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ChapterChapter 2626 -- ElectricElectric FieldField AA PowerPointPowerPoint PresentationPresentation byby PaulPaul E.E. Tippens,Tippens, ProfessorProfessor ofof PhysicsPhysics SouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity

© 2007 Objectives:Objectives: AfterAfter finishingfinishing thisthis unitunit youyou shouldshould bebe ableable to:to:

• Define the electric and explain what determines its magnitude and direction. • Write and apply formulas for the electric field at known distances from point charges. • Discuss electric field lines and the meaning of of space. • Write and apply Gauss's law for fields around surfaces of known charge densities. TheThe ConceptConcept ofof aa FieldField

AA fieldfield isis defineddefined asas aa propertyproperty ofof spacespace inin whichwhich aa materialmaterial objectobject experiencesexperiences aa forceforce..

AboveAbove earth,earth, wewe saysay therethere isis Pm . aa gravitationalgravitational fieldfield atat PP.. F BecauseBecause aa massmass mm experiencesexperiences aa downwarddownward forceforce atat thatthat point.point. No , no field; No field, no force!

TheThe directiondirection ofof thethe fieldfield isis determineddetermined byby thethe forceforce.. TheThe GravitationalGravitational FieldField

ConsiderNoteConsiderNote thatthat pointspointsthethe forceforce AA and andFF isis B B realreal aboveabove,, butbut  B the surface of the earth—just A FF thethe fieldsurfacefield isis justjustof the aa convenientconvenient earth—just  pointspointswayway ofof inin describingdescribing spacespace.. spacespace.. FF TheThe fieldfield atat pointspoints AA oror BB mightmight bebe foundfound from:from: F g  IfIf gg isis knownknown atat m everyevery pointpoint aboveabove thethe earthearth thenthen thethe TheThe magnitudemagnitude andand directiondirection ofof forceforce FF onon aa givengiven thethe fieldfield gg isis dependsdepends onon thethe massmass cancan bebe found.found. weight,weight, whichwhich isis thethe forceforce F.F. TheThe ElectricElectric FieldField 1.1. Now,Now, considerconsider pointpoint PP aa distancedistance rr fromfrom +Q+Q.. F +q +. P E 2. An electric field E exists 2. An electric field E exists r atat PP ifif aa testtest chargecharge +q+q hashas aa forceforce FF atat thatthat point.point. + ++ +Q + 3.3. TheThe directiondirection ofof thethe EE isis ++ thethe samesame asas thethe directiondirection ofof Electric Field aa forceforce onon ++ (pos)(pos) charge.charge. F N 4.4. TheThe magnitudemagnitude ofof EE isis E  ; Units givengiven byby thethe formula:formula: q C FieldField isis PropertyProperty ofof SpaceSpace

ForceForce onon +q+q isis withwith F fieldfield direction.direction. +q +. E -q -. E r ForceForce onon --qq isis F r + + againstagainst fieldfield + + +Q + +Q + +++ direction.direction. +++ Electric Field Electric Field

TheThe fieldfield EE atat aa pointpoint existsexists whetherwhether therethere isis aa chargecharge atat thatthat pointpoint oror not.not. TheThe directiondirection ofof thethe fieldfield isis awayaway fromfrom thethe +Q+Q charge.charge. FieldField NearNear aa NegativeNegative ChargeCharge

ForceForce onon +q+q isis withwith F +q +. E fieldfield direction.direction. -q -. E F r r ForceForce onon --qq isis - - - - againstagainst fieldfield - -- --Q - --Q - - - direction.direction. - - Electric Field Electric Field

NoteNote thatthat thethe fieldfield EE inin thethe vicinityvicinity ofof aa negativenegative chargecharge ––QQ isis towardtoward thethe chargecharge——thethe directiondirection thatthat aa +q+q testtest chargecharge wouldwould move.move. TheThe MagnitudeMagnitude ofof EE--FieldField

TheThe magnitudemagnitude ofof thethe electricelectric fieldfield intensityintensity atat aa pointpoint inin spacespace isis defineddefined asas thethe forceforce perper unitunit chargecharge (N/C)(N/C) thatthat wouldwould bebe experiencedexperienced byby anyany testtest chargecharge placedplaced atat thatthat point.point.

