1. P07 - 1 Workshop: Using Visualization in Teaching Introductory E&M AAPT National Summer Meeting, Edmonton, Alberta, Canada. Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy

2. P07 - 2 MIT Class: Gauss’s Law, Conductors and Capacitors, Shielding

3. P07 - 3 Gauss’s Law The first Maxwell Equation A very useful computational technique This is important!

4. P07 - 4 Gauss’s Law – The Idea The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside

5. P07 - 5 Gauss’s Law – The Equation Electric flux  E (the surface integral of E over closed surface S) is proportional to charge inside the volume enclosed by S

6. P07 - 6 Now the Details

7. P07 - 7 Electric Flux  E Case I: E is constant vector field perpendicular to planar surface S of area A Our Goal: Always reduce problem to this

8. P07 - 8 Case II: E is constant vector field directed at angle  to planar surface S of area A Electric Flux  E

9. P07 - 9 PRS Question: Flux Thru Sheet

10. P07 - 10 PRS: Flux The flux through the planar surface below (positive unit normal to left): +q -q 1.is positive. 2.is negative. 3.is zero. 4.I don’t know 15

11. P07 - 11 PRS Answer: Flux The field lines go from left to right, opposite the assigned normal direction. Hence the flux is negative. Answer: 2. The flux is negative. +q -q

12. P07 - 12 Gauss’s Law Note: Integral must be over closed surface

13. P07 - 13 Open and Closed Surfaces A rectangle is an open surface — it does NOT contain a volume A sphere is a closed surface — it DOES contain a volume

14. P07 - 14 Area Element dA: Closed Surface For closed surface, dA is normal to surface and points outward ( from inside to outside)  E > 0 if E points out  E < 0 if E points in

15. P07 - 15 Case III: E not constant, surface curved Electric Flux  E In practice you’ll never do an integral, although here’s an example of non-constant E

16. P07 - 16 PRS Question: Flux Thru Sphere

17. P07 - 17 PRS: Flux thru Sphere The total flux through the below spherical surface is +q 1.positive (net outward flux) 2.negative (net inward flux) 3.zero. 4.I don’t know 15

18. P07 - 18 PRS Answer: Flux thru Sphere We know this from Gauss’s Law: No enclosed charge  no net flux. Flux in on left cancelled by flux out on right Answer: 3. The total flux is zero +q+q +q+q

19. P07 - 19 Electric Flux: Sphere Point charge Q at center of sphere, radius r E field at surface: Electric flux through sphere:

20. P07 - 20 Arbitrary Gaussian Surfaces True for all surfaces such as S 1, S 2 or S 3 Why? As A gets bigger E gets smaller

21. P07 - 21 Choosing Gaussian Surface Desired E: Punch through surface & be constant. Flux is EA or -EA. Other E: Run along surface. Flux is zero True for ALL surfaces Useful (to calculate E) for SOME surfaces

22. P07 - 22 Symmetry & Gaussian Surfaces SymmetryGaussian Surface SphericalConcentric Sphere CylindricalCoaxial Cylinder PlanarGaussian “Pillbox” Desired E: Punch through surface & be constant So Gauss’s Law good to calculate E field from highly symmetric sources

23. P07 - 23 Applying Gauss’s Law 1.Identify regions in which to calculate E field. 2.Choose Gaussian surfaces S: Symmetry 3.Calculate 4.Calculate q in, charge enclosed by surface S 5.Apply Gauss’s Law to calculate E:

24. P07 - 24 Examples: Spherical Symmetry Cylindrical Symmetry Planar Symmetry

25. P07 - 25 Gauss: Spherical Symmetry +Q uniformly distributed throughout non-conducting solid sphere of radius a. Find E everywhere

26. P07 - 26 Gauss: Spherical Symmetry Symmetry is Spherical Use Gaussian Spheres

27. P07 - 27 Gauss: Spherical Symmetry Region 1: r > a Draw Gaussian Sphere in Region 1 (r > a) Note: r is arbitrary but is the radius for which you will calculate the E field!

