1. Thur. Oct. 8, 2009Physics 208 Lecture 111 Last time… Equipotential lines Capacitance and capacitors

2. Thur. Oct. 8, 2009Physics 208 Lecture 112 Parallel plate capacitor +Q+Q -Q-Q d Geometrical factor determined from electric fields Energy stored in parallel-plate capacitor Energy density Area A

3. Thur. Oct. 8, 2009Physics 208 Lecture 113 An isolated parallel plate capacitor has charge Q and potential V. The plates are pulled apart. Which describes the situation afterwards? A) Charge Q has decreased B) Capacitance C has increased C) Electric field E has increased D) Voltage difference V between plates has increased E) None of these +Q+Q-Q-Q ++++++++ -------- d pull E = (Q/A)/  0  E constant V= Ed  V increases C =  0 A/d  C decreases Quick Quiz Cap. isolated  Q constant

4. Thur. Oct. 8, 2009Physics 208 Lecture 114 An isolated parallel plate capacitor has a charge q. The plates are then pulled further apart. What happens to the energy stored in the capacitor? 1) Increases 2) Decreases 3) Stays the same +q-q ++++++++ -------- d pull Quick Quiz

5. Thur. Oct. 8, 2009Physics 208 Lecture 115 Different geometries of capacitors Parallel plate capacitor Spherical capacitor Cylindrical capacitor +Q -Q L +Q+Q -Q-Q d A

6. Thur. Oct. 8, 2009Physics 208 Lecture 116 Combining Capacitors — Parallel Connect capacitors together with metal wire C1C1 C2C2 C eq Both have same  V Need different charge “Equivalent” capacitor Potential difference  V Total charge

7. Thur. Oct. 8, 2009Physics 208 Lecture 117 Combining Capacitors — Series C1C1 C2C2 C eq VAVA VBVB VmVm VAVA VBVB Q Q -Q Q Q on each is same

8. Thur. Oct. 8, 2009Physics 208 Lecture 118

9. Thur. Oct. 8, 2009Physics 208 Lecture 119 Current in a wire: not electrostatic equilibrium Battery produces E-field in wire Charge moves in response to E-field

10. Thur. Oct. 8, 2009Physics 208 Lecture 1110 Electric Current Electric current = I = amount of charge per unit time flowing through a plane perpendicular to charge motion SI unit: ampere 1 A = 1 C / s Depends on sign of charge: + charge particles: current in direction of particle motion is positive - charge particles: current in direction of particle motion is negative

11. Thur. Oct. 8, 2009Physics 208 Lecture 1111 Quick Quiz An infinite number of positively charged particles are uniformly distributed throughout an otherwise empty infinite space. A spatially uniform positive electric field is applied. The current due to the charge motion A. increases with time B. decreases with time C. is constant in time D. Depends on field Constant force qE Produces constant accel. qE/m Velocity increases v(t)=qEt/m Charge / time crossing plane increases with time

12. Thur. Oct. 8, 2009Physics 208 Lecture 1112 But experiment says… Current constant in time Proportional to voltage R = resistance (unit Ohm =  ) Also written J = current density = I / (cross-section area)  = resistivity = R x (cross-section area) / (length) Resistivity is independent of shape

13. Thur. Oct. 8, 2009Physics 208 Lecture 1113 Charge motion with collisions Wire not empty space, has various fixed objects. Charge carriers accelerate, then collide. After collision, charged particle reaccelerates. Result: average “drift” velocity v d

14. Thur. Oct. 8, 2009Physics 208 Lecture 1114 Current and drift velocity This average velocity called drift velocity Current density J This drift leads to a current Conductivity Electric field

15. Thur. Oct. 8, 2009Physics 208 Lecture 1115 What about Ohm’s law? Current density proportional to electric field Current proportional to current density through geometrical factor Electric field proportional to electric potential through geometrical factor

16. Thur. Oct. 8, 2009Physics 208 Lecture 1116 Resistivity Resistivity SI units Ω-m Independent of sample geometry

17. Thur. Oct. 8, 2009Physics 208 Lecture 1117 Resistors Schematic layout Circuits Physical layout

18. Thur. Oct. 8, 2009Physics 208 Lecture 1118 Quick Quiz Which bulb is brighter? A.A B.B C.Both the same Current through each must be same Conservation of current (Kirchoff’s current law) Charge that goes in must come out

19. Thur. Oct. 8, 2009Physics 208 Lecture 1119 Current conservation I in I out I out = I in I1I1 I2I2 I3I3 I 1 =I 2 +I 3 I2I2 I3I3 I1I1 I 1 +I 2 =I 3

20. Thur. Oct. 8, 2009Physics 208 Lecture 1120 Quick Quiz How does brightness of bulb B compare to that of A? A.B brighter than A B.B dimmer than A C.Both the same Battery maintain constant potential difference Extra bulb makes extra resistance -> less current

21. Thur. Oct. 8, 2009Physics 208 Lecture 1121 Resistors in Series I 1 = I 2 = I Potentials add  V =  V 1 +  V 2 = I R 1 + I R 2 = = I (R 1 +R 2 ) The equivalent resistance R eq = R 1 +R 2 R R = 2R 2 resistors in series: R  L Like summing lengths

22. Thur. Oct. 8, 2009Physics 208 Lecture 1122 Quick Quiz What happens to the brightness of the bulb B when the switch is closed? A.Gets dimmer B.Gets brighter C.Stays same D.Something else Battery is constant voltage, not constant current

23. Thur. Oct. 8, 2009Physics 208 Lecture 1123 Resistors in Parallel  V =  V 1 =  V 2 I = I 1 + I 2 (lower resistance path has higher current) Equivalent Resistance R/2 R R Add areas

24. Thur. Oct. 8, 2009Physics 208 Lecture 1124 Quick Quiz What happens to the brightness of the bulb A when the switch is closed? A.Gets dimmer B.Gets brighter C.Stays same D.Something else