricField

Theelectricfieldexistsaroundallchargedobjects.

Thisisshownbyelectricfieldlines.

Constructedid i ll hidenticallytotheforcelines.

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DrawingElectricFieldLines

Numberoflinesatachargeisproportionaltothe chargethecharge.

ElectricFieldLines

Thechargewith3qmusthavethreetimesasmanylinesasthe charge with +q and 1.5 times the lines for +/ 2q.thechargewith q and1.5timesthelinesfor / 2q.

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ParallelPlatesofCharge

Addingfieldsfromeachpointonsurfacecausesfieldlinesparalleltotheplatestocancel.

PositivelyCharged

NegativelyCharged

Awayfromtheedges,thefieldisparallelandperpendiculartotheplate(s).

ElectricField&Conductors

Inconductorsexcesselectronswillmovewhenanelectricfieldispresent. Forsteadystate(i.e.nomovingcharges),thenetelectricfieldmustbezero.r r r r Thereforethechargesmovetothesurface(s)oftheconductoruntiltheycancel

theelectricfieldinsidetheconductor. Fieldlinescanstartandendonaconductoratrightanglestothesurfaceofthe

conductor.

0q = = =F E F Er r r r

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ElectricFieldLines

18.44 Twoparticlesareinauniformelectricfieldwhosevalueis+2500N/C.Themassandchargeofparticle1arem1 =1.4x105 kgandq1 =7.0C,whilethe

di l f ti l 2 2 6 10 5 k d 18 C I iti llcorrespondingvaluesforparticle2arem2 =2.6x105 kgandq2 =+18C.Initiallytheparticlesareatrest.Theparticlesarebothlocatedonthesameelectricfieldline,butareseparatedfromeachotherbyadistanced.Whenreleased,theyaccelerate,butalwaysremainatthissamedistancefromeachother.Findd.

Exd

+x

q1 =7.0Cm1 =1.4 105 kg

q2 =+18Cm2 =2.6 105 kg

ThedrawingshowsthetwoparticlesintheelectricfieldEx.Theyareseparatedbyadistanced.Iftheparticlesaretostayatthesamedistancefromeachotherafterbeingreleased,theymusthavethesameacceleration,soax,1 =ax,2.AccordingtoNewtonssecondlaw,theaccelerationax ofeachparticleisequaltothenetforceFx actingonitdividedbyitsmassm,or./x xa F m=

ElectricFieldLines

18.44 (cont.)

N t th t ti l 1 t b t th l ft f ti l 2 If ti l 1 t th i ht

Exd

+x

q1 =7.0Cm1 =1.4 105 kg

q2 =+18Cm2 =2.6 105 kg

Notethatparticle1mustbetotheleftofparticle2.Ifparticle1weretotherightofparticle2,theparticleswouldaccelerateinoppositedirectionsandthedistancebetweenthemwouldnotremainconstant.

1 2, 1 1 2

1 21 2, 1

,11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

= = =

1 2, 2 2 2

1 22 2, 2

, 22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

= ++= =

Notethesignsindicatedthedirectionofthevectorpointingfromcharge2tocharge1fortheforceoncharge1.

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ElectricFieldLines

18.44 (cont.)

Th l ti f h h t b th if th di t b t th

Exd

+x

q1 =7.0Cm1 =1.4 105 kg

q2 =+18Cm2 =2.6 105 kg

Theaccelerationofeachchargemust bethesameifthedistancebetweenthechargesremainsunchanged:

Solving for d

1, 2,

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

+=

Solvingford,

( )( )1 2 1 21 2 2 1 6.5 mx

kq q m md

E q m q m+= =

ElectricFlux Flux(fromLatin meaningtoflow) Electricfluxmeasuresamountofelectricfieldpassingthroughasurface.

Er

Er

So to calculate find the normal to

E EA =

FluxthroughsurfaceisAreaofsurfacetimesthefieldflowingthroughthatsurface.

So,tocalculate,findthenormal tothesurface.

Thenfindtheanglebetweenthefieldandthatnormal vector:

( )cosE E A =

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ElectricFlux

( )cosE E A =

Er

So to calculate find the normal to

Thiscanbeexpressedasthevectordotproduct:

E = E Arr

cosE A =E Arr So,tocalculate,findthenormal tothesurface.

