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Chapter 23 Electric Potential Copyright © 2009 Pearson Education, Inc. 23-4 Potential Due to Any Charge Distribution The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous): or Copyright © 2009 Pearson Education, Inc. 23-4 Potential Due to Any Charge Distribution Example 23-8: Potential due to a ring of charge. A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center. Copyright © 2009 Pearson Education, Inc. 23-4 Potential Due to Any Charge Distribution Example 23-9: Potential due to a charged disk. A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center. Copyright © 2009 Pearson Education, Inc. ConcepTest 23.5 Equipotential Surfaces Which of these configurations gives V = 0 at +2mC +2mC +2mC +1mC -2mC +1mC all points on the x axis? x -1mC -2mC x 1) +1mC -1mC 2) 4) all of the above Copyright © 2009 Pearson Education, Inc. -2mC -1mC x 3) 5) none of the above ConcepTest 23.5 Equipotential Surfaces Which of these configurations gives V = 0 at +2mC +2mC +2mC +1mC -2mC +1mC all points on the x axis? x -1mC -2mC x -2mC -1mC 1) x 2) 4) all of the above +1mC -1mC 3) 5) none of the above Only in case (1), where opposite charges lie directly across the x axis from each other, do the potentials from the two charges above the x axis cancel the ones below the x axis. Copyright © 2009 Pearson Education, Inc. 23-5 Equipotential Surfaces An equipotential is a line or surface over which the potential is constant. Electric field lines are perpendicular to equipotentials. The surface of a conductor is an equipotential (E// =0 → ∂V/∂s// = 0) Copyright © 2009 Pearson Education, Inc. 23-5 Equipotential Surfaces Equipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges). Copyright © 2009 Pearson Education, Inc. 23-5 Equipotential Surfaces A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude). Copyright © 2009 Pearson Education, Inc. 23-6 Electric Dipole Potential The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation: Copyright © 2009 Pearson Education, Inc. 23-7 E Determined from V If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating: This is a vector differential equation; here it is in component form: Copyright © 2009 Pearson Education, Inc. 23-7 E Determined from V Example 23-11: E for ring and disk. Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk. Copyright © 2009 Pearson Education, Inc. 23-8 Electrostatic Potential Energy; the Electron Volt The potential energy of a charge in an electric potential is U = qV. To find the electric potential energy of two charges, imagine bringing each in from infinitely far away. The first one takes no work, as there is no field. To bring in the second one, we must do work due to the field of the first one; this means the potential energy of the pair is: Copyright © 2009 Pearson Education, Inc. 23-8 Electrostatic Potential Energy; the Electron Volt One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt: 1 eV = 1.6 × 10-19 J. The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles. Copyright © 2009 Pearson Education, Inc. Summary of Chapter 23 • Electric potential is potential energy per unit charge: • Potential difference between two points: • Potential of a point charge: Copyright © 2009 Pearson Education, Inc. Summary of Chapter 23 • Equipotential: line or surface along which potential is the same. • Electric dipole potential is proportional to 1/r2. • To find the field from the potential: Copyright © 2009 Pearson Education, Inc. Chapter 24 Capacitance, Dielectrics, Electric Energy Storage 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. It is defined for ANY pair of separated conductors at different potentials. Unit of capacitance: the farad (F): 1 F = 1 C/V. 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. 24-2 Determination of Capacitance For a parallel-plate capacitor as shown, the field between the plates is E = σ/ε0 = Q/ε0A. Integrating along a path between the plates gives the potential difference: Vba = Ed = (Q/ε0A)d. This gives the capacitance: Copyright © 2009 Pearson Education, Inc. 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. 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 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. 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 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 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. Summary of Chapter 24 • Capacitor: nontouching conductors carrying equal and opposite charge. • Capacitance: • Capacitance of a parallel-plate capacitor: Copyright © 2009 Pearson Education, Inc. Summary of Chapter 24 • Capacitors in parallel: • Capacitors in series: Copyright © 2009 Pearson Education, Inc. Summary of Chapter 24 • 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: Copyright © 2009 Pearson Education, Inc. Homework Assignment # 4 Chapter 23 – 36 Chapter 24 – 6, 16, 28, 46, 60, 82 Tentative HW # 5: Chapter 25 – 10, 20, 34, 40, 54, 58 Copyright © 2009 Pearson Education, Inc.