Electric Field F N Electric Field E  ; Units  IntensityIntensity EE q C

TheThe directiondirection ofof EE atat aa pointpoint isis thethe samesame asas thethe directiondirection thatthat aa positivepositive chargecharge wouldwould movemove IFIF placedplaced atat thatthat point.point. ExampleExample 1.1. AA +2+2 nCnC chargecharge isis placedplaced atat aa distancedistance rr fromfrom aa +2 nC +q +. P ––88 CC charge.charge. IfIf thethe chargecharge 4000 N E experiencesexperiences aa forceforce ofof 40004000 NN,, E r whatwhat isis thethe electricelectric fieldfield - -- intensityintensity EE atat pointpoint P?P? --Q - –8 C - - First,First, wewe notenote thatthat thethe directiondirection ofof Electric Field EE isis towardtoward ––QQ (down).(down). F 4000 N 12 E  E = 2 x 10 N/C q 2 x 10-9 C Downward

Note:Note: TheThe fieldfield EE wouldwould bebe thethe samesame forfor anyany chargecharge placedplaced atat pointpoint PP.. ItIt isis aa propertyproperty ofof thatthat spacespace.. ExampleExample 2.2. AA constantconstant EE fieldfield ofof 40,00040,000 N/CN/C isis maintainedmaintained betweenbetween thethe twotwo parallelparallel plates.plates. WhatWhat areare thethe magnitudemagnitude andand directiondirection ofof thethe forceforce onon anan electronelectron thatthat passespasses horizontallyhorizontally betweenbetween thethe plates.cplates.c + + + + + + + + + - TheThe EE--fieldfield isis downward,downward, F ee - - ee- andand thethe forceforce onon ee- isis up.up. ee - -. E F E ; FqE ------q -19 4 N FqE(1.6 x 10 C)(4 x 10C )

FF == 6.406.40 xx 1010-15-15 N,N, UpwardUpward TheThe EE--FieldField atat aa distancedistance rr fromfrom aa singlesingle chargecharge QQ

ConsiderConsider aa testtest chargecharge +q+q placedplaced FE at P a distance r from Q. at P a distance r from Q. +q +.. PP The outward force on +q is: rr The outward force on +q is: kQ E  kQq ++ ++ 2 ++Q ++ r F  2 + Q ++ r ++++ TheThe electricelectric fieldfield EE isis therefore:therefore: FkQqr2 kQ E  E  2 qq r ExampleExample 3.3. WhatWhat isis thethe electricelectric fieldfield intensityintensity EE atat pointpoint PP,, aa distancedistance ofof 33 mm fromfrom aa negativenegative chargecharge ofof ––88 nCnC??

E = ? . First,First, findfind thethe magnitude:magnitude: P 9-Nm2 9 r kQ (9 x 102 )(8 x 10 C) E  C 3 m r 22(3 m) -Q -8 nC EE== 8.008.00 N/CN/C

TheThe directiondirection isis thethe samesame asas thethe forceforce onon aa positivepositive chargecharge ifif itit werewere placedplaced atat thethe pointpoint P:P: towardtoward ––QQ..

EE == 8.008.00 N,N, towardtoward --QQ TheThe ResultantResultant ElectricElectric Field.Field. TheThe resultantresultant fieldfield EE inin thethe vicinityvicinity ofof aa numbernumber ofof pointpoint chargescharges isis equalequal toto thethe vectorvector sumsum ofof thethe fieldsfields duedue toto eacheach chargecharge takentaken individually.individually.

ConsiderConsider EE forfor eacheach charge.charge. E2 E1 q1 -  A Vector Sum: Vector Sum: ER

E = E + E + E E3 + E = E11 + E22 + E33 q - 3 q2

Magnitudes are from: kQ Directions are based E  r 2 on positive test charge. ExampleExample 4.4. FindFind thethe resultantresultant fieldfield atat pointpoint AA duedue toto thethe ––33 nCnC chargecharge andand thethe +6+6 nCnC chargecharge arrangedarranged asas shown.shown. -3 nC q1 - EE forfor eacheach qq isis shownshown 3 cm 5 cm withwith directiondirection given.given. E1 +6 nC  + kq12 kq A 4 cm q EE; E2 2 1222 rr12

(9 x 109-Nm2 )(3 x 109 C) (9 x 109-Nm2 )(6 x 109 C) E  C2 E  C2 1 (3 m)2 2 (4 m)2