Thenfindtheanglebetweenthefieldandthatnormalvector:

cos

x x y y z z

E Aor

E A E A E A

=

= + +

E A

E Arr

ElectricFlux

18.48 Arectangularsurface(0.16mx0.38m)isorientedinauniformelectricfieldof580N/C.Whatisthemaximumpossibleelectricfluxthroughthesurface?

Themaximumpossiblefluxoccurswhentheelectricfieldisparalleltothenormaloftherectangularsurface(thatis,whentheanglebetweenthedirectionofthefieldandthedirectionofthenormaliszero).Then,

( ) ( ) ( )2

( cos ) 580 N / C (cos 0 ) 0.16 m 0.38 m

35 N m / C E

E

E A = = =

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GaussLaw ElectricFluxthroughaclosedsurfaceisproportionaltothechargeenclosedbythatsurface:

encE

q =

Thefundamentalconstant0 is called the permittivity of free space. It is related to Coulombs law by , 0 =8.85x1012 C2/(Nm2)

GaussLawismostusefulwhentheelectricfieldontheentiresurfaceisconstant.

Takeapointcharge.Constructaspherecenteredonthepointcharge(allpointsonthesurfaceequidistantfromthecharge) theelectricfieldisconstant.

Since the field is radial it is perpendicular to the concentric spherical surface everywhere so

0

0

14

k =

Sincethefieldisradial,itisperpendiculartotheconcentricsphericalsurfaceeverywhereso=0,cos =1

Thedirectionofthefieldisradially outwardfromthecharge(orinwardtowardsthechargeifqisnegative).

0 0

cosenc encEq qEA ==

( )2 20 0

44

encq qE r Er

= =

GaussLawforaplateofcharge Inthiscasewetakeourclosedsurfacetobeaboxorientedsuchthattwofacesare

paralleltotheplateofchargeandtheothersareperpendiculartotheplate.

Si th l t i fi ld f l t f h i di l t th f f th Sincetheelectricfieldfromaplateofchargeisperpendiculartothesurfaceoftheplate,thesidesthatareperpendiculartotheplatehavenoflux.

Er

Thecharge enclosedbytheboxisthesurfacechargedensitymultipliedbytheareatheboxpassesthroughthe plate: Atheplate:.

Thearea oftheboxwithfluxis2A:

A

0 0 0E

q AEA E = = =

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GaussLawforalineofcharge

18.54 Along,thin,straightwireoflengthL hasapositivechargeQ distributeduniformlyalongit.UseGausslawtofindtheelectricfieldcreatedbythiswireatadi l di tradialdistancer.

Asthehintsuggests,wewilluseacylinderasourGaussiansurface.Theelectricfieldmustpointradially outwardfromthewire.Becauseofthissymmetrythefieldmustbethesamestrength pointingradially outwardonthesurfaceofthecylinderregardlessofthepointonthecylinder,thuswearejustifiedinusingGaussLaw.

Long straight wire Gaussian cylinderGausssiancylinderE

++++++ ++

12

3

L

r

End view

GaussLawforalineofcharge

18.54 (cont.)Letusdefinethefluxthroughthesurface.Notethatthefieldisparalleltothe

d th th l b t th l f th f ( hi h i ll l tendcaps,thustheanglebetweenthenormalfromthesurface(whichisparalleltothewire)isperpendiculartothefield,thus =/2,so,cos()=0fortheends.Forthesideifthecylinder,thefluxis

Thus,fromGaussLaw,

( )2E sideEA E rL = =++++++ ++

Long straight wire Gaussian cylinder

12

3

L

r

( )2encE q QE rL = =

Itiscustomarytorefertochargedensity =Q/L ,

End view

GausssiancylinderE

( )0 0

02QE

rL

=

02E

r=

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Work

LetustakeanobjectwithchargeqinaconstantelectricfieldE. Ifwereleasethischarge,theelectricfieldcausesaforcetomovethecharge.

Letusdefinethefieldas,sothereisonlyonedimensiontoconsider.Letusalsosettheinitialvelocityandpositiontobebothzero.

