# Chapter 24 Capacitance, Dielectrics, Electric Energy Storage

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Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
Chapter 24 opener. Capacitors come in a wide range of sizes and shapes, only a few of which are shown here. A capacitor is basically two conductors that do not touch, and which therefore can store charge of opposite sign on its two conductors. Capacitors are used in a wide variety of circuits, as we shall see in this and later Chapters.

Units of Chapter 24 Capacitors Determination of Capacitance
Capacitors in Series and Parallel Electric Energy Storage Dielectrics Molecular Description of Dielectrics

24-1 Capacitors A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge. Figure Capacitors: diagrams of (a) parallel plate, (b) cylindrical (rolled up parallel plate).

24-1 Capacitors Parallel-plate capacitor connected to battery. (b) is a circuit diagram. Figure (a) Parallel-plate capacitor connected to a battery. (b) Same circuit shown using symbols.

24-1 Capacitors 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.

24-2 Determination of Capacitance
For a parallel-plate capacitor as shown, the field between the plates is E = Q/ε0A. Integrating along a path between the plates gives the potential difference: Vba = Qd/ε0A. This gives the capacitance: Figure Parallel-plate capacitor, each of whose plates has area A. Fringing of the field is ignored.

24-2 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. Solution: a. C = 53 pF. b. Q = CV = 6.4 x C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.

24-2 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. Figure Key on a computer keyboard. Pressing the key reduces the capacitor spacing thus increasing the capacitance which can be detected electronically.

24-2 Determination of Capacitance
Example 24-2: Cylindrical capacitor. A cylindrical capacitor consists of a cylinder (or wire) of radius Rb surrounded by a coaxial cylindrical shell of inner radius Ra. 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. Figure (a) Cylindrical capacitor consists of two coaxial cylindrical conductors. (b) The electric field lines are shown in cross-sectional view. Solution: We need to find the potential difference between the cylinders; we can do this by integrating the field (which was calculated for a long wire already). The field is proportional to 1/R, so the potential is proportional to ln Ra/Rb. Then C = Q/V.

24-2 Determination of Capacitance
Example 24-3: Spherical capacitor. A spherical capacitor consists of two thin concentric spherical conducting shells of radius ra and rb 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. Figure Cross section through the center of a spherical capacitor. The thin inner shell has radius rb and the thin outer shell has radius ra. Solution: As in the previous example, we find the potential difference by integrating the field (that of a point charge) from rb to ra. Then C = Q/V.

24-2 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). Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

24-3 Capacitors in Series and Parallel
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: Figure Capacitors in parallel: Ceq = C1 + C2 + C3 .

24-3 Capacitors in Series and Parallel
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! Figure Capacitors in series: 1/Ceq = 1/C1 + 1/C2 + 1/C3.

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. Solution: First, find the equivalent capacitance of the two capacitors in parallel (2C); then the equivalent of that capacitor in series with the third (2/3 C).

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. Solution: First, find the charge on C1 from Q = CVeq. This charge is split equally between C2 and C3. Now we can calculate the voltages across each capacitor, using V = Q/C.

24-3 Capacitors in Series and Parallel
Example 24-7: Capacitors reconnected. Two capacitors, C1 = 2.2 μF and C2 = 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. Solution: When the capacitors are connected to the voltage source, the charges on them are given by Q = CV: Q1 = 52.8 μC and Q2 = 28.8 μC. After disconnection and reconnection, we know three things: 1. The voltage across each capacitor is the same. 2. The charge on each capacitor is given by Q = CV (using the appropriate values of C and V). 3. The sum of the charges on the two capacitors equals the total charge they started with. This gives three equations (two of the form Q = CV and the charge conservation one), and can be solved for the three unknowns, the two charges and the potential. V = 7.1 V, Q’1 = 16 µC, Q’2 = 8.5 µC.

24-4 Electric Energy Storage
A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor:

24-4 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? Solution: a. U = ½ CV2 = 3.0 J. b. P = U/t = 3000 W.

24-4 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? Solution: Increasing the plate separation decreases the capacitance, but the charge remains the same. Therefore the energy, U = ½ Q2/C, doubles.

24-4 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? Solution: a. U = ½ CV2, and increasing the plate separation decreases the capacitance, so the initial and final energies can be calculated from this. b. The work is equal to the negative of change in energy. c. The energy of the battery is increased by the work done minus the change in potential energy of the capacitor (which is negative).

24-4 Electric Energy Storage
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
Heart defibrillators use electric discharge to “jump-start” the heart, and can save lives. Figure Heart defibrillator.