SignsSigns ofof thethe chargescharges areare usedused onlyonly toto findfind directiondirection ofof EE ExampleExample 4.4. (Cont.)(Cont.)FindFind thethe resultantresultant fieldfield atat pointpoint AA.. TheThe magnitudesmagnitudes are:are: -3 nC (9 x 109-Nm2 )(3 x 109 C) q C2 1 - E1  5 cm (3 m)2 3 cm E 1 +6 nC 2 (9 x 109-Nm )(6 x 109 C)  + E  C2 A 4 cm q 2 2 E2 2 (4 m)

EE1 == 3.003.00 N,N, WestWest EE2 == 3.383.38 N,N, NorthNorth ER Next,Next, wewe findfind vectorvector resultantresultant EER E1 22 E1  EERR 21; tan  E2 E2 ExampleExample 4.4. (Cont.)(Cont.)FindFind thethe resultantresultant fieldfield atat pointpoint AA usingusing vectorvector mathematics.mathematics. E R E1 = 3.00 N, West E 1 E2 = 3.38 N, North  E 2 FindFind vectorvector resultantresultant EER 3.38 N E (3.00 N)22 (3.38 N) 4.52 N; tan  3.00 N  == 48.448.40 NN ofof W;W; oror  == 131.6131.60

Resultant Field: E = 4.52 N; 131.600 Resultant Field: ERR = 4.52 N; 131.6 ElectricElectric FieldField LinesLines ElectricElectric FieldField LinesLines areare imaginaryimaginary lineslines drawndrawn inin suchsuch aa wayway thatthat theirtheir directiondirection atat anyany pointpoint isis thethe samesame asas thethe directiondirection ofof thethe fieldfield atat thatthat point.point.

+ + -- -- +Q + --Q - +++ - -

FieldField lineslines gogo awayaway fromfrom positivepositive chargescharges andand towardtoward negativenegative charges.charges. RulesRulesRules forforfor DrawingDrawingDrawing FieldFieldField LinesLinesLines

1.1. TheThe directiondirection ofof thethe fieldfield lineline atat anyany pointpoint isis thethe samesame asas motionmotion ofof +q+q atat thatthat point.point. 2.2. TheThe spacingspacing ofof thethe lineslines mustmust bebe suchsuch thatthat theythey areare closeclose togethertogether wherewhere thethe fieldfield isis strongstrong andand farfar apartapart wherewhere thethe fieldfield isis weak.weak.

EE1

EE2 - + qq1 qq2 EER ExamplesExamples ofof EE--FieldField LinesLines TwoTwo equalequal butbut TwoTwo identicalidentical oppositeopposite charges.charges. chargescharges (both(both +).+).

NoticeNotice thatthat lineslines leaveleave ++ chargescharges andand enterenter -- charges.charges. Also,Also, EE isis strongeststrongest wherewhere fieldfield lineslines areare mostmost densedense.. TheThe DensityDensity ofof FieldField LinesLines

Gauss’sGauss’s Law:Law: TheThe fieldfield EE atat anyany pointpoint inin spacespace isis proportionalproportional toto thethe lineline densitydensity atat thatthat point.point.

Gaussian Surface Line density  N

rr

A

N   Radius r A LineLine DensityDensity andand SpacingSpacing ConstantConstant ConsiderConsider thethe fieldfield nearnear aa positivepositive pointpoint chargecharge q:q: Then,Then, imagineimagine aa surfacesurface (radius(radius r)r) surroundingsurrounding q.q.

Radius r EE isis proportionalproportional toto N/N/AA andand isis equalequal toto kq/rkq/r2 atat anyany point.point. rr Nkq  E;  E Ar2

Gaussian Surface DefineDefine  asas spacingspacing constant.constant. Then:Then: N 1  E Where is:   A 000 4 k PermittivityPermittivity ofof FreeFree SpaceSpace TheThe proportionalityproportionality constaconstantnt forfor lineline densitydensity isis knownknown asas thethe permittivitypermittivity  andand itit isis defineddefined by:by:

1C2  8.85 x 10-12 0 4Nm k  2 RecallingRecalling thethe relationshiprelationship withwith lineline density,density, wewe have:have: N E or N E A A 00 SummingSumming overover entireentire areaarea N =  EA AA givesgives thethe totaltotal lineslines as:as: N = oo EA ExampleExample 5.5. WriteWrite anan equationequation forfor findingfinding thethe totaltotal numbernumber ofof lineslines NN leavingleaving aa singlesingle positivepositive chargecharge qq.. Radius r DrawDraw sphericalspherical GaussianGaussian surface:surface: rr NEAN00 and  EA SubstituteSubstitute forfor EE andand AA from:from: kq q E  ; A = 4 r2 Gaussian Surface rr224

q 2 NEA00 2 (4 r ) NN == oqAqA == qq 4 r o

TotalTotal numbernumber ofof lineslines isis equalequal toto thethe enclosedenclosed chargecharge q.q. GaussGauss’’ss LawLaw

GaussGaussGauss’s’’ss Law:Law:Law: TheTheThe netnetnet numbernumbernumber ofofof electricelectricelectric fieldfieldfield lineslineslines crossingcrossingcrossing anyanyany closedclosedclosed surfacesurfacesurface ininin ananan outwardoutwardoutward directiondirectiondirection isisis numericallynumericallynumerically equalequalequal tototo thethethe netnetnet totaltotaltotal chargechargecharge withinwithinwithin thatthatthat surface.surface.surface.

NEAq 0

IfIf wewe representrepresent qq asas netnet enclosedenclosed q positivepositive chargecharge,, wewe cancan writewrite EA  rewriterewrite GaussGauss’’ss lawlaw as:as: 0 ExampleExample 6.6. HowHow manymany electricelectric fieldfield lineslines passpass throughthrough thethe GaussianGaussian surfacesurface drawndrawn below.below. Gaussian surface FirstFirst wewe findfind thethe NETNET chargecharge qq enclosedenclosed byby thethe surfacesurface:: -4 C +8 C q2 q1 - + q = (+8 –4 – 1) = +3 C q = (+8 –4 – 1) = +3 C q4 -1 C + q3 - +5 C NEAq0 

NN == +3+3 CC == +3+3 xx 1010-6-6 lineslines ExampleExample 6.6. AA solidsolid spheresphere (R(R == 66 cm)cm) havinghaving netnet chargecharge +8+8 CC isis insideinside aa hollowhollow shellshell (R(R == 88 cm)cm) havinghaving aa netnet chargecharge ofof ––66 CC.. WhatWhat isis thethe electricelectric fieldfield atat aa distancedistance ofof 1212 cmcm fromfrom thethe centercenter ofof thethe solidsolid sphere?sphere?

DrawDraw GaussianGaussian spheresphere atat Gaussian surface radiusradius ofof 1212 cmcm toto findfind E.E. - -6 C NEAq  8cm - - 0 - 6 cm qq == (+8(+8 –– 6)6) == +2+2 CC +8 C - - - q 12 cm - 0 AE qnet ; E 0 A q 2 x 10-6 C E  2-122Nm2 0 (4r ) (8.85 x 10C2 )(4 )(0.12 m) ExampleExample 66 (Cont.)(Cont.) WhatWhat isis thethe electricelectric fieldfield atat aa distancedistance ofof 1212 cmcm fromfrom thethe centercenter ofof thethe solidsolid sphere?sphere? DrawDraw GaussianGaussian spheresphere atat Gaussian surface radiusradius ofof 1212 cmcm toto findfind E.E. - -6 C 8cm - - NEAq  6 cm 0 - +8 C - qq == (+8(+8 –– 6)6) == +2+2 CC - 12 cm - - q 0 AE qnet ; E 0 A

2 C 6 N E 2 1.25 x 10 C EE== 1.251.25 MN/CMN/C 0 (4r ) ChargeCharge onon SurfaceSurface ofof ConductorConductor

SinceSince likelike chargescharges Gaussian Surface just repel,repel, youyou wouldwould inside conductor expectexpect thatthat allall chargecharge wouldwould movemove untiluntil theythey comecome toto rest.rest. ThenThen fromfrom GaussGauss’’ss LawLaw ...... Charged Conductor

SinceSince chargescharges areare atat rest,rest, EE == 00 insideinside conductor,conductor, thus:thus:

NEAq0  or 0 =  q

AllAll chargecharge isis onon surface;surface; NoneNone insideinside ConductorConductor ExampleExample 7.7. UseUse GaussGauss’’ss lawlaw toto findfind thethe EE-- fieldfield justjust outsideoutside thethe surfacesurface ofof aa conductor.conductor. TheThe surfacesurface chargecharge densitydensity == q/Aq/A.. ConsiderConsider qq insideinside thethe E pillboxpillbox.. EE--lineslines throughthrough E3 1 E3 allall areasareas outward.outward. A + + + + +E E+ + 3 +3 0 AEq E + + + +2 + EE--lineslines throughthrough sidessides cancelcancel byby .symmetry. Density 

