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Chapter 24 Capacitance, Dielectrics, Electric Energy Storage Copyright © 2009 Pearson Education, Inc. Units of Chapter 24 • Capacitors • Determination of Capacitance • Capacitors in Series and Parallel • Electric Energy Storage • Dielectrics • Molecular Description of Dielectrics Copyright © 2009 Pearson Education, Inc. 24-1 Capacitors A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge. Copyright © 2009 Pearson Education, Inc. 24-1 Capacitors Parallel-plate capacitor connected to battery. (b) is a circuit diagram. Copyright © 2009 Pearson Education, Inc. 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. Copyright © 2009 Pearson Education, Inc. 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: Copyright © 2009 Pearson Education, Inc. 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. Copyright © 2009 Pearson Education, Inc. 24-2 Determination of Capacitance Capacitors are now made with capacitances of 1 farad or more, but they are not parallelplate 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. Copyright © 2009 Pearson Education, Inc. 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. Copyright © 2009 Pearson Education, Inc. 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. Copyright © 2009 Pearson Education, Inc. 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). Copyright © 2009 Pearson Education, Inc. 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: Copyright © 2009 Pearson Education, Inc. 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! 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. 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. Copyright © 2009 Pearson Education, Inc. 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. Copyright © 2009 Pearson Education, Inc. 24-4 Electric Energy Storage A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor: Copyright © 2009 Pearson Education, Inc. 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? Copyright © 2009 Pearson Education, Inc. 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? Copyright © 2009 Pearson Education, Inc. 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? Copyright © 2009 Pearson Education, Inc. 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! Copyright © 2009 Pearson Education, Inc. 24-4 Electric Energy Storage Heart defibrillators use electric discharge to “jumpstart” the heart, and can save lives. Copyright © 2009 Pearson Education, Inc. 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: Copyright © 2009 Pearson Education, Inc. 24-5 Dielectrics Dielectric strength is the maximum field a dielectric can experience without breaking down. 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. 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. 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 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. Copyright © 2009 Pearson Education, Inc. 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. Copyright © 2009 Pearson Education, Inc. 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: Copyright © 2009 Pearson Education, Inc. Review Questions Copyright © 2009 Pearson Education, Inc. ConcepTest 24.1 Capacitors Capacitor C1 is connected across 1) C1 a battery of 5 V. An identical 2) C2 capacitor C2 is connected across a battery of 10 V. Which one has more charge? 3) both have the same charge 4) it depends on other factors ConcepTest 24.1 Capacitors Capacitor C1 is connected across 1) C1 a battery of 5 V. An identical 2) C2 capacitor C2 is connected across a battery of 10 V. Which one has more charge? 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 What must be done to 1) increase the area of the plates a capacitor in order to 2) decrease separation between the plates increase the amount of 3) decrease the area of the plates charge it can hold (for a constant voltage)? 4) either (1) or (2) 5) either (2) or (3) +Q – Q ConcepTest 24.2a Varying Capacitance I What must be done to 1) increase the area of the plates a capacitor in order to 2) decrease separation between the plates increase the amount of 3) decrease the area of the plates charge it can hold (for a constant voltage)? 4) either (1) or (2) 5) either (2) or (3) +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 C 0 A , that can be d done by either increasing A or decreasing d. ConcepTest 24.2b Varying Capacitance II A parallel-plate capacitor 1) the voltage decreases initially has a voltage of 400 V 2) the voltage increases and stays connected to the 3) the charge decreases battery. If the plate spacing is now doubled, what happens? 4) the charge increases 5) both voltage and charge change +Q – Q ConcepTest 24.2b Varying Capacitance II A parallel-plate capacitor 1) the voltage decreases initially has a voltage of 400 V 2) the voltage increases and stays connected to the 3) the charge decreases battery. If the plate spacing is now doubled, what happens? 4) the charge increases 5) both voltage and charge change Since the battery stays connected, the voltage must remain constant! Since C 0 A , when the spacing d is d doubled, the capacitance C is halved. And since Q = CV, that means the charge must decrease. Follow-up: How do you increase the charge? +Q – Q 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 1) 100 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V 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? Once the battery is disconnected, Q has to remain constant, since no charge can flow either to or from the battery. Since C 0 A , when the spacing d is doubled, the d capacitance C is halved. And since Q = CV, that means the voltage must double. 1) 100 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V +Q – Q ConcepTest 24.3a Capacitors I 1) Ceq = 3/2C What is the equivalent capacitance, 2) Ceq = 2/3C Ceq , of the combination below? 3) Ceq = 3C 4) Ceq = 1/3C 5) Ceq = 1/2C o Ceq o C C C ConcepTest 24.3a Capacitors I 1) Ceq = 3/2C What is the equivalent capacitance, 2) Ceq = 2/3C Ceq , of the combination below? 3) Ceq = 3C 4) Ceq = 1/3C 5) Ceq = 1/2C The 2 equal capacitors in series add o up as inverses, giving 1/2C. These are parallel to the first one, which Ceq add up directly. Thus, the total equivalent capacitance is 3/2C. o C C C ConcepTest 24.3b Capacitors II How does the voltage V1 across 1) V1 = V2 the first capacitor (C1) compare 2) V1 > V2 to the voltage V2 across the 3) V1 < V2 second capacitor (C2)? 4) all voltages are zero C2 = 1.0 mF 10 V C1 = 1.0 mF C3 = 1.0 m ConcepTest 24.3b Capacitors II How does the voltage V1 across 1) V1 = V2 the first capacitor (C1) compare 2) V1 > V2 to the voltage V2 across the 3) V1 < V2 second capacitor (C2)? 4) all voltages are zero 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. C2 = 1.0 mF 10 V C1 = 1.0 mF C3 = 1.0 m Follow-up: What is the current in this circuit? ConcepTest 24.3c How does the charge Q1 on the first Capacitors III 1) Q1 = Q2 2) Q1 > Q2 capacitor (C1) compare to the charge Q2 on the second capacitor 3) Q1 < Q2 (C2)? 4) all charges are zero C2 = 1.0 mF 10 V C1 = 1.0 mF C3 = 1.0 mF ConcepTest 24.3c How does the charge Q1 on the first Capacitors III 1) Q1 = Q2 2) Q1 > Q2 capacitor (C1) compare to the charge Q2 on the second capacitor 3) Q1 < Q2 (C2)? 4) all charges are zero We already know that the C2 = 1.0 mF 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. 10 V C1 = 1.0 mF C3 = 1.0 mF 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 Copyright © 2009 Pearson Education, Inc.