Electric charge and electric field


Published on

1 Like
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Electric charge and electric field

  1. 1. Chapter 16Electric Charge and Electric Field
  2. 2. Demonstration #11. Demonstrate how you can pick up the tissue without touching it in any way with your body.2. What is occurring on the atomic level that lets you do this?
  3. 3. The atomThe atom has positive charge in the nucleus,located in the protons. The positive chargecannot move from the atom unless there is anuclear reaction.The atom has negative charge in the electroncloud on the outside of the atom. Electrons canmove from atom to atom without all that muchdifficulty.
  4. 4. • Demonstration #2• 1. Rub the black rod with the fur. Bring the rod toward the pole of the electroscope. What happens to the vanes?• ---• 2. Come up with an atomic level explanation for your observations.
  5. 5. • Demonstration #3+++• 1. Rub the glass rod with the silk. Bring the rod toward the pole of the electroscope. What happens to the vanes?• 2. Come up with an atomic level explanation for your observations.
  6. 6. • Demonstration #4• 1. What happens when your touch the electroscope with the glass rod?
  7. 7. Sample Problem A certain static discharge delivers -0.5 Coulombs of electrical charge.• How many electrons are in this discharge?
  8. 8. Sample Problem1. How much positive charge resides in two moles ofhydrogen gas (H2)?2. How much negative charge?3. How much net charge?
  9. 9. Sample Problem (end 1)The total charge of a system composed of 1800 particles, all of which are protons or electrons, is 31x10-18 C.How many protons are in the system?How many electrons are in the system?
  10. 10. Coulomb’s Law and Electrical Force Day 2
  11. 11. Sample ProblemA point charge of positive 12.0 μC experiencesan attractive force of 51 mN when it is placed 15cm from another point charge. What is the othercharge?
  12. 12. Sample ProblemCalculate the mass of ball B, which is suspended inmidair.
  13. 13. Sample Problem• What is the force on the 4 μC charge?
  14. 14. Sample Problem end 2• What is the force on the 4 μC charge?
  15. 15. The Electric Field Day 3
  16. 16. Sample Problem end 3A 400 μg styrofoam bead has 600 excess electrons onits surface. What is the magnitude and direction of theelectric field that will suspend the bead in midair?
  17. 17. Superposition Day 4
  18. 18. Sample ProblemA proton traveling at 440 m/s in the +x direction entersan electric field of magnitude 5400 N/C directed in the+y direction. Find the acceleration.
  19. 19. Sample ProblemA particle bearing -5.0 μC is placed at -2.0 cm, and aparticle bearing 5.0 μC is placed at 2.0 cm. What is thefield at the origin?
  20. 20. Sample ProblemA particle bearing 10.0 mC is placed at the origin, and aparticle bearing 5.0 mC is placed at 1.0 m. Where is thefield zero?
  21. 21. Sample Problem end 4What is the charge on the bead? It’s mass is 32 mg.
  22. 22. Electric Potential and Potential Energy Day 5
  23. 23. Chapter 17Electric Potential
  24. 24. Sample ProblemA 3.0 μC charge is moved through a potential differenceof 640 V. What is its potential energy change?
  25. 25. Electrical Potential in Uniform Electric FieldsThe electric potential is related in a simple way to auniform electric field.DV = -EdDV: change in electrical potential (V)E: Constant electric field strength (N/C or V/m)d: distance moved (m)
  26. 26. Sample ProblemAn electric field is parallel to the x-axis. What is itsmagnitude and direction of the electric field if thepotential difference between x =1.0 m and x = 2.5 m isfound to be +900 V?
  27. 27. Sample ProblemWhat is the voltmeter reading between A and B?Between A and C? Assume that the electric field has amagnitude of 400 N/C.
  28. 28. Sample Problem end 5How much work would be done BY THE ELECTRICFIELD in moving a 2 mC charge from A to C? From A toB? from B to C?. How much work would be done by anexternal force in each case? End 5
  29. 29. Electric Field and Shielding Day 6
  30. 30. Sample ProblemIf a proton is accelerated through a potential differenceof -2,000 V, what is its change in potential energy?How fast will this proton be moving if it started at rest?
  31. 31. Sample ProblemA proton at rest is released in a uniform electric field.How fast is it moving after it travels through a potentialdifference of -1200 V? How far has it moved?