28. P07 - 28 Gauss: Spherical Symmetry Region 1: r > a Total charge enclosed q in = +Q

29. P07 - 29 Gauss’s law: Region 2: r < a Total charge enclosed: Gauss: Spherical Symmetry OR

30. P07 - 30 PRS Question: Field Inside Spherical Shell

31. P07 - 31 PRS: Spherical Shell We just saw that in a solid sphere of charge the electric field grows linearly with distance. Inside the charged spherical shell at left (r<a) what does the electric field do? a Q 1.Constant and Zero 2.Constant but Non-Zero 3.Still grows linearly 4.Some other functional form (use Gauss’ Law) 5.Can’t determine with Gauss Law :15

32. P07 - 32 PRS Answer: Spherical Shell Spherical symmetry  Use Gauss’ Law with spherical surface. Any surface inside shell contains no charge  No flux E = 0! Answer: 1. Constant and Zero a Q

33. P07 - 33 Demonstration Field Inside Spherical Shell (Grass Seeds):

34. P07 - 34 Gauss: Cylindrical Symmetry Infinitely long rod with uniform charge density Find E outside the rod.

35. P07 - 35 Gauss: Cylindrical Symmetry Symmetry is Cylindrical Use Gaussian Cylinder Note: r is arbitrary but is the radius for which you will calculate the E field! is arbitrary and should divide out

36. P07 - 36 Gauss: Cylindrical Symmetry Total charge enclosed:

37. P07 - 37 Gauss: Planar Symmetry Infinite slab with uniform charge density  Find E outside the plane

38. P07 - 38 Gauss: Planar Symmetry Symmetry is Planar Use Gaussian Pillbox Note: A is arbitrary (its size and shape) and should divide out Gaussian Pillbox

39. P07 - 39 Gauss: Planar Symmetry NOTE: No flux through side of cylinder, only endcaps Total charge enclosed: ++++++++++++++++++++++++

40. P07 - 40 PRS Question: Slab of Charge

41. P07 - 41 PRS: Slab of Charge A positively charged, semi-infinite (in x & y) flat slab has thickness 2d. z-axis is perp. to the sheet, with center at z = 0. 2d  z = 0 z At the plane’s center (z = 0), E 1.points in the positive z-direction 2.points in the negative z-direction 3.is zero 4.I don’t know 15

42. P07 - 42 PRS Answer: Slab of Charge Symmetry tell us this – the amount of charge above and below the center of the plane is equal hence the fields cancel. Another way of thinking about this: Since you can’t tell which way the field would point it must be 0. Answer: 3. E(z=0) is zero 2d  z = 0 z

43. P07 - 43 Group Problem: Charge Slab Infinite slab with uniform charge density  Thickness is 2d (from x=-d to x=d). Find E for x > 0 (how many regions is that?)

44. P07 - 44 Potential from E

45. P07 - 45 Potential for Uniformly Charged Non-Conducting Solid Sphere From Gauss’s Law Use Region 1: r > a Point Charge!

46. P07 - 46 Potential for Uniformly Charged Non-Conducting Solid Sphere Region 2: r < a

47. P07 - 47 Potential for Uniformly Charged Non-Conducting Solid Sphere

48. P07 - 48 Group Problem: Charge Slab Infinite slab with uniform charge density  Thickness is 2d (from x=-d to x=d). If V=0 at x=0 (definition) then what is V(x) for x>0?

49. P07 - 49 Conductors

50. P07 - 50 Conductors and Insulators Conductor: Charges are free to move Electrons weakly bound to atoms Example: metals Insulator: Charges are NOT free to move Electrons strongly bound to atoms Examples: plastic, paper, wood

51. P07 - 51 Conductors Conductors have free charges  E must be zero inside the conductor  Conductors are equipotential objects Neutral Conductor -------- ++++++++ E

52. P07 - 52 Conductors in Equilibrium Conductors are equipotential objects: 1) E = 0 inside 2) E perpendicular to surface 3) Net charge inside is 0