Lettheobjectmovethroughadistanced.Theequationsofmotionare

qqm

= =F E a E

21x at at= =

E=E x

So,afteratimet,theobjectmoves,2

x at atx d=

2122

qE d m qEdm

= =

21 2 /2

d at t d a= =/

2 2

t a

d ada a

== =

ConservationofEnergy

Thelefthandsideisthekineticenergytheparticlehasaftermovingadistanced.

212

m qEd=

Thatmeans,therighthandsidemustrepresentsomeformofpotentialenergytheobjecthadpriortobeingreleased.

TypesofEnergyalreadycovered: Translationalkineticenergy Rotationalkineticenergy Gravitationalpotentialenergy Elastic(spring)potentialenergy

ElectricPotentialEnergy Cutnell &Johnsonuse(EPE),however,itisoftenmoreconvenienttouseU.

2 2212

1 )12

(2total

E kx Em h PI g Em= + + + +

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ElectricPotentialEnergy

Likegravitationalpotentialenergy,EPE isdefinedwithrespecttotheenergyatsomelocation.

Importanceliesinthedifferencebetweenelectricpotentialenergybetweentwopoints.

ExampleProblem

19.4 Aparticlehasachargeof+1.5CandmovesfrompointAtopointB,adistanceof0.20m.Theparticleexperiencesaconstantelectricforce,anditsmotionisl th li f ti f th f Th diff b t th ti l l t ialongthelineofactionoftheforce.ThedifferencebetweentheparticleselectricpotentialenergyatAandBisUA UB =+9.0x104 J.Findthemagnitudeanddirectionoftheelectricforcethatactsontheparticleandthemagnitudeanddirectionoftheelectricfieldthattheparticleexperiences.Theworkdonemustbeequaltothechangeinpotentialenergy:

(from A to B)4

39.0 10 J 4.5 10 N

A B A B

A B

Work U U F x U UU UF

= = = = =

(fromA toB)

But,weknowthatforaparticleinanelectricfield:F=qE

( ) 4.5 10 N0.20mF x

( )( )4

36

9.0 10 J 3.0 10 N / C1.5 10 C 0.20 m

A B A BU U U UqE Ex q x

E

= = = =

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ElectricPotential

Theelectricpotentialenergyofanobjectdependsonthechargeofanobject.

( )EPE Onceagain,wecanfindameasurementindependentoftheobjectscharge: TheSIunitforV ismeasuredinVolts(=Joules/Coulomb)

AcommonunitofenergyistheelectronVolt.

Specifically,Howmuchpotentialenergydoesanelectrongetifmovedthrougha

( )EPEVq

=

potentialof1Volt.

191eV 1.602 10 J=

ElectricPotential

LiketheElectricfield,theelectricpotentialisindependent ofanytestchargeplacedtomeasureaforce.

Unliketheelectricfield,thepotentialisascaler,notavector.

Byusingcalculusmethodsitcanbeshownthattheelectricpotentialduetoapointchargeis

N h h l f h l i i l d d h i f h h

kqVr

= Notethatthevalueoftheelectricpotentialdependsonthesign ofthecharge.

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ElectricPotentialforapointcharge

Example:WhatwasthepotentialdifferenceduetoachargeqA = 10.0 nC, at the distances r0 = 10.0 cm and r1 = 20.0 cm ?qA 10.0nC,atthedistancesr0 10.0cmandr1 20.0cm?

0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

= = =

( )( )9 2 2 5 1 18.99 10 N m / C 10 C 0.1m 0.2 m450V

V

V

= =


Example:AchargeqA isheldfixed.AsecondchargeqB isplaced a distance r0 away and is released. What is the velocityplacedadistancer0 awayandisreleased.WhatisthevelocityofqB whenitisadistancer1 away?

