2. Do Now (9/23/13): What is the voltage of a proton moving at a constant speed of 3 m/s over 1 s in an electric field of 300 N/C?What is the voltage of a proton moving at a constant speed of 3 m/s over 1 s in an electric field of 300 N/C? What does the word “capacity” mean to you?What does the word “capacity” mean to you?

3. Chapter 26A - Capacitance A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

4. Objectives: After completing this module, you should be able to: Define capacitance in terms of charge and voltage, and calculate the capacitance for a parallel plate capacitor given separation and area of the plates.Define capacitance in terms of charge and voltage, and calculate the capacitance for a parallel plate capacitor given separation and area of the plates. Define dielectric constant and apply to calculations of voltage, electric field intensity, and capacitance.Define dielectric constant and apply to calculations of voltage, electric field intensity, and capacitance. Find the potential energy stored in capacitors.Find the potential energy stored in capacitors.

5. Maximum Charge on a Conductor Earth Battery Conductor - - - - - - - - - - - - - - e-e- e-e- A battery establishes a difference of potential that can pump electrons e - from a ground (earth) to a conductor There is a limit to the amount of charge that a conductor can hold without leaking to the air. There is a certain capacity for holding charge.

6. Capacitance The capacitance C of a conductor is defined as the ratio of the charge Q on the conductor to the potential V produced. Earth Battery Conductor - - - - - - - - - - - - - - e-e- e-e- Capacitance: Q, V

7. Capacitance in Farads One farad (F) is the capacitance C of a conductor that holds one coulomb of charge for each volt of potential. Example: When 40  C of charge are placed on a con- ductor, the potential is 8 V. What is the capacitance? C = 5  F

8. Parallel Plate Capacitance d Area A +Q -Q You will recall from Gauss’ law that E is also: For these two parallel plates: Q is charge on either plate. A is area of plate. And

9. Parallel Plate Capacitance Capacitance:Capacitance:

10. Permitivity of free space “Epsilon-naught”“Epsilon-naught” 8.854 x 10 -12 C 2 /N m 28.854 x 10 -12 C 2 /N m 2

11. Practice: Work on your homeworkWork on your homework Work on the bonusWork on the bonus Be ready for an exit question!!Be ready for an exit question!!

12. Example 3. The plates of a parallel plate capacitor have an area of 0.4 m 2 and are 3 mm apart in air. What is the capacitance? 3 mm d A 0.4 m 2 C = 1.18 nF

13. Capacitance of Spherical Conductor +Q r E and V at surface. At surface of sphere: Recall: And: Capacitance: Capacitance, C

14. Example 1: What is the capacitance of a metal sphere of radius 8 cm? r = 0.08 m Capacitance, C +Q r Capacitance: C = 4   r C = 8.90 x 10 -12 F Note: The capacitance depends only on physical para- meters (the radius r) and is not determined by either charge or potential. This is true for all capacitors.

15. Q = 3.56 nC Total Charge on Conductor: Example 1 (Cont.): What charge Q is needed to give a potential of 400 V? r = 0.08 m Capacitance, C +Q r C = 8.90 x 10 -12 F Note: The farad (F) and the coulomb (C) are extremely large units for static electricity. The SI prefixes micro , nano n, and pico p are often used.

16. Dielectric Strength The dielectric strength of a material is that electric intensity E m for which the material becomes a conductor. (Charge leakage.) r Q Dielectric E m varies considerably with physical and environmental conditions such as pressure, humidity, and surfaces. For air: E m = 3 x 10 6 N/C for spherical surfaces and as low as 0.8 x 10 6 N/C for sharp points.

17. Example 2: What is the maximum charge that can be placed on a spherical surface one meter in diameter? (R = 0.50 m) r Q E m = 3 x 10 6 N/C Maximum Q Air Maximum charge in air: Q m = 83.3  C This illustrates the large size of the coulomb as a unit of charge in electrostatic applications.

18. Capacitance and Shapes The charge density on a surface is significantly affected by the curvature. The density of charge is greatest where the curvature is greatest. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Leakage (called corona discharge) often occurs at sharp points where curvature r is greatest.

19. Parallel Plate Capacitance d Area A +Q -Q You will recall from Gauss’ law that E is also: For these two parallel plates: Q is charge on either plate. A is area of plate. And

20. Example 3. The plates of a parallel plate capacitor have an area of 0.4 m 2 and are 3 mm apart in air. What is the capacitance? 3 mm d A 0.4 m 2 C = 1.18 nF

21. Applications of Capacitors + + + + + + + - - - - - - - A Variable Capacitor Changing Area d Changing d Microphone A microphone converts sound waves into an electrical signal (varying voltage) by changing d. The tuner in a radio is a variable capacitor. The changing area A alters capacitance until desired signal is obtained.

