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Copyright © 2009 Pearson Education, Inc. Chapter 24 Capacitance, Dielectrics, Electric Energy Storage.

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Presentation on theme: "Copyright © 2009 Pearson Education, Inc. Chapter 24 Capacitance, Dielectrics, Electric Energy Storage."— Presentation transcript:

1 Copyright © 2009 Pearson Education, Inc. Chapter 24 Capacitance, Dielectrics, Electric Energy Storage

2 Copyright © 2009 Pearson Education, Inc. Capacitors Determination of Capacitance Capacitors in Series and Parallel Electric Energy Storage Dielectrics Molecular Description of Dielectrics Units of Chapter 24

3 Copyright © 2009 Pearson Education, Inc. A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge Capacitors

4 Copyright © 2009 Pearson Education, Inc. Parallel-plate capacitor connected to battery. (b) is a circuit diagram Capacitors

5 Copyright © 2009 Pearson Education, Inc. When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage: The quantity C is called the capacitance. Unit of capacitance: the farad ( F ): 1 F = 1 C/V Capacitors

6 Copyright © 2009 Pearson Education, Inc Determination of Capacitance For a parallel-plate capacitor as shown, the field between the plates is E = Q/ ε 0 A. Integrating along a path between the plates gives the potential difference: V ba = Qd/ ε 0 A. This gives the capacitance:

7 Copyright © 2009 Pearson Education, Inc Determination of Capacitance Example 24-1: Capacitor calculations. (a) Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d.

8 Copyright © 2009 Pearson Education, Inc Determination of Capacitance Capacitors are now made with capacitances of 1 farad or more, but they are not parallel- plate capacitors. Instead, they are activated carbon, which acts as a capacitor on a very small scale. The capacitance of 0.1 g of activated carbon is about 1 farad. Some computer keyboards use capacitors; depressing the key changes the capacitance, which is detected in a circuit.

9 Copyright © 2009 Pearson Education, Inc Determination of Capacitance Example 24-2: Cylindrical capacitor. A cylindrical capacitor consists of a cylinder (or wire) of radius R b surrounded by a coaxial cylindrical shell of inner radius R a. Both cylinders have length l which we assume is much greater than the separation of the cylinders, so we can neglect end effects. The capacitor is charged (by connecting it to a battery) so that one cylinder has a charge +Q (say, the inner one) and the other one a charge –Q. Determine a formula for the capacitance.

10 Copyright © 2009 Pearson Education, Inc Determination of Capacitance Example 24-3: Spherical capacitor. A spherical capacitor consists of two thin concentric spherical conducting shells of radius r a and r b as shown. The inner shell carries a uniformly distributed charge Q on its surface, and the outer shell an equal but opposite charge –Q. Determine the capacitance of the two shells.

11 Copyright © 2009 Pearson Education, Inc Determination of Capacitance Example 24-4: Capacitance of two long parallel wires. Estimate the capacitance per unit length of two very long straight parallel wires, each of radius R, carrying uniform charges +Q and –Q, and separated by a distance d which is large compared to R ( d >> R ).

12 Copyright © 2009 Pearson Education, Inc. Capacitors in parallel have the same voltage across each one. The equivalent capacitor is one that stores the same charge when connected to the same battery: 24-3 Capacitors in Series and Parallel

13 Copyright © 2009 Pearson Education, Inc. Capacitors in series have the same charge. In this case, the equivalent capacitor has the same charge across the total voltage drop. Note that the formula is for the inverse of the capacitance and not the capacitance itself! 24-3 Capacitors in Series and Parallel

14 Copyright © 2009 Pearson Education, Inc Capacitors in Series and Parallel Example 24-5: Equivalent capacitance. Determine the capacitance of a single capacitor that will have the same effect as the combination shown.

15 Copyright © 2009 Pearson Education, Inc Capacitors in Series and Parallel Example 24-6: Charge and voltage on capacitors. Determine the charge on each capacitor and the voltage across each, assuming C = 3.0 μF and the battery voltage is V = 4.0 V.

16 Copyright © 2009 Pearson Education, Inc Capacitors in Series and Parallel Example 24-7: Capacitors reconnected. Two capacitors, C 1 = 2.2 μF and C 2 = 1.2 μF, are connected in parallel to a 24-V source as shown. After they are charged they are disconnected from the source and from each other and then reconnected directly to each other, with plates of opposite sign connected together. Find the charge on each capacitor and the potential across each after equilibrium is established.

