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Copyright © 2009 Pearson Education, Inc. May Term in Guatemala GDS 3559/STS 3500: Engineering Public Health: An Interdisciplinary Exploration of Community.

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Presentation on theme: "Copyright © 2009 Pearson Education, Inc. May Term in Guatemala GDS 3559/STS 3500: Engineering Public Health: An Interdisciplinary Exploration of Community."— Presentation transcript:

1 Copyright © 2009 Pearson Education, Inc. May Term in Guatemala GDS 3559/STS 3500: Engineering Public Health: An Interdisciplinary Exploration of Community Development in Guatemala Contact Kent Wayland (kaw6r@virginia.edu), Eric Anderson (ewa3a@virginia.edu), or David R. Burt, MD, at drb5p@virginia.edukaw6r@virginia.eduewa3a@virginia.edudrb5p@virginia.edu for more information! Want to Learn how engineering affects community health? Interested in Public Health? Global Health? Medicine?

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

3 Copyright © 2009 Pearson Education, Inc. Recap:

4 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

5 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

6 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

7 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

8 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

9 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

10 Copyright © 2009 Pearson Education, Inc. 24-3 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.

11 Copyright © 2009 Pearson Education, Inc. 24-3 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.

12 Copyright © 2009 Pearson Education, Inc. 24-3 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.

13 Copyright © 2009 Pearson Education, Inc. 24-3 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.

14 Copyright © 2009 Pearson Education, Inc. 24-3 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.

15 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?

16 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?

17 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 )?

18 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?

19 Copyright © 2009 Pearson Education, Inc. A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor. Consider two sheets of charge, one positive and one negative on the surface of conductor b. 24-4 Electric Energy Storage Work required to pull the negatively charged sheet to conductor a is:

20 Copyright © 2009 Pearson Education, Inc. 24-4 Electric Energy Storage National Ignition Facility (NIF) Lawrence Livermore National Laboratory

21 Copyright © 2009 Pearson Education, Inc. NIF Laser system driven by 4000 300 μF capacitors which store a total of 422 MJ. They take 60 s to charge and are discharged in 400 μs. 1)What is the potential difference across each capacitor? 2)What is the power delivered during the discharge?

22 Copyright © 2009 Pearson Education, Inc. NIF Laser system driven by 4000 300 μF capacitors which store a total of 422 MJ. They take 60 s to charge and are discharged in 400 μs. 1)What is the potential difference across each capacitor? 2)What is the power delivered during the discharge? Solution: 1)U = CV 2 /2 →V = (2U/C) 1/2 →V = [2(422x10 6 )/4000/300x10 -6 ] ½ = 26.5 kV 2)P = W/t = U/t = 422x10 6 /400x10 -6 ~ 10 12 W = 1000 GW! cf. 1.0-1.5 GW for power plant

23 Copyright © 2009 Pearson Education, Inc. 24-4 Electric Energy Storage Conceptual Example: 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. a) How has the energy stored in this capacitor changed? b) Would the answer change if the battery were not disconnected?

24 Copyright © 2009 Pearson Education, Inc. 24-4 Electric Energy Storage

25 Copyright © 2009 Pearson Education, Inc. Heart defibrillators use electric discharge to “jump- start” the heart, and can save lives. 24-4 Electric Energy Storage

26 Copyright © 2009 Pearson Education, Inc. The energy density, u, 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

27 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:

28 Copyright © 2009 Pearson Education, Inc. Dielectric strength is the maximum field a dielectric can experience without breaking down. 24-5 Dielectrics

29 Copyright © 2009 Pearson Education, Inc. 24-5 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.

30 Copyright © 2009 Pearson Education, Inc. 24-5 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.

31 Copyright © 2009 Pearson Education, Inc. 24-5 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.

32 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. 24-6 Molecular Description of Dielectrics

33 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

34 Copyright © 2009 Pearson Education, Inc. Capacitor: nontouching conductors carrying equal and opposite charge. Capacitance: Capacitance of a parallel-plate capacitor: Summary of Chapter 24

35 Copyright © 2009 Pearson Education, Inc. Summary of Chapter 24 Capacitors in parallel: Capacitors in series:

36 Copyright © 2009 Pearson Education, Inc. Energy density in electric field: A dielectric is an insulator. Dielectric constant gives ratio of total field to external field. For a parallel-plate capacitor: Summary of Chapter 24

37 Copyright © 2009 Pearson Education, Inc. Questions?


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