1. Capacitance Chapter 25

2. Capacitance A capacitor consists of two isolated conductors (the plates) with charges +q and -q. Its capacitance C is defined from where V is the potential difference between the plates.

3. A parallel-plate capacitor, made up of two plates of area A separated by a distance d. The charges on the facing plate surfaces have the same magnitude q but opposite signs As the field lines show, the electric field due to the charged plates is uniform in the central region between the plates. The field is not uniform at the edges of the plates, as indicated by the “fringing” of the field lines there. Capacitance Capacitance depends only on the geometry of the plates and NOT on their charge or potential difference.

4. When a circuit with a battery, an open switch, and an uncharged capacitor is completed by closing the switch, conduction electrons shift, leaving the capacitor plates with opposite charges. Capacitance - Charging

5. 4.1 L L 4.1b ++ –– L L 4.1c L L 4.1a L L ++ – + - + - + - + -

6. To relate the electric field E between the plates of a capacitor to the charge q on either plate, we shall use Gauss’ law: Calculating electric field A charged parallel-plate capacitor. A Gaussian surface encloses the charge on the positive plate. The integration is taken along a path extending directly from the negative plate to the positive plate. Calculating the Capacitance We draw a Gaussian surface that encloses just the charge q on the positive plate. E is uniform and parallel to dA, so

7. the potential difference between the plates of a capacitor is related to the field E by Letting V represent the difference V f = V i, we can then recast the above equation as: Calculating potential difference A charged parallel-plate capacitor. A Gaussian surface encloses the charge on the positive plate. The integration is taken along a path extending directly from the negative plate to the positive plate. Calculating the Capacitance We will apply this general expression to 4 particular cases.

8. We assume that the plates of our parallel-plate capacitor are so large and so close together that we can neglect the fringing of the electric field at the edges of the plates, taking E to be constant throughout the region between the plates. where A is the area of the plate. And therefore, A charged parallel-plate capacitor. A Gaussian surface encloses the charge on the positive plate. The integration is taken along a path extending directly from the negative plate to the positive plate. Parallel-Plate Capacitor Calculating the Capacitance

9. Electric Potential – Charged Sphere

10. Capacitance +Q –Q

11. Concept Check - Capacitors 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 the most charge? 1. C 1 2. C 2 3. both have the same charge 4. it depends on other factors +Q –Q

12. Concept Check - Capacitors 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 the most charge? 1. C 1 2. C 2 3. both have the same charge 4. it depends on other factors Q = C V greater voltagemost chargeC2 Since Q = C V and the two capacitors are identical, the one that is connected to the greater voltage has the most charge, which is C2 in this case. +Q –Q

13. Concept Check – Capacitors (2) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? 1. increase the area of 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) +Q –Q

14. Concept Check – Capacitors (2) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? 1. increase the area of 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) Q = C V increase its capacitance increasing Adecreasing d Since Q = C V, 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. +Q –Q

15. Concept Check – Capacitors (3) A parallel-plate capacitor initially is at a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? 1. the voltage decreases 2. the voltage increases 3. the charge decreases 4. the charge increases 5. both voltage and charge change +Q –Q

16. Concept Check – Capacitors (3) A parallel-plate capacitor initially is at a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? 1. the voltage decreases 2. the voltage increases 3. the charge decreases 4. the charge increases 5. both voltage and charge change Since the battery stays connected, the voltage must remain constant ! Q = C V 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 = C V, that means the charge must decrease. +Q –Q

17. Concept Check – Capacitors (4) A parallel-plate capacitor initially is at a voltage 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? 1. 100 V 2. 200 V 3. 400 V 4. 800 V 5. 1600 V +Q –Q

18. Concept Check – Capacitors (4) A parallel-plate capacitor initially is at a voltage 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? 1. 100 V 2. 200 V 3. 400 V 4. 800 V 5. 1600 V Once the battery is disconnected, Q has to remain constant Q = C Vvoltage 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 = C V, that means the voltage must double. +Q –Q

19. Figure shows, in cross section, a cylindrical capacitor of length L formed by two coaxial cylinders of radii a and b. We assume that L >> b so that we can neglect the fringing of the electric field that occurs at the ends of the cylinders. Each plate contains a charge of magnitude q. Here, charge and the field magnitude E is related as follows, Solving for E field: Cylindrical Capacitor A cross section of a long cylindrical capacitor, showing a cylindrical Gaussian surface of radius r (that encloses the positive plate) and the radial path of integration. This figure also serves to illustrate a spherical capacitor in a cross section through its center. Calculating the Capacitance

20. Cylindrical Capacitor A cross section of a long cylindrical capacitor, showing a cylindrical Gaussian surface of radius r (that encloses the positive plate) and the radial path of integration. This figure also serves to illustrate a spherical capacitor in a cross section through its center. Calculating the Capacitance

21. Cylindrical Capacitor A cross section of a long cylindrical capacitor, showing a cylindrical Gaussian surface of radius r (that encloses the positive plate) and the radial path of integration. This figure also serves to illustrate a spherical capacitor in a cross section through its center. Calculating the Capacitance

22. For spherical capacitor the capacitance is: Capacitance of an isolated sphere: Others… Answer: (a) decreases (b) increases (c) increases A cross section of a long cylindrical capacitor, showing a cylindrical Gaussian surface of radius r (that encloses the positive plate) and the radial path of integration. This figure also serves to illustrate a spherical capacitor in a cross section through its center. Calculating the Capacitance

23. Sample Problem 25.01

24. Capacitors in Parallel and in Series

25. Capacitors in Parallel

26. Capacitors in Parallel and in Series

27. Capacitors in Series

28. Concept Check – Capacitors in Combination What is the equivalent capacitance, C eq, of the combination below? 1. C eq = 3/2 C 2. C eq = 2/3 C 3. C eq = 3 C 4. C eq = 1/3 C 5. C eq = 1/2 C o o C C C C eq

