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Electric fields


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Electric Fields

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Electric fields

  1. 1. Chapter 18Electric Forces andElectric Fields
  2. 2. 18.1 The Origin of ElectricityThe electrical nature of matter is inherentin atomic structure.kg10673.1 27−×=pmkg10675.1 27−×=nmkg1011.9 31−×=emC1060.1 19−×=ecoulombs
  3. 3. 18.1 The Origin of ElectricityIn nature, atoms are normallyfound with equal numbers of protonsand electrons, so they are electricallyneutral.By adding or removing electronsfrom matter it will acquire a netelectric charge with magnitude equalto e times the number of electronsadded or removed, N.Neq =
  4. 4. 18.1 The Origin of ElectricityExample 1 A Lot of ElectronsHow many electrons are there in one coulomb of negative charge?Neq =1819-1025.6C101.60C00.1×=×==eqN
  5. 5. 18.2 Charged Objects and the Electric ForceThe body that loses electrons has an excess of positive charge, whilethe body that gains electrons has an excess of negative charge.It is possible to transfer electric chargefrom one object to another.
  6. 6. 18.2 Charged Objects and the Electric ForceLAW OF CONSERVATION OF ELECTRIC CHARGEDuring any process, the net electric charge of an isolated system remainsconstant (is conserved).
  7. 7. 18.2 Charged Objects and the Electric ForceLike charges repel and unlikecharges attract each other.
  8. 8. 18.2 Charged Objects and the Electric Force
  9. 9. 18.3 Conductors and InsulatorsNot only can electric charge exist on an object, but it can also movethrough and object.Substances that readily conduct electric charge are called electricalconductors.Materials that conduct electric charge poorly are called electricalinsulators.
  10. 10. 18.4 Charging by Contact and by InductionCharging by contact.
  11. 11. 18.4 Charging by Contact and by InductionCharging by induction.
  12. 12. 18.4 Charging by Contact and by InductionThe negatively charged rod induces a slight positive surface chargeon the plastic.
  13. 13. 18.5 Coulomb’s Law
  14. 14. 18.5 Coulomb’s LawCOULOMB’S LAWThe magnitude of the electrostatic force exerted by one point chargeon another point charge is directly proportional to the magnitude of thecharges and inversely proportional to the square of the distance betweenthem.221rqqkF =( ) 229CmN1099.841 ⋅×== ok πε( )2212mNC1085.8 ⋅×= −οε
  15. 15. 18.5 Coulomb’s LawExample 3 A Model of the Hydrogen AtomIn the Bohr model of the hydrogen atom, the electron is in orbit about thenuclear proton at a radius of 5.29x10-11m. Determine the speed of theelectron, assuming the orbit to be circular.221rqqkF =
  16. 16. 18.5 Coulomb’s Law( )( )( )N1022.8m1029.5C1060.1CmN1099.8 8211219229221 −−−×=××⋅×==rqqkFrmvmaF c2==( )( ) sm1018.2kg109.11m1029.5N1022.8 631-118×=×××==−−mFrv
  17. 17. 18.5 Coulomb’s LawExample 4 Three Charges on a LineDetermine the magnitude and direction of the net force on q1.
  18. 18. 18.5 Coulomb’s Law( )( )( )( )N7.2m20.0C100.4C100.3CmN1099.826622922112 =××⋅×==−−rqqkF( )( )( )( )N4.8m15.0C100.7C100.3CmN1099.826622923113 =××⋅×==−−rqqkF5.7NN4.8N7.21312 +=+−=+= FFF
  19. 19. 18.5 Coulomb’s Law
  20. 20. 18.6 The Electric FieldThe positive charge experiences a force which is the vector sum of theforces exerted by the charges on the rod and the two spheres.This test charge should have a small magnitude so it doesn’t affectthe other charge.
  21. 21. 18.6 The Electric FieldExample 6 A Test ChargeThe positive test charge has a magnitude of3.0x10-8C and experiences a force of 6.0x10-8N.(a) Find the force per coulomb that the test chargeexperiences.(b) Predict the force that a charge of +12x10-8Cwould experience if it replaced the test charge.CN0.2C100.3N100.688=××= −−oqF(a)(b) ( )( ) N1024C100.12CN0.2 88 −−×=×=F
  22. 22. 18.6 The Electric FieldDEFINITION OF ELECRIC FIELDThe electric field that exists at a point is the electrostatic force experiencedby a small test charge placed at that point divided by the charge itself:oqFE=SI Units of Electric Field: newton per coulomb (N/C)
  23. 23. 18.6 The Electric FieldIt is the surrounding charges that create the electric field at a given point.
  24. 24. 18.6 The Electric FieldExample 7 An Electric Field Leads to a ForceThe charges on the two metal spheres and the ebonite rod create an electricfield at the spot indicated. The field has a magnitude of 2.0 N/C. Determinethe force on the charges in (a) and (b)
