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Electric Fieldfield AA Powerpointpowerpoint Presentationpresentation Byby Paulpaul E.E 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 field and explain what determines its magnitude and direction. • Write and apply formulas for the electric field intensity at known distances from point charges. • Discuss electric field lines and the meaning of permittivity 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 force, 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.
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