2. September 29, 2008 CAPACITORS

3. How did you do? A. Great B. OK C. Poor D. Really bad E. I absolutely flunked!

4. Calendar of the Day Exams will be returned within a week. If you did badly in the exam you need to have a plan to succeed. Let me know if you want any help with this. Quiz on Friday – Potential or Capacitance. WebAssign will appear shortly if it hasn’t done so already. There is a WA on board for potential. Quizzes are in the bin on the third floor through the double doors.

5. Two +q charges are separated by a distance d. What is the potential at a point midway between the charges on the line connecting them A. Zero B. Kq/d C. Kq/d D. 2Kq/d E. 4kq/d

6. Capacitors

7. A simple Capacitor Remove the battery Charge Remains on the plates. The battery did WORK to charge the plates That work can be recovered in the form of electrical energy – Potential Difference WIRES TWO PLATES Battery WIRES

8. INSIDE THE DEVICE

9. Two Charged Plates (Neglect Fringing Fields) d Air or Vacuum Area A - Q +Q E V=Potential Difference Symbol ADDED CHARGE

10. Where is the charge? d Air or Vacuum Area A - Q +Q E V=Potential Difference ------------ ++++++++++++ AREA=A  =Q/A

11. One Way to Charge: Start with two isolated uncharged plates. Take electrons and move them from the + to the – plate through the region between. As the charge builds up, an electric field forms between the plates. You therefore have to do work against the field as you continue to move charge from one plate to another.

12. Capacitor

13. More on Capacitors d Air or Vacuum Area A - Q +Q E V=Potential Difference Gaussian Surface Same result from other plate!

14. DEFINITION - Capacity The Potential Difference is APPLIED by a battery or a circuit. The charge q on the capacitor is found to be proportional to the applied voltage. The proportionality constant is C and is referred to as the CAPACITANCE of the device.

15. UNITS A capacitor which acquires a charge of 1 coulomb on each plate with the application of one volt is defined to have a capacitance of 1 FARAD One Farad is one Coulomb/Volt

16. The two metal objects in the figure have net charges of +79 pC and -79 pC, which result in a 10 V potential difference between them. (a) What is the capacitance of the system? [7.9] pF (b) If the charges are changed to +222 pC and -222 pC, what does the capacitance become? [7.9] pF (c) What does the potential difference become? [28.1] V

17. NOTE Work to move a charge from one side of a capacitor to the other is qEd. Work to move a charge from one side of a capacitor to the other is qV Thus qV=qEd E=V/d As before

18. Continuing… The capacitance of a parallel plate capacitor depends only on the Area and separation between the plates. C is dependent only on the geometry of the device!

19. Units of  0

20. Simple Capacitor Circuits Batteries Apply potential differences Capacitors Wires Wires are METALS. Continuous strands of wire are all at the same potential. Separate strands of wire connected to circuit elements may be at DIFFERENT potentials.

21. Size Matters! A Random Access Memory stores information on small capacitors which are either charged (bit=1) or uncharged (bit=0). Voltage across one of these capacitors ie either zero or the power source voltage (5.3 volts in this example). Typical capacitance is 55 fF (femto=10 -15 ) Probably less these days! Question: How many electrons are stored on one of these capacitors in the +1 state?

22. Small is better in the IC world!

23. October 1, 2008 Cap-II

24. Note: I do not have the grades yet. Probably by Friday. Quiz on Friday … Potential or Capacitors. Watch WebAssign for new stuff.

25. Last Time We defined capacitance: C=q/V Q=CV We showed that C=  0 A/d And E=V/d

26. TWO Types of Connections SERIES PARALLEL

27. Parallel Connection V C Equivalent =C E

28. Series Connection V C 1 C 2 q -q The charge on each capacitor is the same !

29. Series Connection Continued V C 1 C 2 q -q

30. More General

31. Example C 1 C 2 V C3C3 C1=12.0  f C2= 5.3  f C3= 4.5  d (12+5.3)pf series (12+5.3)pf

32. More on the Big C We move a charge dq from the (-) plate to the (+) one. The (-) plate becomes more (-) The (+) plate becomes more (+). dW=Fd=dq x E x d +q -q E=  0 A/d +dq

