ElectricElectric PotentialPotential AA PowerPointPowerPoint PresentationPresentation byby PaulPaul E.E. Tippens,Tippens, ProfessorProfessor ofof PhysicsPhysics SouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity

© 2007 Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: • Understand an apply the concepts of electric , , and electric potential difference. • Calculate the required to move a known charge from one point to another in an electric created by point charges. • Write and apply relationships between the , potential difference, and plate separation for parallel plates of equal and opposite charge. Review:Review: WorkWork andand EnergyEnergy

WorkWork isis defineddefined asas thethe productproduct ofof displacementdisplacement dd andand aa parallelparallel appliedapplied forceforce F.F.

WorkWork == Fd;Fd; Units:Units: 11 JJ == 11 NN mm

PotentialPotential EnergyEnergy UU isis defineddefined asas thethe abilityability toto dodo workwork byby virtuevirtue ofof positionposition oror conditioncondition.. ()(Joules)

KineticKinetic EnergyEnergy KK isis defineddefined asas thethe abilityability toto dodo workwork byby virtuevirtue ofof motionmotion ()(velocity).. (Also(Also inin joules)joules) SignsSigns forfor WorkWork andand EnergyEnergy

WorkWork ((FdFd)) isis positivepositive ifif anan appliedapplied forceforce FF isis inin thethe samesame directiondirection asas thethe displacementdisplacement dd.. TheThe forceforce FF doesdoes positivepositive work.work. B F TheThe forceforce mgmg doesdoes negativenegative workwork. m d TheThe P.E.P.E. atat BB relativerelative toto AA isis mg positivepositive becausebecause thethe fieldfield cancan dodo A positivepositive workwork ifif mm isis released.released. P.E.P.E. atat AA relativerelative toto BB isis negativenegative;; outsideoutside forceforce neededneeded toto movemove mm.. GravitationalGravitational WorkWork andand EnergyEnergy Consider work against g to move m B from A to B, a vertical height h. F Work = Fh = mgh m h g At level B, the U is: mg A U = mgh (gravitational) The external does positive work; the g does negative work.

TheThe externalexternal forceforce FF againstagainst thethe g-fieldg-field increasesincreases thethe potentialpotential energy.energy. IfIf releasedreleased thethe fieldfield givesgives workwork back.back. ElectricalElectrical WorkWork andand EnergyEnergy An external force F moves +q from A to B against the field force qE. B + + + +

Work = Fd = (qE)d Fe d At level B, the potential energy U is: +q + qE E U = qEd (Electrical) A ---- The E-field does negative work; External force does positive work.

TheThe externalexternal forceforce FF againstagainst thethe E-fieldE-field increasesincreases thethe potentialpotential energy.energy. IfIf releasedreleased thethe fieldfield givesgives workwork back.back. WorkWork andand NegativeNegative ChargesCharges Suppose a negative charge –q is moved against E from A to B. B + + + + Work by E = qEd qE d At A, the potential energy U is: -q E U = qEd (Electrical) A ---- No external force is required !

TheThe E-fieldE-field doesdoes positivepositive workwork onon –q–q decreasingdecreasing thethe potentialpotential energy.energy. IfIf releasedreleased fromfrom BB nothingnothing happens.happens. WorkWork toto MoveMove aa ChargeCharge Work to move +q from A to B. rb + +   F   kqQ + + F  +Q +   At A: a 2 ++ qE ra ra kqQ At B: Fb  2 rb kqQ Avg. Force: Favg  : ra -rb rrab kQq 11 Work Fd() r  r Work kQq rr ab  ab rrba AbsoluteAbsolute PotentialPotential EnergyEnergy AbsoluteAbsolute P.E.P.E. isis relativerelative toto  r b ItIt isis workwork toto bringbring + ++   F   +Q +   +q from infinity to +++ qE +q from infinity to ra pointpoint nearnear QQ——i.e.,i.e., fromfrom  toto rrb 0 11 11 kQq Work kQqWork kQq rrba rrbb Absolute Potential kQq U  Energy: r ExampleExample 1.1. WhatWhat isis thethe potentialpotential energyenergy ifif aa +2+2 nCnC chargecharge movesmoves fromfrom  toto pointpoint AA,, 88 cmcm awayaway fromfrom aa +6+6 CC charge?charge? TheThe P.E.P.E. willwill bebe positivepositive atat A pointpoint AA,, becausebecause thethe fieldfield cancan  +2 nC dodo ++ workwork ifif qq isis released.released. 8 cm +Q PotentialPotential kQq U  +6 C Energy:Energy: r (9 x 109-Nm2 )( 6 x 106- C)(+2 x 109 C) U  C2 (0.08 m)