22. Dielectric Materials Most capacitors have a dielectric material between their plates to provide greater dielectric strength and less probability for electrical discharge. The separation of dielectric charge allows more charge to be placed on the plates—greater capacitance C > C o. + + + + + + - - - - - - Air CoCoCoCo EoEoEoEo + + + + + + - - - - - - - + C > C o E < E o + + + + + + - - - - - - - + - + Dielectric reduced E

23. Smaller plate separation without contact.Smaller plate separation without contact. Increases capacitance of a capacitor.Increases capacitance of a capacitor. Higher voltages can be used without breakdown.Higher voltages can be used without breakdown. Often it allows for greater mechanical strength.Often it allows for greater mechanical strength. Smaller plate separation without contact.Smaller plate separation without contact. Increases capacitance of a capacitor.Increases capacitance of a capacitor. Higher voltages can be used without breakdown.Higher voltages can be used without breakdown. Often it allows for greater mechanical strength.Often it allows for greater mechanical strength. Advantages of Dielectrics

24. Insertion of Dielectric + + + + + + C o V o E o   +Q -Q + + +Q -Q Dielectric Air Permittivity increases.  >  o Capacitance increases. C > C o Voltage decreases. V < V o Field decreases. E < E o Insertion of a dielectric Same Q Q = Q o C V E 

25. Dielectric Constant, K The dielectric constant K for a material is the ratio of the capacitance C with this material as compared with the capacitance C o in a vacuum. Dielectric constant: K = 1 for Air K can also be given in terms of voltage V, electric field intensity E, or permittivity  :

26. The Permittivity of a Medium The capacitance of a parallel plate capacitor with a dielectric can be found from: The constant  is the permittivity of the medium which relates to the density of field lines.

27. Example 4: Find the capacitance C and the charge Q if connected to 200-V battery. Assume the dielectric constant is K = 5.0. 2 mm d A 0.5 m 2      5(8.85 x 10 -12 C/Nm 2 )    44.25 x 10 -12 C/Nm 2 C = 11.1 nF Q if connected to V = 200 V? Q if connected to V = 200 V? Q = CV = (11.1 nF)(200 V) Q = 2.22  C

28. Example 4 (Cont.): Find the field E between the plates. Recall Q = 2.22  C; V = 200 V.  44.25 x 10 -12 C/Nm 2 E = 100 N/C Since V = 200 V, the same result is found if E = V/d is used to find the field. 2 mm d A 0.5 m 2   200 V

29. Example 5: A capacitor has a capacitance of 6  F with air as the dielectric. A battery charges the capacitor to 400 V and is then disconnected. What is the new voltage if a sheet of mica (K = 5) is inserted? What is new capacitance C ? V = 80.0 V C = Kc o = 5(6  F) C = 30  F V o = 400 V Mica, K = 5 Air dielectric Mica dielectric

30. Example 5 (Cont.): If the 400-V battery is reconnected after insertion of the mica, what additional charge will be added to the plates due to the increased C? Q 0 = C 0 V 0 = (6  F)(400 V)  Q = 9.60 mC V o = 400 V Mica, K = 5 Air C o = 6 Air C o = 6  F Mica C = 30 Mica C = 30  F Q 0 = 2400  C Q = CV = (30  F)(400 V) Q = 12,000  C  Q = 12,000  C – 2400  C  Q = 9600  C

31. Energy of Charged Capacitor The potential energy U of a charged capacitor is equal to the work (qV) required to charge the capacitor. If we consider the average potential difference from 0 to V f to be V/2: Work = Q(V/2) = ½QV

32. Example 6: In Ex-4, we found capacitance to be 11.1 nF, the voltage 200 V, and the charge 2.22  C. Find the potential energy U. U = 222  J Verify your answer from the other formulas for P.E. C = 11.1 nF 200 V Q = 2.22  C U = ? Capacitor of Example 5.

33. Energy Density for Capacitor Energy density u is the energy per unit volume (J/m 3 ). For a capacitor of area A and separation d, the energy density u is found as follows: Energy Density u for an E-field: A d Energy Density u:

34. Summary of Formulas

35. CONCLUSION: Chapter 25 Capacitance