  32. 32. Sample ProblemDraw field lines for the charge configuration below. Thefield is 600 V/m, and the plates are 2 m apart. Labeleach plate with its properpotential, and draw and label 3equipotential surfaces between the plates. You mayignore edge effects.-- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -+++++++++++++++++++++++++++++
  33. 33. Sample ProblemDraw a negative point charge of -Q and its associatedelectric field. Draw 4 equipotential surfaces such that DVis the same between the surfaces, and draw them at thecorrect relative locations. What do you observe aboutthe spacing between the equipotential surfaces?
  34. 34. Objectives: After finishing this unit, you should be able to: • Explain and demonstrate the first law of electro- statics and discuss charging by contact and by induction.• Write and apply Coulomb’s Law and apply it to problems involving electric forces.• Define the electron, the coulomb, and the microcoulomb as units of electric charge.
  35. 35. Objectives: After finishing this unit you should be able 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 Gausss law for fields around surfaces of known charge densities.
  36. 36. 16.1 Static Electricity; Electric Charge and Its Conservation11.1 When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod. Electrons negative move from positive fur to the -- -- ++++ rubber rod. The rod is said to be negatively charged because of an excess of electrons. The fur is said to be positively charged because of a deficiency of electrons.
  37. 37. Two Negative Charges Repel1. Charge the rubber rod by rubbing against fur.2. Transfer electrons from rod to each pith ball. The two negative charges repel each other. The two negative charges repel each other.
  38. 38. The Two Types of Charge Rubber glass Attraction fur silkNote that the negatively charged (green) ball isattracted to the positively charged (red) ball. Opposite Charges Attract! Opposite Charges Attract!
  39. 39. 16.2 Electric Charge in the AtomAtom:Nucleus (small,massive, positivecharge)Electron cloud (large,very low density,negative charge) 11.2
  40. 40. 16.2 Electric Charge in the AtomAtom is electrically neutral.Rubbing charges objects by moving electronsfrom one to the other.
  41. 41. 16.2 Electric Charge in the AtomPolar molecule: neutral overall, but charge notevenly distributed
  42. 42. 16.3 Insulators and ConductorsConductor: Insulator:Charge flows freely Almost no charge flowsMetals Most other materials Some materials are semiconductors.
  43. 43. 16.4 Induced Charge; the ElectroscopeCharging Spheres by Induction Induction --- - - - - Electrons ++ ++ -- Repelled Uncharged Spheres Separation of Charge--- - - ++ ++ -- + + + - - - -- + - Isolation of Spheres Charged by Induction 11.3
  44. 44. Induction for a Single Sphere Induction --- - - + ---- + -- ++ --Uncharged Sphere Separation of Charge--- - - -- - - - - ++ - ++ - + + + + Electrons move Charged by Induction to ground.
  45. 45. The Quantity of Charge The quantity of charge (q) can be defined in terms of the number of electrons, but the Coulomb (C) is a better unit for later work. A temporary definition might be as given below: The Coulomb: 1 C = 6.25 x 1018 electrons The Coulomb: 1 C = 6.25 x 1018 electronsWhich means that the charge on a single electron is: 1 electron: e-- = -1.6 x 10-19 C 1 electron: e = -1.6 x 10-19 C
  46. 46. Units of ChargeThe coulomb (selected for use with electriccurrents) is actually a very large unit for staticelectricity. Thus, we often encounter a need touse the metric prefixes. 1 µC = 1 x 10-6 C 1 µC = 1 x 10-6 C 1 nC = 1 x 10-9 C 1 nC = 1 x 10-9 C 1 pC = 1 x 10-12 C 1 pC = 1 x 10-12 C
  47. 47. 16.4 Induced Charge; the ElectroscopeThe electroscopecan be used fordetecting charge:
  48. 48. 16.4 Induced Charge; the ElectroscopeThe electroscope can be charged either byconduction or by induction.
  49. 49. 16.4 Induced Charge; the ElectroscopeThe charged electroscope can then be used todetermine the sign of an unknown charge.
  50. 50. 16.5 Coulomb’s LawExperiment shows that the electric forcebetween two charges is proportional to theproduct of the charges and inverselyproportional to the distance between them.
  51. 51. 11.4 Coulomb’s LawThe force of attraction or repulsion between two The force of attraction or repulsion between twopoint charges is directly proportional to the product point charges is directly proportional to the productof the two charges and inversely proportional to the of the two charges and inversely proportional to thesquare of the distance between them. square of the distance between them. F - q q’ + r qq F∝ 2F F r q q’ - -
  52. 52. 16.5 Coulomb’s Law The force is along the line connecting thecharges, and is attractive if the charges areopposite, and repulsive if they are the same.