17 Copyright © 2009 Pearson Education, Inc. A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor: 24-4 Electric Energy Storage

18 Copyright © 2009 Pearson Education, Inc Electric Energy Storage Example 24-8: Energy stored in a capacitor. A camera flash unit stores energy in a 150-μF capacitor at 200 V. (a) How much electric energy can be stored? (b) What is the power output if nearly all this energy is released in 1.0 ms?

19 Copyright © 2009 Pearson Education, Inc Electric Energy Storage Conceptual Example 24-9: Capacitor plate separation increased. A parallel-plate capacitor carries charge Q and is then disconnected from a battery. The two plates are initially separated by a distance d. Suppose the plates are pulled apart until the separation is 2d. How has the energy stored in this capacitor changed?

20 Copyright © 2009 Pearson Education, Inc Electric Energy Storage Example 24-10: Moving parallel capacitor plates. The plates of a parallel-plate capacitor have area A, separation x, and are connected to a battery with voltage V. While connected to the battery, the plates are pulled apart until they are separated by 3x. (a) What are the initial and final energies stored in the capacitor? (b) How much work is required to pull the plates apart (assume constant speed)? (c) How much energy is exchanged with the battery?

21 Copyright © 2009 Pearson Education, Inc. The energy density, defined as the energy per unit volume, is the same no matter the origin of the electric field: The sudden discharge of electric energy can be harmful or fatal. Capacitors can retain their charge indefinitely even when disconnected from a voltage source – be careful! 24-4 Electric Energy Storage

22 Copyright © 2009 Pearson Education, Inc. Heart defibrillators use electric discharge to “jump- start” the heart, and can save lives Electric Energy Storage

23 Copyright © 2009 Pearson Education, Inc. A dielectric is an insulator, and is characterized by a dielectric constant K. Capacitance of a parallel-plate capacitor filled with dielectric: 24-5 Dielectrics Using the dielectric constant, we define the permittivity:

24 Copyright © 2009 Pearson Education, Inc. Dielectric strength is the maximum field a dielectric can experience without breaking down Dielectrics

25 Copyright © 2009 Pearson Education, Inc Dielectrics Here are two experiments where we insert and remove a dielectric from a capacitor. In the first, the capacitor is connected to a battery, so the voltage remains constant. The capacitance increases, and therefore the charge on the plates increases as well.

26 Copyright © 2009 Pearson Education, Inc Dielectrics In this second experiment, we charge a capacitor, disconnect it, and then insert the dielectric. In this case, the charge remains constant. Since the dielectric increases the capacitance, the potential across the capacitor drops.

27 Copyright © 2009 Pearson Education, Inc Dielectrics Example 24-11: Dielectric removal. A parallel-plate capacitor, filled with a dielectric with K = 3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m 2 and are separated by d = 4.0 mm. (a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor.

28 Copyright © 2009 Pearson Education, Inc. The molecules in a dielectric, when in an external electric field, tend to become oriented in a way that reduces the external field Molecular Description of Dielectrics

29 Copyright © 2009 Pearson Education, Inc. This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. This reorientation of the molecules results in an induced charge – there is no net charge on the dielectric, but the charge is asymmetrically distributed. The magnitude of the induced charge depends on the dielectric constant: 24-6 Molecular Description of Dielectrics

30 Copyright © 2009 Pearson Education, Inc. Review Questions

31 Capacitor C 1 is connected across a battery of 5 V. An identical capacitor C 2 is connected across a battery of 10 V. Which one has more charge? C 1 1) C 1 C 2 2) C 2 3) both have the same charge 4) it depends on other factors ConcepTest 24.1Capacitors ConcepTest 24.1 Capacitors

32 Q = CV greater voltage charge C 2 Since Q = CV and the two capacitors are identical, the one that is connected to the greater voltage has more charge, which is C 2 in this case. Capacitor C 1 is connected across a battery of 5 V. An identical capacitor C 2 is connected across a battery of 10 V. Which one has more charge? C 1 1) C 1 C 2 2) C 2 3) both have the same charge 4) it depends on other factors ConcepTest 24.1Capacitors ConcepTest 24.1 Capacitors

33 increase the area of the plates 1) increase the area of the plates decrease separation between the plates 2) decrease separation between the plates 3) decrease the area of the plates 4) either (1) or (2) 5) either (2) or (3) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? +Q+Q –Q–Q ConcepTest 24.2aVarying Capacitance I ConcepTest 24.2a Varying Capacitance I