Weusetheconservationofenergy:

2 2 20 0 1 1

1 1 12 2 2

Energy m U m U m U= + + = +

( )21 0 1 0 112 Bm U U q V V= = ( )1 0 12q V Vm=

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( )1 0 12q V Vm= Findtheelectricpotentialatr0 andr1:

0 10 1

A Akq kqV Vr r

= =

122 1 1A A A Bkq kq kq qq

m r r m r r = =

m

Or,

0 1 0 1m r r m r r

1 2 /q V m=


Forasinglecharge,wesetthepotentialatinfinitytozero.kqV =

Sowhatisthepotentialenergycontainedintwocharges? Bringinthechargesfrominfinity. Thefirstchargeisfreeasitdoesnotexperienceanyforces. Thesecondmovesthroughthefieldcreatedbythefirst. So,theenergyinthesystemmustbe

Vr

=

2

1 1 1

12 12

rU q V V V Vkq kq kqVr r

= = = =

1 2

12

kq qUr

=

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PotentialEnergyofacollectionofcharges

Letsaddathirdchargetothisgroup.

2 1 3 1 3 2

1 21 2

13 23

U q V q V q Vkq kqV Vr r

= + + = =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

= + +

Eachsuccessivechargeexperiencesthefield(i.e.potential)createdbythepreviouscharges.Becarefulnottodoublecount!

ExampleProblem

19.13 Anelectronandaprotonareinitiallyveryfarapart(effectivelyaninfinitedistanceapart).Theyarethenbroughttogethertoformahydrogenatom,inwhichth l t bit th t t di t f 5 29 10 11 Wh t i ththeelectronorbitstheprotonatanaveragedistanceof5.29x1011 m.Whatisthechangeintheelectricpotentialenergy?Thefirstchargecomesinforfree,thesecondchargeexperiencesthefieldofthefirst,so,

( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m /C 1 602 10 C

rU q V V V Vkq kq kqVr r

= = = =

( ) ( )( )19 1118

8.99 10 N m /C 1.602 10 C1.602 10 C

5.29 10 m4.35 10 J

U

U

= =

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ElectricFieldandPotential

LetachargebemovedbyaconstantelectricfieldfrompointsAtoB(letd be that distance).

W F d W qE d= =

Now,weknowthattheworkdonemustbeequaltothenegativechangeinelectricpotentialenergy:

Butpotentialenergyissimply:

W F d W qE d= =

ABU qE d =

Thus:

ABq V qE d =

VEd

=


Thisimpliesthatthefieldlinesareperpendiculartolines(surfaces)ofequal

VEd

= potential.

Thisequationonlygivesthecomponentoftheelectricfieldalongthedisplacementd.

Totrulyfindtheelectricvectorfield,youmustfindthecomponentsineachdirection:

V V V x y z

V V VE E Ex y z

= = =

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Ifthepotentialsatthefollowingpoints(x,y)areV(1,1) = 30 V, V(0.9,1)=15V, and V(1,0.9)=40 V. What is theV(1,1) 30V,V(0.9,1) 15V,andV(1,0.9) 40V.Whatisthe(approximate)electricfieldat(1,1)?

x y zV V VE E Ex y z

= = = 30V 15V 150 V / m

1.0m 0.9mxE = =

30V 40V 100 V / m1.0m 0.9my

E = =

( ) ( )2 21

150V 100V 180V

100Vtan 34150VE

E

= + =

= = o

x

E

(1,1)

y

Example

Problem19.34 Determinetheelectric field between each of theelectricfieldbetweeneachofthepoints.

VEx

= 5.0V 5.0V 0 V / m

0.20m 0.0mABE = =

3.0V 5.0V 10.0 V / m0.40m 0.20mBC

E = =1.0V 3.0V 5.0 V / m

0.80m 0.40mCDE = =

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Equipotential Surfaces

Electricfieldlinesareperpendiculartoequipotential surfaces.

Electricfieldlinesareperpendiculartosurfacesofconductors.

Thus,surfacesofconductorsareequipotential surfaces.

Theinteriorofconductorshavenoelectricfieldandarethereforeequipotential.

TheElectricpotentialofaconductorisconstantthroughouttheconductor.

Example19.68 Anelectronisreleasedatthenegativeplateofa

parallelplatecapacitorandacceleratestothepositiveplate(seethedrawing),(a)Astheelectrongainskineticenergy,doesitselectricpotentialenergyincreaseordecrease?Why?decrease? Why?