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

24-5 Dielectrics Dielectric strength is the maximum field a dielectric can experience without breaking down.

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. Figure 24-15a. Dielectric inserted into a capacitor, voltage held constant.

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. Figure 24-15b. Dielectric inserted into a capacitor, charge held constant.

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 m2 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. Solution: a. C = Kε0A/d = 3.0 x 10-8 F. Q = CV = 3.0 x 10-6 C. E = V/d = 25 kV/m. U = ½ CV2 = 1.5 x 10-4 J. B. Now C = 8.8 x 10-9 F, Q = 3.0 x 10-6 C (no change), V = 340 V, E = 85 kV/m, U = 5.1 x 10-4 J. The increase in energy comes from the work it takes to remove the dielectric.

24-6 Molecular Description of Dielectrics
The molecules in a dielectric, when in an external electric field, tend to become oriented in a way that reduces the external field. Figure Molecular view of the effects of a dielectric.

24-6 Molecular Description of Dielectrics
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:

Review Questions

ConcepTest 24.1 Capacitors
Capacitor C1 is connected across a battery of 5 V. An identical capacitor C2 is connected across a battery of 10 V. Which one has more charge? 1) C1 2) C2 3) both have the same charge 4) it depends on other factors Answer: 2

ConcepTest 24.1 Capacitors
Capacitor C1 is connected across a battery of 5 V. An identical capacitor C2 is connected across a battery of 10 V. Which one has more charge? 1) C1 2) C2 3) both have the same charge 4) it depends on other factors Since Q = CV and the two capacitors are identical, the one that is connected to the greater voltage has more charge, which is C2 in this case.

ConcepTest 24.2a Varying Capacitance I
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) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? +Q –Q Answer: 4

ConcepTest 24.2a Varying Capacitance I
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) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? +Q –Q 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.

ConcepTest 24.2b Varying Capacitance II
1) the voltage decreases 2) the voltage increases 3) the charge decreases 4) the charge increases 5) both voltage and charge change 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? +Q –Q Answer: 3

ConcepTest 24.2b Varying Capacitance II
1) the voltage decreases 2) the voltage increases 3) the charge decreases 4) the charge increases 5) both voltage and charge change 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? 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 Follow-up: How do you increase the charge?

ConcepTest 24.2c Varying Capacitance III
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? +Q –Q Answer: 4

ConcepTest 24.2c Varying Capacitance III
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? +Q –Q 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.

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

ConcepTest 24.3a Capacitors I
1) Ceq = 3/2C 2) Ceq = 2/3C 3) Ceq = 3C 4) Ceq = 1/3C 5) Ceq = 1/2C What is the equivalent capacitance, Ceq , of the combination below? o C Ceq 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 24.3b Capacitors II
1) V1 = V2 2) V1 > V2 3) V1 < V2 4) all voltages are zero How does the voltage V1 across the first capacitor (C1) compare to the voltage V2 across the second capacitor (C2)? C1 = 1.0 mF C3 = 1.0 mF C2 = 1.0 mF 10 V Answer: 2

ConcepTest 24.3b Capacitors II
1) V1 = V2 2) V1 > V2 3) V1 < V2 4) all voltages are zero How does the voltage V1 across the first capacitor (C1) compare to the voltage V2 across the second capacitor (C2)? The voltage across C1 is 10 V. The combined capacitors C2 + C3 are parallel to C1. The voltage across C2 + C3 is also 10 V. Since C2 and C3 are in series, their voltages add. Thus the voltage across C2 and C3 each has to be 5 V, which is less than V1. C1 = 1.0 mF C3 = 1.0 mF C2 = 1.0 mF 10 V Follow-up: What is the current in this circuit?

ConcepTest 24.3c Capacitors III
1) Q1 = Q2 2) Q1 > Q2 3) Q1 < Q2 4) all charges are zero How does the charge Q1 on the first capacitor (C1) compare to the charge Q2 on the second capacitor (C2)? C1 = 1.0 mF C3 = 1.0 mF C2 = 1.0 mF 10 V Answer: 2

ConcepTest 24.3c Capacitors III
1) Q1 = Q2 2) Q1 > Q2 3) Q1 < Q2 4) all charges are zero How does the charge Q1 on the first capacitor (C1) compare to the charge Q2 on the second capacitor (C2)? We already know that the voltage across C1 is 10 V and the voltage across both C2 and C3 is 5 V each. Since Q = CV and C is the same for all the capacitors, we have V1 > V2 and therefore Q1 > Q2. C1 = 1.0 mF C3 = 1.0 mF C2 = 1.0 mF 10 V

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