33. So….

34. Not All Capacitors are Created Equal Parallel Plate Cylindrical Spherical

35. Spherical Capacitor

36. Calculate Potential Difference V (-) sign because E and ds are in OPPOSITE directions.

37. Continuing… Lost (-) sign due to switch of limits.

38. A Thunker If a drop of liquid has capacitance 1.00 pF, what is its radius? STEPS Assume a charge on the drop. Calculate the potential See what happens

39. In the drawing below, find the equivalent capacitance of the combination. Assume that C 1 = 8 µF, C 2 = 4 µF, and C 3 = 3 µF. 5.67µF

40. In the diagram, the battery has a potential difference of 10 V and the five capacitors each have a capacitance of 20 µF. What is the charge on ( a) capacitor C 1 and (b) capacitor C 2 ?

41. In the figure, capacitors C 1 = 0.8 µF and C 2 = 2.8 µF are each charged to a potential difference of V = 104 V, but with opposite polarity as shown. Switches S 1 and S 2 are then closed. (a) What is the new potential difference between points a and b? 57.8 V What are the new charges on each capacitor? (b)46.2µC (on C 1 ) (c)162µC (on C 2 )

42. Anudder Thunker Find the equivalent capacitance between points a and b in the combination of capacitors shown in the figure. V(a  b) same across each

43. DIELECTRIC

44. Polar Materials (Water)

45. Apply an Electric Field Some LOCAL ordering Larger Scale Ordering

46. Adding things up.. - + Net effect REDUCES the field

47. Non-Polar Material

48. Effective Charge is REDUCED

49. We can measure the C of a capacitor (later) C 0 = Vacuum or air Value C = With dielectric in place C=  C 0 (we show this later)

50. How to Check This Charge to V 0 and then disconnect from The battery. C0C0 V0V0 Connect the two together V C 0 will lose some charge to the capacitor with the dielectric. We can measure V with a voltmeter (later).

51. Checking the idea.. V Note: When two Capacitors are the same (No dielectric), then V=V 0 /2.

53. Messing with Capacitors + V - + V - +-+-+-+- The battery means that the potential difference across the capacitor remains constant. For this case, we insert the dielectric but hold the voltage constant, q=CV since C   C 0 q    C 0 V THE EXTRA CHARGE COMES FROM THE BATTERY! Remember – We hold V constant with the battery.

54. Another Case We charge the capacitor to a voltage V 0. We disconnect the battery. We slip a dielectric in between the two plates. We look at the voltage across the capacitor to see what happens.

55. No Battery +-+-+-+- q0q0 qq q 0 =C 0 V o When the dielectric is inserted, no charge is added so the charge must be the same. V0VV0V

56. Another Way to Think About This There is an original charge q on the capacitor. If you slide the dielectric into the capacitor, you are adding no additional STORED charge. Just moving some charge around in the dielectric material. If you short the capacitors with your fingers, only the original charge on the capacitor can burn your fingers to a crisp! The charge in q=CV must therefore be the free charge on the metal plates of the capacitor.

57. A Closer Look at this stuff.. Consider this virgin capacitor. No dielectric experience. Applied Voltage via a battery. C0C0 ++++++++++++ ------------------ V0V0 q -q

58. Remove the Battery ++++++++++++ ------------------ V0V0 q -q The Voltage across the capacitor remains V 0 q remains the same as well. The capacitor is fat (charged), dumb and happy.

59. Slip in a Dielectric Almost, but not quite, filling the space ++++++++++++ ------------------ V0V0 q -q - - - - + + + -q’ +q’ E0E0 E E’ from induced charges Gaussian Surface

60. A little sheet from the past.. ++++++ ------ q -q -q’ +q’ 0 2xE sheet 0

61. Some more sheet…

62. A Few slides back No Battery +-+-+-+- q0q0 qq q=C 0 V o When the dielectric is inserted, no charge is added so the charge must be the same. V0VV0V

63. From this last equation

64. Another look +-+- VoVo

65. Add Dielectric to Capacitor Original Structure Disconnect Battery Slip in Dielectric +-+- VoVo +-+- +-+- V0V0 Note: Charge on plate does not change!

66. What happens? +-+-  i   i  oooo Potential Difference is REDUCED by insertion of dielectric. Charge on plate is Unchanged! Capacitance increases by a factor of  as we showed previously

67. SUMMARY OF RESULTS

68. APPLICATION OF GAUSS’ LAW

69. New Gauss for Dielectrics