53. P07 - 53 Conductors in Equilibrium: Free Charges Move To Surface Put net charge inside conductor It moves to get away from other charges Java applet link

54. P07 - 54 Conductors in Equilibrium Conductors are equipotential objects: 1) E = 0 inside 2) E perpendicular to surface 3) Net charge inside is 0 4) Excess charge on surface

55. P07 - 55 Capacitors and Capacitance Our first of 3 standard electronics devices (Capacitors, Resistors & Inductors) Start Charging

56. P07 - 56 Capacitors: Store Electric Charge Capacitor:Two isolated conductors Equal and opposite charges ±Q Potential difference  V between them. Units: Coulombs/Volt or Farads C is Always Positive

57. P07 - 57 Parallel Plate Capacitor

58. P07 - 58 Parallel Plate Capacitor Oppositely charged plates: Charges move to inner surfaces to get close Link to Capacitor Applet

59. P07 - 59 Calculating E (Gauss’s Law) Note: We only “consider” a single sheet! Doesn’t the other sheet matter?

60. P07 - 60 Alternate Calculation Method + + + + + + + Top Sheet: - - - - - - - Bottom Sheet:

61. P07 - 61 Parallel Plate Capacitor C depends only on geometric factors A and d

62. P07 - 62 Demonstration: Big Capacitor

63. P07 - 63 Group Problem: Spherical Shells These two spherical shells have equal but opposite charge. Find E everywhere Find V everywhere (assume V(  ) = 0)

64. P07 - 64 Spherical Capacitor Two concentric spherical shells of radii a and b Gauss’s Law  E ≠ 0 only for a < r < b, where it looks like a point charge: What is E?

65. P07 - 65 Spherical Capacitor For an isolated spherical conductor of radius a: Is this positive or negative? Why?

66. P07 - 66 Capacitance of Earth For an isolated spherical conductor of radius a: A Farad is REALLY BIG! We usually use pF (10 -12 ) or nF (10 -9 )

67. P07 - 67 PRS Questions: Changing C Dimensions

68. P07 - 68 PRS: Changing Dimensions A parallel-plate capacitor has plates with equal and opposite charges ±Q, separated by a distance d, and is not connected to a battery. The plates are pulled apart to a distance D > d. What happens? 1.V increases, Q increases 2.V decreases, Q increases 3.V is the same, Q increases 4.V increases,Q is the same 5.V decreases, Q is thesame 6.V is the same, Q is the same 7.V increases, Q decreases 8.V decreases, Q decreases 9.V is the same,Q decreases :20

69. P07 - 69 PRS Answer: Changing Dimensions With no battery connected to the plates the charge on them has no possibility of changing. In this situation, the electric field doesn’t change when you change the distance between the plates, so: V = E d As d increases, V increases. Answer: 4. V increases, Q is the same

70. P07 - 70 PRS: Changing Dimensions A parallel-plate capacitor has plates with equal and opposite charges ±Q, separated by a distance d, and is connected to a battery. The plates are pulled apart to a distance D > d. What happens? 20 1.V increases, Q increases 2.V decreases,Q increases 3.V is the same, Q increases 4.V increases, Q is the same 5.V decreases,Q is the same 6.V is the same, Q is the same 7.V increases,Q decreases 8.V decreases,Q decreases 9.V is the same, Q decreases

71. P07 - 71 PRS Answer: Changing Dimensions With a battery connected to the plates the potential V between them is held constant In this situation, since V = E d As d increases, E must decrease. Since the electric field is proportional to the charge on the plates, Q must decrease as well. Answer: 9. V is the same, Q decreases

72. P07 - 72 Demonstration: Changing C Dimensions

73. P07 - 73 Energy Stored in Capacitor

74. P07 - 74 Energy To Charge Capacitor 1.Capacitor starts uncharged. 2.Carry +dq from bottom to top. Now top has charge q = +dq, bottom -dq 3.Repeat 4.Finish when top has charge q = +Q, bottom -Q +q -q

75. P07 - 75 Work Done Charging Capacitor At some point top plate has +q, bottom has –q Potential difference is  V = q / C Work done lifting another dq is dW = dq  V +q -q