(b)ThedifferenceintheelectronselectricpotentialenergybetweenthepositiveandnegativeplatesisUpos Uneg .Howisthisdifferencerelatedtothechargeontheelectron(e)andtothedifferenceintheelectricpotentialbetweentheplates?The change in the electrons electric potential energy is equal to the charge on the

Theelectricpotentialenergydecreases.TheelectricforceFisaconservativeforce,sothetotalenergy(kineticenergypluselectricpotentialenergy)remainsconstantastheelectronmovesacrossthecapacitor.Thus,astheelectronacceleratesanditskineticenergyincreases,itselectricpotentialenergydecreases.

Thechangeintheelectron selectricpotentialenergyisequaltothechargeontheelectron(e)timesthepotentialdifferencebetweentheplates.

(c)Howisthepotentialdifferencerelatedtotheelectricfieldwithinthecapacitorandthedisplacementofthepositiveplaterelativetothenegativeplate?

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Example19.68 (cont.)(c)Howisthepotentialdifferencerelatedto

theelectricfieldwithinthecapacitorandthedisplacementofthepositiveplaterelativetothenegativeplate?

Problem Theplatesofaparallelplatecapacitorareseparatedbyadistanceof1.2cm,

TheelectricfieldEisrelatedtothepotentialdifferencebetweentheplatesandthedisplacements by

Notethat(Vpos Vneg)ands arepositivenumbers,sotheelectricfieldisanegativenumber,denotingthatitpointstotheleftinthedrawing.

( )pos negV VEs

=

andtheelectricfieldwithinthecapacitorhasamagnitudeof2.1x106 V/m.Anelectronstartsfromrestatthenegativeplateandacceleratestothepositiveplate.Whatisthekineticenergyoftheelectronjustastheelectronreachesthepositiveplate?Useconservationofenergy:

Total energy at Total energy atpositive plate negative plate

pos pos neg n pos negeg pos negeV eVKE U KE U KE KE + = ++ = +1442443 1442443

Example19.68 (cont.)

Buttheelectronstartsfromrestonthepositiveplate,so,h ki i h i i l i

pospos neg negKE KEeV eV+ = +

thekineticenergyatthepositiveplateiszero.

But so( )pos negV VE ( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

=

=

But,,so,( )Es

=

( ) ( )( )( )19 615

1.602 10 C 2.16 10 V / m 0.012m

4.0 10 J

neg

neg

KE

K

s

E

e E

=

=

=

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LinesofEqualPotential

Linesofequalelevationonamap show equal potentialmapshowequalpotentialenergy(mgh).

Example

19.29 Twoequipotential surfacessurrounda+1.5x108 Cpointcharge.Howfaristhe190Vsurfacefromthe75.0Vsurface?Th l t i t ti l V t di t f i t h i V k / ThTheelectricpotentialV atadistancer fromapointchargeq isV=kq/r.Thepotentialisthesameatallpointsonasphericalsurfacewhosedistancefromthechargeisr =kq/V.Theradialdistancer75fromthechargetothe75.0Vequipotential surfaceisr75 =kq/V75,andthedistancetothe190Vequipotentialsurfaceisr190 =kq/V190.Thedistancebetweenthesetwosurfacesis

75 19075 190 75 190

1 1kq kqr r kqV V V V

= =

( )29 875 190 2N m 1 18.99 10 1.50 10 C 1.1 m75.0 V 190 VCr r = + =

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Capacitors

Acapacitorconsistsoftwoconductorsatdifferentpotentialsbeingusedtostorecharge.

Commonly,thesetaketheshapeofparallelplates,concentricspheresorconcentriccylinders,butanyconductorswilldo.

q=CV,whereCisthecapacitance,measuredinSIunitsofFarads(Coulomb/VoltorC2/J),Visthevoltagedifferencebetweentheconductorsandq isthechargeoneachconductor(oneispositive,theothernegative).

Capacitanceisdependsonlyuponthegeometryoftheconductors.