UU == 1.351.35 mJmJ Positive potential energy SignsSigns forfor PotentialPotential EnergyEnergy

ConsiderConsider PointsPoints A,A, B,B, andand C.C. A B   ForFor +2+2 nCnC atat A:A: UU == +1.35+1.35 mJmJ 8 cm 12 cm Questions: +Q  C 4 cm If +2 nC moves from A to B, +6 C does field E do + or – work? Moving Does P.E. increase or decrease? positive q +2 nC

TheThe fieldfield EE doesdoes positivepositive work,work, thethe P.E.P.E. decreases.decreases. If +2 nC moves from A to C (closer to +Q), the field E does negative work and P.E. increases. ExampleExample 2.2. WhatWhat isis thethe changechange inin potentialpotential energyenergy ifif aa +2+2 nCnC chargecharge movesmoves fromfrom  toto BB??

A B PotentialPotential kQq   Energy:Energy: U  r 8 cm 12 cm +Q From Ex-1: U = + 1.35 mJ A +6 C

(9 x 199-Nm2 )( 6 x 106- C)(+2 x 109 C) U C2 0.900 mJ B (0.12 m)

U = UB –UA = 0.9 mJ – 1.35 mJ UU == -0.450-0.450 mJmJ

Note that P.E. has decreased as work is done by E. MovingMoving aa NegativeNegative ChargeCharge ConsiderConsider PointsPoints A,A, B,B, andand C.C. A B   SupposeSuppose aa negativenegative --qq isis moved.moved. 8 cm 12 cm Questions: +Q  C 4 cm If -q moves from A to B, does +6 C field E do + or – work? Does Moving P.E. increase or decrease? negative q -

TheThe fieldfield EE doesdoes negativenegative work,work, thethe P.E.P.E. increases.increases. What happens if we move a –2 nC charge from A to B instead of a +2 nC charge. This example follows . . . ExampleExample 3.3. WhatWhat isis thethe changechange inin potentialpotential energyenergy ifif aa --22 nCnC chargecharge movesmoves fromfrom  toto BB??

PotentialPotential kQq A B Energy:Energy: U   r 8 cm 12 cm From Ex-1: UA = -1.35 mJ +Q (Negative(Negative duedue toto –– charge)charge) +6 C (9 x 199-Nm2 )(6 x 106- C)(-2 x 109 C) U C2 0.900 mJ B (0.12 m)

UB –UA = -0.9 mJ – (-1.35 mJ) UU == +0.450+0.450 mJmJ

AA –– chargecharge movedmoved awayaway fromfrom aa ++ chargecharge gainsgains P.E.P.E. PropertiesProperties ofof SpaceSpace

Electric Field An electric field is a property of allowing prediction of the . E force on a charge at that point. r F E ; FqE + + q +Q + +++ The field E exist independently of E is a Vector the charge q and is found from:

kQ Electric Field: E  r 2 ElectricElectric PotentialPotential ElectricElectric potentialpotential isis anotheranother propertyproperty ofof spacespace allowingallowing usus toto predictpredict thethe U P. V  P.E.P.E. ofof anyany chargecharge qq atat aa point.point. q r ElectricElectric U + + VUqV; +Q + Potential:Potential: q +++ The units are: joulesjoules perper coulombcoulomb (J/C)(J/C) Potential

ForFor example,example, ifif thethe potentialpotential isis 400400 J/CJ/C atat pointpoint PP,, aa ––22 nCnC chargecharge atat thatthat pointpoint wouldwould havehave P.E.P.E. ::

U = qV = (-2 x 10-9C)(400 J/C); UU == -800-800 nJnJ TheThe SISI UnitUnit ofof PotentialPotential ()(Volt) FromFrom thethe definitiondefinition ofof electricelectric potentialpotential asas P.E.P.E. perper unitunit chargecharge,, wewe seesee thatthat thethe unitunit mustmust bebe J/CJ/C.. WeWe redefineredefine thisthis unitunit asas thethe voltvolt (V).(V). U 1 V  ;  1 volt = q 1

AA potentialpotential ofof oneone voltvolt atat aa givengiven pointpoint meansmeans thatthat aa chargecharge ofof oneone coulombcoulomb placedplaced atat thatthat pointpoint willwill experienceexperience aa potentialpotential energyenergy ofof oneone joulejoule.. CalculatingCalculating ElectricElectric PotentialPotential

ElectricElectric PotentialPotential EnergyEnergy andand Potential:Potential: kQ P. V  kQq U r UV; r rq + + +Q + kQq +++ SubstitutingSubstituting forfor  r  kQ V  Potential U,U, wewe findfind V:V: qr

kQ TheThe potentialpotential duedue toto aa positivepositive chargecharge isis V  positive;positive; TheThe potentialpotential duedue toto aa negativenegative r chargecharge isis positive.positive. (Use(Use signsign ofof charge.)charge.) ExampleExample 4:4: FindFind thethe potentialpotential atat aa distancedistance ofof 66 cmcm fromfrom aa ––55 nCnC charge.charge.