29. Concept Check – Capacitors in Combination What is the equivalent capacitance, C eq, of the combination below? 1. C eq = 3/2 C 2. C eq = 2/3 C 3. C eq = 3 C 4. C eq = 1/3 C 5. C eq = 1/2 C seriesinverses1/2 C paralleldirectly 3/2 C The 2 equal capacitors in series add up as inverses, giving 1/2 C. These are parallel to the first one, which add up directly. Thus, the total equivalent capacitance is 3/2 C. o o C C C C eq

30. Concept Check – Capacitors in Combination (2) How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )? 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

31. Concept Check – Capacitors in Combination (2) How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )? 1. V 1 = V 2 2. V 1 > V 2 3. V 1 < V 2 4. all voltages are zero 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 half of V 1. C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V

32. Concept Check – Capacitors in Combination (3) How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )? 1. Q 1 = Q 2 2. Q 1 > Q 2 3. Q 1 < Q 2 4. all voltages are zero C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V

33. Concept Check – Capacitors in Combination (3) How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )? 1. Q 1 = Q 2 2. Q 1 > Q 2 3. Q 1 < Q 2 4. all voltages are zero Q = CV sameV 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, if V 1 > V 2 then Q 1 > Q 2. C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V

34. Sample Problem 25.02

35. Sample 25.03

36. Capacitors in Combination

37. The electric potential energy U of a charged capacitor, is equal to the work required to charge the capacitor. This energy can be associated with the capacitor’s electric field E. Every electric field, in a capacitor or from any other source, has an associated stored energy. In vacuum, the energy density u (potential energy per unit volume) in a field of magnitude E is Energy Stored in an Electric Field

38. Concept Check – Capacitors (5) Consider a simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D > d. The electrostatic energy stored in the capacitor is 1. greater than 2. the same as 3. smaller than before the plates were pulled apart.

39. Concept Check – Capacitors (5) Consider a simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D > d. The electrostatic energy stored in the capacitor is 1. greater than 2. the same as 3. smaller than before the plates were pulled apart. Since the oppositely charged plates are attracted to each other, work must be done to increase the separation between them. Thus, the electrostatic energy stored in the capacitor increases when the plates are separated.

40. (b) If the charge on the capacitor plates is maintained, as in this case by isolating the capacitor, the effect of a dielectric is to reduce the potential difference between the plates. The scale shown is that of a potentiometer, a device used to measure potential difference (here, between the plates). A capacitor cannot discharge through a potentiometer. (a)If the potential difference between the plates of a capacitor is maintained, as by the presence of battery B, the effect of a dielectric is to increase the charge on the plates. If the space between the plates of a capacitor is completely filled with a dielectric material, the capacitance C in vacuum (or, effectively, in air) is multiplied by the material’s dielectric constant κ, (Greek kappa) which is a number greater than 1. Capacitor with a Dielectric

42. An Atomic View (a) Molecules with a permanent electric dipole moment, showing their random orientation in the absence of an external electric field. (b) An electric field is applied, producing partial alignment of the dipoles. Thermal agitation prevents complete alignment. Polar Dielectrics Nonpolar Dielectrics Capacitor with a Dielectric

43. Capacitors and Dielectrics

46. Concept Check – Capacitance (6) A dielectric is inserted between the plates of a capacitor. The system is then charged and the dielectric is removed. The electrostatic energy stored in the capacitor is 1. greater than 2. the same as 3. smaller than it would have been if the dielectric were left in place.

47. Concept Check – Capacitance (6) A dielectric is inserted between the plates of a capacitor. The system is then charged and the dielectric is removed. The electrostatic energy stored in the capacitor is 1. greater than 2. the same as 3. smaller than it would have been if the dielectric were left in place. The dielectric is polarized when it is between the charged plates. The positively charged dielectric surface is attracted to the negatively charged plate, and the negatively charged dielectric surface is attracted to the positively charged plate. Thus, you must do work to remove the dielectric. As a result, the electrostatic energy stored in the capacitor increases when the dielectric is removed.

48. When a dielectric is present, Gauss’ law may be generalized to where q is the free charge. Any induced surface charge is accounted for by including the dielectric constant k inside the integral. A parallel-plate capacitor (a) without and (b) with a dielectric slab inserted. The charge q on the plates is assumed to be the same in both cases. Inserting a dielectric into a capacitor causes induced charge to appear on the faces of the dielectric and weakens the electric field between the plates. The induced charge is less than the free charge on the plates. Note: The flux integral now involves κE, not just E. The vector ε 0 κE is sometimes called the electric displacement D, so that the above equation can be written in the form Dielectrics and Gauss’ Law

49. Capacitor and Capacitance The capacitance of a capacitor is defined as: Determining Capacitance Parallel-plate capacitor: Cylindrical Capacitor: Spherical Capacitor: Isolated sphere: Eq. 25-1 Eq. 25-9 Eq. 25-14 Capacitor in parallel and series In parallel: In series Eq. 25-19 Eq. 25-17 Eq. 25-18 Eq. 25-20 Potential Energy and Energy Density Electric Potential Energy (U): Energy density (u) Eq. 25-21&22 Eq. 25-25 Summary

50. Capacitance with a Dielectric If the space between the plates of a capacitor is completely filled with a dielectric material, the capacitance C is increased by a factor κ, called the dielectric constant, which is characteristic of the material. Gauss’ Law with a Dielectric When a dielectric is present, Gauss’ law may be generalized to Eq. 25-36 Summary