76. P07 - 76 So work done to move dq is: Total energy to charge to q = Q: Work Done Charging Capacitor +q -q

77. P07 - 77 Energy Stored in Capacitor Since Where is the energy stored???

78. P07 - 78 Energy Stored in Capacitor Parallel-plate capacitor: Energy stored in the E field!

79. P07 - 79 PRS Question: Changing C Dimensions Energy Stored

80. P07 - 80 PRS: Changing Dimensions A parallel-plate capacitor, disconnected from a battery, has plates with equal and opposite charges, separated by a distance d. Suppose the plates are pulled apart until separated by a distance D > d. How does the final electrostatic energy stored in the capacitor compare to the initial energy? 1.The final stored energy is smaller 2.The final stored energy is larger 3.Stored energy does not change. 20

81. P07 - 81 PRS Answer: Changing Dimensions As you pull apart the capacitor plates you increase the amount of space in which the E field is non-zero and hence increase the stored energy. Where does the extra energy come from? From the work you do pulling the plates apart. Answer: 2. The stored energy increases

82. P07 - 82 Conductors as Shields

83. P07 - 83 PRS Question: Point Charge Inside Conductor

84. P07 - 84 PRS: Point Charge in Conductor A point charge +Q is placed inside a neutral, hollow, spherical conductor. As the charge is moved around inside, the electric field outside +Q 15 1.is zero and does not change 2.is non-zero but does not change 3.is zero when centered but changes 4.is non-zero and changes 5.I don’t know

85. P07 - 85 PRS Answer: Q in Conductor E = 0 in conductor  -Q on inner surface Charge conserved  +Q on outer surface E = 0 in conductor  No “communication” between –Q & +Q  + Q uniformly distributed Answer: 2. is non-zero but does not change

86. P07 - 86 Hollow Conductors Charge placed INSIDE induces balancing charge ON INSIDE

87. P07 - 87 Hollow Conductors Charge placed OUTSIDE induces charge separation ON OUTSIDE

88. P07 - 88 PRS Questions: Point Charge Inside Conductor

89. P07 - 89 PRS Setup What happens if we put Q in the center of these nested (concentric) spherical conductors?

90. P07 - 90 PRS: Hollow Conductors A point charge +Q is placed at the center of the conductors. The induced charges are: 1.Q(I1) = Q(I2) = -Q; Q(O1) = Q(O2)= +Q 2.Q(I1) = Q(I2) = +Q; Q(O1) = Q(O2)= -Q 3.Q(I1) = -Q; Q(O1) = +Q; Q(I2) = Q(O2)= 0 4.Q(I1) = -Q; Q(O2)= +Q;Q(O1) = Q(I2)= 0 15

91. P07 - 91 PRS Answer: Hollow Conductors Looking in from each conductor, the total charge must be zero (this gives the inner surfaces as –Q). But the conductors must remain neutral (which makes the outer surfaces have induced charge +Q). Answer: 1. The inner faces are negative, the outer faces are positive.

92. P07 - 92 PRS: Hollow Conductors A point charge +Q is placed at the center of the conductors. The potential at O1 is: 1.Higher than at I1 2.Lower than at I1 3.The same as at I1 15

93. P07 - 93 PRS Answer: Hollow Conductors A conductor is an equipotential surface. O1 and I1 are on the same conductor, hence at the same potential Answer: 3. O1 and I1 are at the same potential

94. P07 - 94 PRS: Hollow Conductors A point charge +Q is placed at the center of the conductors. The potential at O2 is: 1.Higher than at I1 2.Lower than at I1 3.The same as at I1 15

95. P07 - 95 PRS Answer: Hollow Conductors As you move away from the positive point charge at the center, the potential decreases. Answer: 2. O2 is lower than I1 V r

96. P07 - 96 PRS: Hollow Conductors A point charge +Q is placed at the center of the conductors. If a wire is used to connect the two conductors, then current (positive charge) will flow 1.from the inner to the outer conductor 2.from the outer to the inner conductor 3.not at all 15