34 Q = CV increase its capacitance increasing Adecreasing d Since Q = CV, in order to increase the charge that a capacitor can hold at constant voltage, one has to increase its capacitance. Since the capacitance is given by, that can be done by either increasing A or decreasing d. increase the area of the plates 1) increase the area of the plates decrease separation between the plates 2) decrease separation between the plates 3) decrease the area of the plates 4) either (1) or (2) 5) either (2) or (3) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? +Q+Q –Q–Q ConcepTest 24.2aVarying Capacitance I ConcepTest 24.2a Varying Capacitance I

35 +Q+Q –Q–Q A parallel-plate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? the voltage decreases 1) the voltage decreases the voltage increases 2) the voltage increases the charge decreases 3) the charge decreases the charge increases 4) the charge increases both voltage and charge change 5) both voltage and charge change ConcepTest 24.2bVarying Capacitance II ConcepTest 24.2b Varying Capacitance II

36 Since the battery stays connected, the voltage must remain constant! Q = CV charge must decrease Since the battery stays connected, the voltage must remain constant! Since, when the spacing d is doubled, the capacitance C is halved. And since Q = CV, that means the charge must decrease. +Q+Q –Q–Q A parallel-plate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? the voltage decreases 1) the voltage decreases the voltage increases 2) the voltage increases the charge decreases 3) the charge decreases the charge increases 4) the charge increases both voltage and charge change 5) both voltage and charge change ConcepTest 24.2bVarying Capacitance II ConcepTest 24.2b Varying Capacitance II Follow-up: How do you increase the charge?

37 A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage? 100 V 1) 100 V 200 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V +Q+Q –Q–Q ConcepTest 24.2cVarying Capacitance III ConcepTest 24.2c Varying Capacitance III

38 Once the battery is disconnected, Q has to remain constant Q = CV voltage must double Once the battery is disconnected, Q has to remain constant, since no charge can flow either to or from the battery. Since, when the spacing d is doubled, the capacitance C is halved. And since Q = CV, that means the voltage must double. A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage? 100 V 1) 100 V 200 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V +Q+Q –Q–Q ConcepTest 24.2cVarying Capacitance III ConcepTest 24.2c Varying Capacitance III

39 ConcepTest 24.3aCapacitors I ConcepTest 24.3a Capacitors I o o C C C C eq 1) C eq = 3/2C 2) C eq = 2/3C 3) C eq = 3C 4) C eq = 1/3C 5) C eq = 1/2C What is the equivalent capacitance,, of the combination below? What is the equivalent capacitance, C eq, of the combination below?

40 series inverses1/2C parallel directly 3/2C The 2 equal capacitors in series add up as inverses, giving 1/2C. These are parallel to the first one, which add up directly. Thus, the total equivalent capacitance is 3/2C. ConcepTest 24.3aCapacitors I ConcepTest 24.3a Capacitors I o o C C C C eq 1) C eq = 3/2C 2) C eq = 2/3C 3) C eq = 3C 4) C eq = 1/3C 5) C eq = 1/2C What is the equivalent capacitance,, of the combination below? What is the equivalent capacitance, C eq, of the combination below?

41 ConcepTest 24.3bCapacitors II ConcepTest 24.3b Capacitors II 1) V 1 = V 2 2) V 1 > V 2 3) V 1 < V 2 4) all voltages are zero C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )?

42 ConcepTest 24.3bCapacitors II ConcepTest 24.3b Capacitors II 1) V 1 = V 2 2) V 1 > V 2 3) V 1 < V 2 4) all voltages are zero C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V The voltage across C 1 is 10 V. The combined capacitors C 2 + C 3 are parallel to C 1. The voltage across C 2 + C 3 is also 10 V. Since C 2 and C 3 are in series, their voltages add. Thus the voltage across C 2 and C 3 each has to be 5 V, which is less than V 1. How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )? Follow-up: ? Follow-up: What is the current in this circuit?

43 ConcepTest 24.3cCapacitors III ConcepTest 24.3c Capacitors III C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )?

44 ConcepTest 24.3cCapacitors III ConcepTest 24.3c Capacitors III C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V Q = CVsame have V 1 > V 2 Q 1 > Q 2 We already know that the voltage across C 1 is 10 V and the voltage across both C 2 and C 3 is 5 V each. Since Q = CV and C is the same for all the capacitors, we have V 1 > V 2 and therefore Q 1 > Q 2. 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )?

45 Copyright © 2009 Pearson Education, Inc. Ch. 24 Homework Read Ch. 24 and begin looking over Ch. 25. Group problems #’s 2, 12, 22 HW problems #’s 3, 5, 7, 11, 13, 17, 23, 25, 29, 35, 41, 47, 59


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