9-Nm2 9 kQ 9 x 10C2  ( 5 x 10 C) P q = –4 C V  . r (0.06 m) r 6 cm Negative V at - - - VVP == -750-750 VV - Q - Point P : P - - Q = -5 nC WhatWhat wouldwould bebe thethe P.E.P.E. ofof aa ––44 CC chargecharge placedplaced atat thisthis pointpoint P?P?

UU == qVqV == ((--44 xx 1010-6 C)(C)(--750750 V);V); UU == 3.003.00 mJmJ

SinceSince P.E.P.E. isis positive,positive, EE willwill dodo ++ workwork ifif qq isis released.released. PotentialPotential ForFor MultipleMultiple ChargesCharges

TheThe ElectricElectric PotentialPotential VV inin thethe vicinityvicinity ofof aa numbernumber ofof chargescharges isis equalequal toto thethe algebraicalgebraic sumsum ofof thethe potentialspotentials duedue toto eacheach charge.charge.

kQ12 kQ kQ3 r1 V  Q1 -  A A rrr123 r2 r3 + kQ Q3 - Q V  2  r

PotentialPotential isis ++ oror –– basedbased onon signsign ofof thethe chargescharges Q.Q. ExampleExample 5:5: TwoTwo chargescharges QQ1= = +3+3 nCnC andand QQ2 == --55 nCnC areare separatedseparated byby 88 cmcm.. CalculateCalculate thethe electricelectric potentialpotential atat pointpoint AA..

kQ kQ B  V 12 A 2 cm rr12 Q 9-Nm2 9 1 + +3 nC kQ 9 x 102 ( 3 x 10 C) 1 C 450 V r1 (0.06 m) 6 cm

9-Nm2 9 9 x 102 ( 5 x 10 C) kQ2 C A  2250 V 2 cm r (0.02 m) 2 -

V = 450 V – 2250 V; V = -1800 V Q2 = -5 nC A VAA = -1800 V ExampleExample 55 (Cont.):(Cont.): CalculateCalculate thethe electricelectric potentialpotential atat pointpoint BB forfor samesame charges.charges. kQ kQ 12 B  VB  rr12 2 cm 9-Nm2 9 Q1 + +3 nC kQ 9 x 102 ( 3 x 10 C) 1 C 1350 V r (0.02 m) 1 6 cm

9-Nm2 9 9 x 102 ( 5 x 10 C) kQ2 C A  450 V 2 cm r (0.10 m) 2 - V = 1350 V – 450 V; V = +900 V Q2 = -5 nC B VBB = +900 V ExampleExample 55 (Cont.):(Cont.): DiscussDiscuss meaningmeaning ofof thethe potentialspotentials justjust foundfound forfor pointspoints AA andand BB..

ConsiderConsider PointPoint A:A: VVA == -1800-1800 VV A B  2 cm ForFor everyevery coulombcoulomb ofof positivepositive chargecharge Q placedplaced atat pointpoint A,A, thethe potentialpotential energyenergy 1 + +3 nC willwill bebe –1800–1800 J.J. (Negative(Negative P.E.)P.E.) 6 cm TheThe fieldfield holdsholds onon toto thisthis positivepositive A  charge.charge. AnAn externalexternal forceforce mustmust dodo 2 cm +1800+1800 JJ ofof workwork toto removeremove eacheach - Q = -5 nC coulombcoulomb ofof ++ chargecharge toto infinity.infinity. 2 ExampleExample 55 (Cont.):(Cont.): DiscussDiscuss meaningmeaning ofof thethe potentialspotentials justjust foundfound forfor pointspoints AA andand BB..

ConsiderConsider PointPoint B:B: V = +900 V B  VBB = +900 V 2 cm ForFor everyevery coulombcoulomb ofof positivepositive chargecharge Q1 + +3 nC placedplaced atat pointpoint B,B, thethe potentialpotential energyenergy willwill bebe +900+900 J.J. (Positive(Positive P.E.)P.E.) 6 cm A  ForFor everyevery coulombcoulomb ofof positivepositive charge,charge, 2 cm thethe fieldfield EE willwill dodo 900900 JJ ofof positivepositive - work in removing it to infinity. work in removing it to infinity. Q2 = -5 nC PotentialPotential DifferenceDifference

TheThe potentialpotential differencedifference betweenbetween twotwo pointspoints AA andand BB isis thethe workwork perper unitunit positivepositive chargecharge donedone byby electricelectric forcesforces inin movingmoving aa smallsmall testtest chargecharge fromfrom thethe pointpoint ofof higherhigher potentialpotential toto thethe pointpoint ofof lowerlower potential.potential.