97. P07 - 97 PRS Answer: Hollow Conductors Positive charges always flow “downhill” – from high to low potential. Since the inner conductor is at a higher potential the charges will flow from the inner to the outer conductor. Answer: 1. Current flows outward

98. P07 - 98 PRS: Hollow Conductors You connect the “charge sensor’s” red lead to the inner conductor and black lead to the outer conductor. What does it actually measure? 15 1.Charge on I1 2.Charge on O1 3.Charge on I2 4.Charge on O2 5.Charge on O1 – Charge on I2 6.Average charge on inner – ave. on outer 7.Potential difference between outer & inner 8.I don’t know

99. P07 - 99 PRS Answer: Hollow Conductors So what is the “charge axis?” From the capacitance and potential difference it can calculate Q = C  V which is charge on O1 and negative charge on I2 Answer: 7. “Charge Sensor” measures potential difference between outer & inner conductor

100. P07 - 100 Demonstration: Conductive Shielding

101. P07 - 101 Visualization: Inductive Charging

102. P07 - 102 Experiment 2: Faraday Ice Pail

103. P07 - 103 Last Time: Capacitors

104. P07 - 104 Capacitors: Store Electric Energy Parallel Plate Capacitor: To calculate: 1) Put on arbitrary ±Q 2) Calculate E 3) Calculate  V

105. P07 - 105 Capacitors in Series & Parallel  In series, V adds: In parallel, Q adds: 

106. P07 - 106 Demonstration: Dissectible Capacitor

107. P07 - 107 Flow of Charge New Topics: Current, Current Density, Resistance, Ohm’s Law

108. P07 - 108 Current: Flow Of Charge Units of Current: Coulomb/second = Ampere Average current I av : Charge  Q flowing across area A in time  t Instantaneous current: differential limit of I av

109. P07 - 109 How Big is an Ampere? Household Electronics Battery Powered Household Service Lightning Bolt To hurt you To throw you To kill you Fuse/Circuit Breaker ~1 A ~100 mA (1-10 A-Hr) 100 A 10 to 100 kA 40 (5) mADC(AC) 60 (15) mADC(AC) 0.5 (0.1) ADC(AC) 15-30 A

110. P07 - 110 Direction of The Current Direction of current is direction of flow of pos. charge or, opposite direction of flow of negative charge

111. P07 - 111 Current Density J J: current/unit area points in direction of current

112. P07 - 112 PRS Question: Current Density

113. P07 - 113 PRS: Current Density A current I = 200 mA flows in the above wire. What is the magnitude of the current density J? 20 cm 10 cm 5 cm 1.J = 40 mA/cm 2.J = 20 mA/cm 3.J = 10 mA/cm 4.J = 1 mA/cm 2 5.J = 2 mA/cm 2 6.J = 4 mA/cm 2 7.I don’t know :15

114. P07 - 114 PRS Answer: Current Density The area that matters is the cross-sectional area that the current is punching through – the 50 cm 2 area shaded grey. So: J = I/A = 200 mA/50 cm 2 = 4 mA/cm 2 Answer: 6. J = 4 mA/cm 2 20 cm 10 cm 5 cm

115. P07 - 115 Why Does Current Flow? If an electric field is set up in a conductor, charge will move (making a current in direction of E) Note that when current is flowing, the conductor is not an equipotential surface (and E inside ≠ 0)!

116. P07 - 116 Microscopic Picture Drift speed is velocity forced by applied electric field in the presence of collisions. It is typically 4x10 -5 m/sec, or 0.04 mm/second! To go one meter at this speed takes about 10 hours! How Can This Be?