  25. 25. 18.6 The Electric Field( )( ) N1036C100.18CN0.2 88 −−×=×== EqF o(a)(b) ( )( ) N1048C100.24CN0.2 88 −−×=×== EqF o
  26. 26. 18.6 The Electric FieldElectric fields from different sourcesadd as vectors.
  27. 27. 18.6 The Electric FieldExample 10 The Electric Field of a Point ChargeThe isolated point charge of q=+15μC isin a vacuum. The test charge is 0.20mto the right and has a charge qo=+15μC.Determine the electric field at point P.oqFE=221rqqkF =
  28. 28. 18.6 The Electric Field( )( )( )( )N7.2m20.0C1015C1080.0CmN1099.82662292=××⋅×==−−rqqkF oCN104.3C100.80N7.2 66-×=×==oqFE
  29. 29. 18.6 The Electric Field2rqkE =The electric field does not depend on the test qrqqkqFE12==Point charge q:
  30. 30. 18.6 The Electric FieldExample 11 The Electric Fields from Separate Charges May CancelTwo positive point charges, q1=+16μC and q2=+4.0μC are separated in avacuum by a distance of 3.0m. Find the spot on the line between the chargeswhere the net electric field is zero.2rqkE =
  31. 31. 18.6 The Electric Field( ) ( )( )2626m0.3C100.4C1016dkdk−×=× −−21 EE =2rqkE =( ) 22m0.30.2 dd =−m0.2+=d
  32. 32. 18.6 The Electric FieldConceptual Example 12 Symmetry and theElectric FieldPoint charges are fixes to the corners of a rectangle in twodifferent ways. The charges have the same magnitudesbut different signs.Consider the net electric field at the center of the rectanglein each case. Which field is stronger?
  33. 33. 18.6 The Electric FieldTHE PARALLEL PLATE CAPACITORParallel platecapacitoroo AqEεσε==( )2212mNC1085.8 ⋅×= −οεcharge density
  34. 34. 18.7 Electric Field LinesElectric field lines or lines of force provide a map of the electric fieldin the space surrounding electric charges.
  35. 35. 18.7 Electric Field LinesElectric field lines are always directed away from positive charges andtoward negative charges.
  36. 36. 18.7 Electric Field LinesElectric field lines always begin on a positive chargeand end on a negative charge and do not stop inmidspace.
  37. 37. 18.7 Electric Field LinesThe number of lines leaving a positive charge or entering anegative charge is proportional to the magnitude of the charge.
  38. 38. 18.7 Electric Field Lines
  39. 39. 18.7 Electric Field LinesConceptual Example 13 Drawing ElectricField LinesThere are three things wrong with part (a) ofthe drawing. What are they?
  40. 40. 18.8 The Electric Field Inside a Conductor: ShieldingAt equilibrium under electrostatic conditions, anyexcess charge resides on the surface of a conductor.At equilibrium under electrostatic conditions, theelectric field is zero at any point within a conductingmaterial.The conductor shields any charge within it fromelectric fields created outside the condictor.
  41. 41. 18.8 The Electric Field Inside a Conductor: ShieldingThe electric field just outside the surface of a conductor is perpendicular tothe surface at equilibrium under electrostatic conditions.
  42. 42. 18.8 The Electric Field Inside a Conductor: ShieldingConceptual Example 14 A Conductor inan Electric FieldA charge is suspended at the center ofa hollow, electrically neutral, sphericalconductor. Show that this charge induces(a) a charge of –q on the interior surface and(b) a charge of +q on the exterior surface ofthe conductor.
  43. 43. 18.9 Gauss’ Law( )224 rqrkqE oπε==( )oAqE ε=oqEAε=EAE =Φflux,Electric
  44. 44. 18.9 Gauss’ Law( )∑ ∆=Φ AEE φcos
  45. 45. 18.9 Gauss’ LawGAUSS’ LAWThe electric flux through a Gaussiansurface is equal to the net chargeenclosed in that surface divided bythe permittivity of free space:( )oQAEεφ =∆∑ cosSI Units of Electric Flux: N·m2/C
  46. 46. 18.9 Gauss’ LawExample 15 The Electric Field of a Charged Thin Spherical ShellA positive charge is spread uniformly over the shell. Find the magnitudeof the electric field at any point (a) outside the shell and (b) inside theshell.( )oQAEεφ =∆∑ cos
  47. 47. 18.9 Gauss’ Law( ) ( )( )240coscosrEAEAEAEEπφ=∆=∆=∆=Φ∑∑∑( )oQrEεπ =24
  48. 48. 18.9 Gauss’ Law( )oQrEεπ =24(a) Outside the shell, the Gaussiansurface encloses all of the charge.orqEεπ 24=(b) Inside the shell, the Gaussiansurface encloses no charge.0=E
  49. 49. 18.9 Gauss’ Law
  50. 50. 18.10 Copiers and Computer Printers
  51. 51. 18.10 Copiers and Computer Printers
  52. 52. 18.10 Copiers and Computer Printers