2. Capacitance AP Physics B

3. Capacitors Consider two separated conductors, like two parallel plates, with external leads to attach to other circuit elements. Such a device is called a capacitor. 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.

4. Capacitance The capacitance (C) of a conductor is defined as the ratio of the charge (Q) on the conductor to the potential (V) produced. Capacitance:

5. 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

6. 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

7. Example 2. 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

8. 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.

9. 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

10. Example 3: In a capacitor, we found its 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 3.

11. Electrical Circuit Symbols Electrical circuits often contain two or more capacitors grouped together and attached to an energy source, such as a battery. The following symbols are often used: + Capacitor + - - + - - + - + - Ground Battery - +

12. Capacitors in Series Capacitors or other devices connected along a single path are said to be connected in series. See circuit below: Series connection of capacitors. “+ to – to + …” Charge inside dots is induced. Battery C1C1 C2C2 C3C3 + + - - + + + + - - - -

13. Charge on Capacitors in Series Since inside charge is only induced, the charge on each capacitor is the same. Charge is same: series connection of capacitors. Q = Q 1 = Q 2 =Q 3 Battery C1C1 C2C2 C3C3 + + - - + + + + - - - - Q1Q1 Q2Q2 Q3Q3

14. Voltage on Capacitors in Series Since the potential difference between points A and B is independent of path, the battery voltage V must equal the sum of the voltages across each capacitor. Total voltage V Series connection Sum of voltages V = V 1 + V 2 + V 3 Battery C1C1 C2C2 C3C3 + + - - + + + + - - - - V1V1 V2V2 V3V3 AB

15. Equivalent Capacitance: Series V = V 1 + V 2 + V 3 Q 1 = Q 2 = Q 3 + + - - + + + + - - - - C1C1 C2C2 C3C3 V1V1 V2V2 V3V3 Equivalent C e for capacitors in series:

16. Example 1. Find the equivalent capacitance of the three capacitors connected in series with a 24-V battery. + + - - + + + + - - - - 2  F C1C1 C2C2 C3C3 24 V 4  F6  F C e for series: C e = 1.09  F

17. Example 1 (Cont.): The equivalent circuit can be shown as follows with single C e. + + - - + + + + - - - - 2  F C1C1 C2C2 C3C3 24 V 4  F 6  F 1.09  F CeCe 24 V C e = 1.09  F Note that the equivalent capacitance C e for capacitors in series is always less than the least in the circuit. (1.09 < 2 Note that the equivalent capacitance C e for capacitors in series is always less than the least in the circuit. (1.09  F < 2  F)

18. 1.09  F CeCe 24 V + + - - + + + + - - - - 2  F C1C1 C2C2 C3C3 24 V 4  F 6  F C e = 1.09  F Q T = C e V = (1.09  F)(24 V); Q T  = 26.2  C For series circuits: Q T = Q 1 = Q 2 = Q 3 Q 1 = Q 2 = Q 3 = 26.2  C Example 1 (Cont.): What is the total charge and the charge on each capacitor?

19. + + - - + + + + - - - - 2  F C1C1 C2C2 C3C3 24 V 4  F 6  F V T  = 24 V Note: V T = 13.1 V + 6.55 V + 4.37 V = 24.0 V Example 1 (Cont.): What is the voltage across each capacitor?

20. Short Cut: Two Series Capacitors The equivalent capacitance C e for two series capacitors is the product divided by the sum. 3  F6  F + + - - + + - - C1C1 C2C2Example: C e = 2  F

21. Parallel Circuits Capacitors which are all connected to the same source of potential are said to be connected in parallel. See below: Parallel capacitors: “+ to +; - to -” C2C2 C3C3 C1C1 ++ -- ++ -- ++ -- Charges: Q T = Q 1 + Q 2 + Q 3 Voltages: V T = V 1 = V 2 = V 3

22. Equivalent Capacitance: Parallel Q = Q 1 + Q 2 + Q 3 Equivalent C e for capacitors in parallel: Equal Voltages: CV = C 1 V 1 + C 2 V 2 + C 3 V 3 Parallel capacitors in Parallel: C2C2 C3C3 C1C1 ++ -- ++ -- ++ -- C e = C 1 + C 2 + C 3

23. Example 2. Find the equivalent capacitance of the three capacitors connected in parallel with a 24-V battery. C e for parallel: C e = 12  F C2C2 C3C3 C1C1 2  F4  F6  F 24 V Q = Q 1 + Q 2 + Q 3 V T = V 1 = V 2 = V 3 C e = (2 + 4 + 6)  F Note that the equivalent capacitance C e for capacitors in parallel is always greater than the largest in the circuit. (12 > 6 Note that the equivalent capacitance C e for capacitors in parallel is always greater than the largest in the circuit. (12  F > 6  F)

24. Example 2 (Cont.) Find the total charge Q T and charge across each capacitor. C e = 12  F C2C2 C3C3 C1C1 2  F4  F6  F 24 V Q = Q 1 + Q 2 + Q 3 V 1 = V 2 = V 3 = 24 V Q 1 = (2  F)(24 V) = 48  C Q 1 = (4  F)(24 V) = 96  C Q 1 = (6  F)(24 V) = 144  C Q T = C e V Q T = (12  F)(24 V) Q T = 288  C

25. Example 3. Find the equivalent capacitance of the circuit drawn below. C1C1 4  F 3  F 6  F 24 V C2C2 C3C3 C1C1 4  F 2  F 24 V C 3,6 CeCe 6  F 24 V C e = 4  F + 2  F C e = 6  F

26. Example 3 (Cont.) Find the total charge Q T. C1C1 4  F 3  F 6  F 24 V C2C2 C3C3 C e = 6  F Q = CV = (6  F)(24 V) Q T = 144  C C1C1 4  F 2  F 24 V C 3,6 CeCe 6  F 24 V

27. Example 3 (Cont.) Find the charge Q 4 and voltage V 4 across the the 4  F capacitor  C1C1 4  F 3  F 6  F 24 V C2C2 C3C3 V 4 = V T = 24 V Q 4 = (4  F)(24 V) Q 4 = 96  C The remainder of the charge: (144  C – 96  C) is on EACH of the other capacitors. (Series) Q 3 = Q 6 = 48  C This can also be found from Q = C 3,6 V 3,6 = (2  F)(24 V)

28. Example 3 (Cont.) Find the voltages across the 3 and 6-  F capacitors  C1C1 4  F 3  F 6  F 24 V C2C2 C3C3 Note: V 3 + V 6 = 16.0 V + 8.00 V = 24 V Q 3 = Q 6 = 48  C Use these techniques to find voltage and capacitance across each capacitor in a circuit.

29. Summary: Series Circuits Q = Q 1 = Q 2 = Q 3 V = V 1 + V 2 + V 3 For two capacitors at a time:

30. Summary: Parallel Circuits Q = Q 1 + Q 2 + Q 3 V = V 1 = V 2 =V 3 For complex circuits, reduce the circuit in steps using the rules for both series and parallel connections until you are able to solve problem.

31. AP Physics HW 3/26 Read Chapter 16.3 and 16.5 (that’s it!)Read Chapter 16.3 and 16.5 (that’s it!) Do Problems…#63, 65, 67, 69, 71, 93, 95Do Problems…#63, 65, 67, 69, 71, 93, 95 Finish ALL other HW problems setsFinish ALL other HW problems sets Yosemite $ DUE.Yosemite $ DUE. T-Shirt Design?...25 points up for grabs!T-Shirt Design?...25 points up for grabs!