Potential Difference: V = V -V Potential Difference: VABAB = VAA -VBB

Work = q(V –V ) Work BY E-field WorkABAB = q(VAA –VBB ) Work BY E-field

TheThe positivepositive andand negativenegative signssigns ofof thethe chargescharges maymay bebe usedused mathematicallymathematically toto givegive appropriateappropriate signs.signs. ExampleExample 6:6: WhatWhat isis thethe potentialpotential differencedifference betweenbetween pointspoints AA andand BB.. WhatWhat workwork isis donedone byby thethe EE--fieldfield ifif aa +2+2 CC chargecharge isis movedmoved fromfrom AA toto B?B?  B 2 cm V = -1800 V V = +900 V VAA = -1800 V VBB = +900 V Q1 + +3 nC 6 cm VV == VV –– VV == --18001800 VV –– 900900 VV AB A B A  NoteNote pointpoint BB isis atat 2 cm V = -2700 V -5 nC VABAB = -2700 V higherhigher potential.potential. Q2 -

-6 WorkWorkAB == q(Vq(VA –– VVB )) == (2(2 xx 1010 CC )()(--27002700 V)V)

WorkWork == -5.40-5.40 mJmJ EE--fieldfield doesdoes negativenegative work.work. Thus, an external force was required to move the charge. ExampleExample 66 (Cont.):(Cont.): NowNow supposesuppose thethe +2+2 CC chargecharge isis movedmoved fromfrom backback fromfrom BB toto AA??  B 2 cm V = -1800 V V = +900 V VAA = -1800 V VBB = +900 V Q1 + +3 nC 6 cm VV == VV –– VV == 900900 VV –– ((--18001800 V)V) BA B A A  ThisThis pathpath isis fromfrom 2 cm V = +2700 V VBABA = +2700 V highhigh toto lowlow potential.potential. Q2 - -5 nC

-6 WorkWorkBA == q(Vq(VB –– VVA )) == (2(2 xx 1010 CC )(+2700)(+2700 V)V)

WorkWork == +5.40+5.40 mJmJ EE--fieldfield doesdoes positivepositive work.work. The work is done BY the E-field this ! ParallelParallel PlatesPlates ConsiderConsider TwoTwo parallelparallel platesplates ofof equalequal V + + + + andand oppositeopposite charge,charge, aa distancedistance dd apart.apart. A +q E ConstantConstant EE field:field: FF == qEqE F = qE WorkWork == FdFd == ((qE)dqE)d VB ----

Also,Also, WorkWork == q(Vq(VA –– VVB )) V = Ed SoSo that:that: qVqVAB == qEdqEd andand VABAB = Ed

TheThe potentialpotential differencedifference betweenbetween twotwo oppositelyoppositely chargedcharged parallelparallel platesplates isis thethe productproduct ofof EE andand dd.. ExampleExample 7:7: TheThe potentialpotential differencedifference betweenbetween twotwo parallelparallel platesplates isis 800800 VV.. IfIf theirtheir separsepar-- ationation isis 33 mmmm,, whatwhat isis thethe fieldfield EE?? V VA + + + + VEdE ;  +q E d F = qE 80 V ---- E 26,700 V/m VB 0.003 m

TheThe E-fieldE-field expressedexpressed inin voltsvolts perper metermeter (V/m)(V/m) isis knownknown asas thethe potentialpotential gradientgradient andand isis equivalentequivalent toto thethe N/C.N/C. TheThe voltvolt perper metermeter isis thethe betterbetter unitunit forfor currentcurrent ,electricity, thethe N/CN/C isis betterbetter .electrostatics. SummarySummary ofof FormulasFormulas kQq U ElectricElectric PotentialPotential UV ;  EnergyEnergy andand PotentialPotential rq

kQ ElectricElectric PotentialPotential NearNear V  MultipleMultiple charges:charges:  r

Work = q(V –V ) Work BY E-field WorkABAB = q(VAA –VBB ) Work BY E-field V OppositelyOppositely ChargedCharged VEdE ;  ParallelParallel Plates:Plates: d CONCLUSION:CONCLUSION: ChapterChapter 2525 ElectricElectric PotentialPotential