TheThe fieldfield isis zerozero insideinside thethe conductor,conductor, soso EE2 == 00 00 q  oE E1 AA ++ oE E2 AA == qq E  00A  ExampleExample 77 (Cont.)(Cont.) FindFind thethe fieldfield justjust outsideoutside thethe surfacesurface ifif == q/Aq/A == +2+2 C/mC/m2.. RecallRecall thatthat sideside fieldsfields E E3 1 E3 cancelcancel andand insideinside A + + + + fieldfield isis zero,zero, soso thatthat +E E+ + 3 +3 + E2 q  + + + + E1  00A  Surface

2 x 10-6 C/m 2 E  -12 Nm2 EE == 226,000226,000 N/CN/C 8.85 x 10 C2 FieldField BetweenBetween ParallelParallel PlatesPlates EqualEqual andand oppositeopposite charges.charges. + - E1 FieldField EE andand EE toto right.right. + - 1 2 E DrawDraw GaussianGaussian pillboxespillboxes Q1 + 2 - Q2 + - onon eacheach insideinside surface.surface. E1 + - GaussGauss’’ss LawLaw forfor eithereither boxbox E2 givesgives samesame fieldfield (E(E1 == EE2). ). q   0 AEq E    00A  LineLine ofof ChargeCharge

A 2r FieldField duedue toto AA1 andand AA2 1 CancelCancel outout duedue toto r symmetry.symmetry. A L E 0 AEq q q EAArL; (2 )   A L 2 0 q q  E  ;  = E  2L0rL 20r ExampleExample 8:8: TheThe ElectricElectric fieldfield atat aa distancedistance ofof 1.51.5 mm fromfrom aa lineline ofof chargecharge isis 55 xx 10104 N/C.N/C. WhatWhat isis thethe linearlinear densitydensity ofof thethe line?line?

r  E  L   20rE E 20r

q   EE == 55 xx 10104 N/CN/C rr == 1.51.5 mm L  2 (8.85 x 10-12 C2 )(1.5 m)(5 x 104 N/C) Nm2

4.17 C/m ConcentricConcentric CylindersCylinders OutsideOutside isis likelike chargedcharged longlong wire:wire: b + + + + + + Gaussian surface a + + + + + + + + + -6 C + + + r + + a + + rb b a 12 cm a r  r1 2 b

For    For a E  ab E  r > rb rb > r > ra 2 r 20r 0 ExampleExample 9.9. TwoTwo concentricconcentric cylinderscylinders ofof radiiradii 33 andand 66 cmcm.. InnerInner linearlinear chargecharge densitydensity isis +3+3 C/mC/m andand outerouter isis --55 C/mC/m.. FindFind EE atat distancedistance ofof 44 cmcm fromfrom center.center. DrawDraw GaussianGaussian surfacesurface -7 C/m + + + + + + betweenbetween cylinders.cylinders. a = 3 + + + + + + + +  cm + + + E  b ++ + rr + + 20r b=6 cm 3C/m E  +5 C/m 20 (0.04 m)

EE ==1.381.38 xx 101066 N/C, N/C, RadiallyRadially outout r r + + + + + + + + C/m C/m 3 cm  6 cm  = -7 b= a +5 N/C, Radially inward N/C, Radially inward 5 5 Next, find E at a distance of Next, find E at a distance of

 r 0   5.00 x 10 5.00 x 10  0 ab  2

 

(3 5)C/m E = E =  2(0.075 m) 2(0.075 E  both cylinders. both cylinders. E Gaussian outside of Gaussian outside of Example 8 (Cont.) 7.5 cm from center (outside both cylinders.) Example 8 (Cont.) 7.5 cm from center (outside both cylinders.) SummarySummary ofof FormulasFormulas

The Electric Field FkQ N The Electric Field E  Units are IntensityIntensity EE:: qr2 C

TheThe ElectricElectric FieldField kQ E   2 Vector Sum NearNear severalseveral charges:charges: r

Gauss’sGauss’s LawLaw forfor q 0 EA q;  ChargeCharge distributions.distributions. A CONCLUSION:CONCLUSION: ChapterChapter 2424 TheThe ElectricElectric FieldField