Copyright © 2009 Pearson Education, Inc. Chapter 23 Electric Potential

Copyright © 2009 Pearson Education, Inc 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 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 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. One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt: 1 eV = 1.6 × J. The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles Electrostatic Potential Energy; the Electron Volt

Copyright © 2009 Pearson Education, Inc Electrostatic Potential Energy; the Electron Volt Example 23-12: Disassembling a hydrogen atom. Calculate the work needed to “disassemble” a hydrogen atom. Assume that the proton and electron are initially separated by a distance equal to the “average” radius of the hydrogen atom in its ground state, × m, and that they end up an infinite distance apart from each other.

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. Equipotential: line or surface along which potential is the same. Electric dipole potential is proportional to 1/r 2. To find the field from the potential: Summary of Chapter 23

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

Copyright © 2009 Pearson Education, Inc. A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge Capacitors

Copyright © 2009 Pearson Education, Inc. Parallel-plate capacitor connected to battery. (b) is a circuit diagram Capacitors

Copyright © 2009 Pearson Education, Inc. 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 Capacitors

Copyright © 2009 Pearson Education, Inc Determination of Capacitance For a parallel-plate capacitor as shown, the field between the plates is E = σ/ε 0 = Q/ ε 0 A. Integrating along a path between the plates gives the potential difference: V ba = Ed = (Q/ ε 0 A)d. This gives the capacitance:

Copyright © 2009 Pearson Education, Inc 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 Determination of Capacitance Example 24-2: Cylindrical capacitor. A cylindrical capacitor consists of a cylinder (or wire) of radius R b surrounded by a coaxial cylindrical shell of inner radius R a. 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.

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

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

Copyright © 2009 Pearson Education, Inc Determination of Capacitance Example 24-3: Spherical capacitor. A spherical capacitor consists of two thin concentric spherical conducting shells of radius r a and r b 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. 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

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

Copyright © 2009 Pearson Education, Inc 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 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 Capacitors in Series and Parallel Example 24-7: Capacitors reconnected. Two capacitors, C 1 = 2.2 μF and C 2 = 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.

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?

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?

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 )?

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?

ConcepTest 24.3cCapacitors III ConcepTest 24.3c Capacitors III C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )?

ConcepTest 24.3cCapacitors III ConcepTest 24.3c Capacitors III C 1 = 1.0  F C 3 = 1.0  F C 2 = 1.0  F 10 V Q = CVsame have V 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, we have V 1 > V 2 and therefore Q 1 > Q 2. 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )?

Copyright © 2009 Pearson Education, Inc. A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor: 24-4 Electric Energy Storage

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

Copyright © 2009 Pearson Education, Inc Electric Energy Storage National Ignition Facility (NIF) Lawrence Livermore National Laboratory

Copyright © 2009 Pearson Education, Inc. NIF Laser system driven by 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?

Copyright © 2009 Pearson Education, Inc. NIF Laser system driven by 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 ~ W = 1000 GW! cf GW for power plant

Copyright © 2009 Pearson Education, Inc 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. 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

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:

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

Copyright © 2009 Pearson Education, Inc 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 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 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.

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 Molecular Description of Dielectrics

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

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

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

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