117. P07 - 117 Conductivity and Resistivity  : conductivity  : resistivity Ability of current to flow depends on density of charges & rate of scattering Two quantities summarize this:

118. P07 - 118 Microscopic Ohm’s Law  and  depend only on the microscopic properties of the material, not on its shape

119. P07 - 119 Demonstrations: Water Temperature Effects on 

120. P07 - 120 PRS Question: Resistance?

121. P07 - 121 PRS: Resistance When a current flows in a wire of length L and cross sectional area A, the resistance of the wire is 15 1.Proportional to A; inversely proportional to L. 2.Proportional to both A and L. 3.Proportional to L; inversely proportional to A. 4.Inversely proportional to both L and A 5.Do Not Know

122. P07 - 122 PRS Answer: Resistance The longer the wire the higher the resistance. The bigger the cross-sectional area of the wire, the more ways that current can flow through it, so the lower the resistance. So, if resistivity is , then 3. Proportional to L; inversely proportional to A.

123. P07 - 123 Why Does Current Flow? Instead of thinking of Electric Field, think of potential difference across the conductor

124. P07 - 124 Ohm’s Law What is relationship between  V and current?

125. P07 - 125 Ohm’s Law R has units of Ohms (  ) = Volts/Amp

126. P07 - 126 How Big is an Ohm? Short Copper Wire Notebook paper (thru) Typical resistors You (when dry) You (when wet) Internally (hand to foot) milliohms (m  ) ~1 G   to 100 M  100 k  1 k  500  Stick your wet fingers in an electrical socket: You’re dead!

127. P07 - 127 Appendix: Dielectrics

128. P07 - 128 Demonstration: Dielectric in Capacitor

129. P07 - 129 Dielectrics A dielectric is a non-conductor or insulator Examples: rubber, glass, waxed paper When placed in a charged capacitor, the dielectric reduces the potential difference between the two plates HOW???

130. P07 - 130 Molecular View of Dielectrics Polar Dielectrics : Dielectrics with permanent electric dipole moments Example: Water

131. P07 - 131 Molecular View of Dielectrics Non-Polar Dielectrics Dielectrics with induced electric dipole moments Example: CH 4

132. P07 - 132 Dielectric in Capacitor Potential difference decreases because dielectric polarization decreases Electric Field!

133. P07 - 133 Dielectric Constant  Dielectric weakens original field by a factor  Dielectric constants Vacuum1.0 Paper 3.7 Pyrex Glass 5.6 Water 80 Dielectric Constant

134. P07 - 134 Dielectric in a Capacitor Q 0 = constant after battery is disconnected Upon inserting a dielectric:

135. P07 - 135 Dielectric in a Capacitor V 0 = constant when battery remains connected Upon inserting a dielectric:

136. P07 - 136 PRS Questions: Dielectric in a Capacitor

137. P07 - 137 PRS: Dielectric A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant  in inserted between the plates. The charge stored in the capacitor + + + + - - - -  1.Increases 2.Decreases 3.Stays the Same 15 Seconds Remaining

138. P07 - 138 PRS Answer: Dielectric Since the capacitor is disconnected from a battery there is no way for the amount of charge on it to change. Answer: 3. Charge stays the same + + + + - - - - 

139. P07 - 139 PRS: Dielectric A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant  in inserted between the plates. The energy stored in the capacitor + + + + - - - -  1.Increases 2.Decreases 3.Stays the Same :15

140. P07 - 140 PRS Answer: Dielectric The dielectric reduces the electric field and hence reduces the amount of energy stored in the field. The easiest way to think about this is that the capacitance is increased while the charge remains the same so U = Q 2 /2C Also from energy density: Answer: 2. Energy stored decreases

141. P07 - 141 PRS: Dielectric A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant  in inserted between the plates. The force on the dielectric + + + + - - - -  1.pulls in the dielectric 2.pushes out the dielectric 3.is zero 15

142. P07 - 142 PRS Answer: Dielectric We just saw that the energy is reduced by the introduction of a dielectric. Since systems want to reduce their energy, the dielectric will be sucked into the capacitor. Alternatively, since opposing charges are induced on the dielectric surfaces close to the plates, the attraction between these will lead to the attractive force. Answer: 1. The dielectric is pulled in

143. P07 - 143 Group: Partially Filled Capacitor What is the capacitance of this capacitor?

144. P07 - 144 Gauss’s Law with Dielectrics