  53. 53. 16.5 Coulomb’s LawUnit of charge: coulomb, CThe proportionality constant in Coulomb’slaw is then: Charges produced by rubbing are typically around a microcoulomb:
  54. 54. 16.5 Coulomb’s LawCharge on the electron:Electric charge is quantized in unitsof the electron charge.
  55. 55. 16.5 Coulomb’s LawThe proportionality constant k can also bewritten in terms of , the permittivity of freespace: (16-2)
  56. 56. Problem-Solving Strategies1. Read, draw, and label a sketch showing all given information in appropriate SI units.2. Do not confuse sign of charge with sign of forces. Attraction/Repulsion determines the direction (or sign) of the force.3. Resultant force is found by considering force due to each charge independently. Review module on vectors, if necessary.4. For forces in equilibrium: ΣFx = 0 = ΣFy = 0.
  57. 57. 16.5 Coulomb’s LawCoulomb’s law strictly applies only to point charges.Superposition: for multiple point charges, the forceson each charge from every other charge can becalculated and then added as vectors.
  58. 58. Summary of Formulas: Like Charges Repel; Unlike Charges Attract. Like Charges Repel; Unlike Charges Attract. kqq N ⋅m 2 F= 2 k =9 x 109 r C2 1 µC = 1 x 10-6 C C 1 µ = 1 x 10-6 C 1 nC = 1 x 10-9 C 1 nC = 1 x 10-9 C1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C 1 electron: e-- = -1.6 x 10-19 C 1 electron: e = -1.6 x 10-19 C
  59. 59. 16.6 Solving Problems Involving Coulomb’s Law and VectorsThe net force on a charge is the vectorsum of all the forces acting on it.
  60. 60. 16.6 Solving Problems InvolvingCoulomb’s Law and VectorsVector addition review:
  61. 61. 16.7 The Electric FieldThe electric field is theforce on a small charge,divided by the charge: (16-3)
  62. 62. 16.7 The Electric FieldFor a point charge: (16-4a) (16-4b) 11.5
  63. 63. 16.7 The Electric FieldForce on a point charge in an electric field: (16-5)Superposition principle for electric fields:
  64. 64. 16.7 The Electric FieldProblem solving in electrostatics: electric forces and electric fields1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions2. Calculate forces using Coulomb’s law3. Add forces vectorially to get result
  65. 65. 16.8 Field LinesThe electric field can be represented by fieldlines. These lines start on a positive chargeand end on a negative charge.
  66. 66. 16.8 Field LinesThe number of field lines starting (ending)on a positive (negative) charge isproportional to the magnitude of the charge.The electric field is stronger where the fieldlines are closer together.
  67. 67. 16.8 Field Lines Examples of E-Field Lines Two equal but Two identical opposite charges. charges (both +).Notice that lines leave + charges and enter - charges.Also, E is strongest where field lines are most dense.
  68. 68. 16.8 Field Lines The electric field between two closely spaced, oppositely charged parallel plates is constant.
  69. 69. 16.8 Field LinesSummary of field lines:1. Field lines indicate the direction of the field; the field is tangent to the line.2. The magnitude of the field is proportional to the density of the lines.3. Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.
  70. 70. 16.9 Electric Fields and ConductorsThe static electric field inside a conductor iszero – if it were not, the charges would move.The net charge on a conductor is on itssurface.
  71. 71. 16.9 Electric Fields and Conductors The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.
  72. 72. 16.10 Gauss’s Law Electric flux: 11.6 (16-7) Electric flux through an area is proportional to the total number of field lines crossing the area.
  73. 73. 16.10 Gauss’s LawFlux through a closed surface:
  74. 74. 16.10 Gauss’s LawThe net number of field lines through thesurface is proportional to the chargeenclosed, and also to the flux, givingGauss’s law: (16-9)This can be used to find the electric fieldin situations with a high degree ofsymmetry.
  75. 75. The Density of Field Lines Gauss’s Law: The field E at any point in spaceGauss’s Law: The field E at any point in space is proportional to the line density σ at that point.is proportional to the line density σ at that point. Gaussian Surface Line density σ ∆N r ∆A ∆N σ= Radius r ∆A
  76. 76. Gauss’s Law Gauss’s Law: The net number of electric field Gauss’s Law: The net number of electric field lines crossing any closed surface in an outward lines crossing any closed surface in an outward direction is numerically equal to the net total direction is numerically equal to the net total charge within that surface. charge within that surface. N = Σε 0 EA = ΣqIf we represent q as net enclosed qpositive charge, we can write ΣEA =rewrite Gauss’s law as: ε0