ParallelPlateCapacitor

Fieldinaparallelplatecapacitoris0

qEA=

But,theaconstantelectricfieldcanbedefinedintermsofelectricpotential:

Substitutingintotheequationforcapacitance,

VE V Edd

= =

0

0

0

q CV CEd

Aqq C d CA d

= = = =

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OtherCapacitors

Twoconcentricsphericalshells4

Cylindricalcapacitorhasacapacitanceperunitlength: 2C

041 1

C

a b

=

02

ln

CbLa

=

Energystoredinacapacitor

Abatteryconnectedtotwoplatesofacapacitordoesworkcharging the capacitor.chargingthecapacitor.

Eachtimeasmallamountofchargeisaddedtothecapacitor,thechargeincreases,sothevoltagebetweentheplatesincreases.

V

qq

q/c

q

V=q/c

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Energystoredinacapacitor

PotentialEnergyisU =qV

Weseethataddingupeachsmallincreaseinchargeq x V isjusttheareaofthetriangle.

Suchthat,2

2

12

12 2

U qV q CV

qU U CVV

= =

= =V

So the energy density (i e energy per unit

qq

q/c

q

V=q/c ( )( )

20

20

1212

AU Edd

U E Ad

= =

Sotheenergydensity(i.e.energy perunitvolume)mustbe

Ad istheareaofthecapacitor!2

012

Uu EAd

= =

Example

Howmuchenergyisstoredinaparallelplatecapacitormadeof 10.0 cm diameter circles kept 2.0 mm apart when aof10.0cmdiametercircleskept2.0mmapartwhenapotentialdifferenceof120Visapplied?

Findthecapacitance:2

0 0A rCd d = =

Theenergystoredis( ) ( ) ( )

( )2 212 2 22 2

2 08.85 10 C / (N m ) 0.05m 120V1

2 2 2 0.002mr VU CVd

= = =92.09 10 JU =

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ElectricPolarizationinMatter

Electricchargedistributioninsomemoleculesis not uniformisnotuniform.

Example:Water bothHatomsareononeside.

InthepresenceofanElectricField,waterwillalignopposite tothefield.


Intheabsenceofafield.l l ll bMoleculeswillbe

orientedrandomly.

Inanelectricfield,theyline up opposite to thelineupoppositetothefield.

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Theresultingelectricfieldisthesumoftheexternalfieldandthefieldsgeneratedbythemolecules.

Theelectricfieldstrengththroughthematterisreduced.

Thereductionofthefielddependsontheenergyinthemolecules,thestrengthofthefield each one makes and the density of thefieldeachonemakesandthedensityofthemolecules.

Thiscanbemeasuredandthematerialissaidtohaveadielectricconstant .

Dielectrics

Thisdielectricconstantchangesthevalueofthepermittivityinspace()

q q Coulombslawinadielectricis

Where isrelatedto0 byaconstant:

Thedielectricconstantforavacuumis =1.

1 22

12

4q q

r=F r

0 =

Allothermaterialshaveadielectricconstantvaluegreaterthan1.

Wateratabout20Chas=80.4. Airatabout20Chas=1.00054.

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Dielectrics

0 = Howdoesthisaffectthecapacitanceofparallelplates?

0

0

d

d

AACd d

C C

= =

=

DielectricBreakdown

Ifthefieldinthedielectricbecomestoot (i lt i t t) th t i lstrong(i.e.voltageistoogreat),thematerialbeginstoconductcurrent thisiscalleddielectricbreakdown.

When this happens the charge flows throughWhenthishappensthechargeflowsthroughthematerialtotheotherplateandneutralizesthecapacitor.

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Example19.57 Twohollowmetalspheresareconcentricwitheachother.Theinnersphere

hasaradiusof0.1500mandapotentialof85.0V.Theradiusoftheoutersphereis0.1520manditspotentialis82.0V.IftheregionbetweenthespheresisfilledwithTeflon,findtheelectricenergycontainedinthisspace.Theelectricenergystoredintheregionbetweenthemetalspheresis

Thevolumebetweenthespheresisthedifferenceinthevolumesofthespheres:

2 20 0

1 1 ( )2 2

u E E VolumU e ==

( )3 3 3 32 2 2 24 4 4( ) 3 3 3Volume r r r r = = SinceE =V/sands=r2 r1,so,E =V/(r2 r1)

( )2 3 3 80 2 22 1

1 4 1.2 10 J2 3

V r rr

Ur

= =

Electric Charges

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