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Chapter 23 Electric Potential Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference The electrostatic force is conservative – potential energy can be defined. Change in electric potential energy is negative of work done by electric force: Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference Electric potential is defined as potential energy per unit charge: Unit of electric potential: the volt (V): 1 V = 1 J/C. Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference Only changes in potential can be measured, allowing free assignment of V = 0: Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference Conceptual Example 23-1: A negative charge. Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change? Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference Analogy between gravitational and electrical potential energy: Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured: Copyright © 2009 Pearson Education, Inc. 23-1 Electrostatic Potential Energy and Potential Difference Example 23-2: Electron in CRT. Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference Vb – Va = Vba = +5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration? Copyright © 2009 Pearson Education, Inc. 23-2 Relation between Electric Potential and Electric Field The general relationship between a conservative force and potential energy: Substituting the potential difference and the electric field: Copyright © 2009 Pearson Education, Inc. 23-2 Relation between Electric Potential and Electric Field The simplest case is a uniform field: Copyright © 2009 Pearson Education, Inc. 23-2 Relation between Electric Potential and Electric Field Example 23-3: Electric field obtained from voltage. Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is 0.050 m, calculate the magnitude of the electric field in the space between the plates. Copyright © 2009 Pearson Education, Inc. 23-2 Relation between Electric Potential and Electric Field Example 23-4: Charged conducting sphere. Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q. Copyright © 2009 Pearson Education, Inc. 23-2 Relation between Electric Potential and Electric Field The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field: Copyright © 2009 Pearson Education, Inc. 23-2 Relation between Electric Potential and Electric Field Example 23-5: Breakdown voltage. In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize O2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3.0 × 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 Copyright © 2009 Pearson Education, Inc. 23-3 Electric Potential Due to Point Charges To find the electric potential due to a point charge, we integrate the field along a field line: Copyright © 2009 Pearson Education, Inc. 23-3 Electric Potential Due to Point Charges Setting the potential to zero at r = ∞ gives the general form of the potential due to a point charge: Copyright © 2009 Pearson Education, Inc. 23-3 Electric Potential Due to Point Charges Example 23-6: Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a charge q = 3.00 μC from a great distance away (take r = ∞) to a point 0.500 m from a charge Q = 20.0 µC? Copyright © 2009 Pearson Education, Inc. 23-3 Electric Potential Due to Point Charges Example 23-7: Potential above two charges. Calculate the electric potential (a) at point A in the figure due to the two charges shown, and (b) at point B. 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. 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. Copyright © 2009 Pearson Education, Inc. 23-5 Equipotential Surfaces Example 23-10: Point charge equipotential surfaces. For a single point charge with Q = 4.0 × 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to V1 = 10 V, V2 = 20 V, and V3 = 30 V. 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. 23-8 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, 0.529 × 10-10 m, and that they end up an infinite distance apart from each other. Copyright © 2009 Pearson Education, Inc. 23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope A cathode ray tube contains a wire cathode that, when heated, emits electrons. A voltage source causes the electrons to travel to the anode. Copyright © 2009 Pearson Education, Inc. 23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope The electrons can be steered using electric or magnetic fields. Copyright © 2009 Pearson Education, Inc. 23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope Televisions and computer monitors (except for LCD and plasma models) have a large cathode ray tube as their display. Variations in the field steer the electrons on their way to the screen. Copyright © 2009 Pearson Education, Inc. 23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope An oscilloscope displays an electrical signal on a screen, using it to deflect the beam vertically while it sweeps horizontally. Copyright © 2009 Pearson Education, Inc. Review Questions Copyright © 2009 Pearson Education, Inc. ConcepTest 23.1a Electric Potential Energy I A) proton A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which feels the larger electric force? B) electron C) both feel the same force D) neither – there is no force E) they feel the same magnitude force but opposite direction Electron electron - + Proton proton E ConcepTest 23.1a Electric Potential Energy I A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which feels the larger electric force? A) proton B) electron C) both feel the same force D) neither – there is no force E) they feel the same magnitude force but opposite direction Since F = qE and the proton and electron have the same charge in magnitude, they both experience the same force. However, Electron electron - the forces point in opposite directions because the proton and electron are oppositely charged. + Proton proton E ConcepTest 23.1b Electric Potential Energy II A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? A) proton B) electron C) both feel the same acceleration D) neither – there is no acceleration E) they feel the same magnitude acceleration but opposite direction Electron electron - + Proton proton E ConcepTest 23.1b Electric Potential Energy II A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? A) proton B) electron C) both feel the same acceleration D) neither – there is no acceleration E) they feel the same magnitude acceleration but opposite direction Since F = ma and the electron is much less massive than the proton, the electron experiences the larger acceleration. Electron electron - + Proton proton E ConcepTest 23.1c Electric Potential Energy III A) proton A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. When it strikes the opposite plate, which one has more KE? B) electron C) both acquire the same KE D) neither – there is no change of KE E) they both acquire the same KE but with opposite signs Electron electron - + Proton proton E ConcepTest 23.1c Electric Potential Energy III A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. When it strikes the opposite plate, which one has more KE? A) proton B) electron C) both acquire the same KE D) neither – there is no change of KE E) they both acquire the same KE but with opposite signs Since PE = qV and the proton and electron have the same charge in magnitude, they both have the same electric potential energy Electron electron - initially. Because energy is conserved, they both must have the same kinetic energy after they reach the opposite plate. + Proton proton E ConcepTest 23.2 Work and Potential Energy Which group of charges took more work to bring together from a very large initial distance apart? +2 d +1 +1 d +1 Both took the same amount of work. d d +1 ConcepTest 23.2 Work and Potential Energy Which group of charges took more work to bring together from a very large initial distance apart? d +2 +1 d +1 +1 Both took the same amount of work. The work needed to assemble a collection of charges is the same as the total PE of those charges: Q1Q 2 PE k r added over all pairs For case 1: d d +1 only 1 pair (2)(1) 2 PE k k d d For case 2: there are 3 (1)(1) 1 PE 3k 3k d d pairs ConcepTest 23.3a Electric Potential I A) V > 0 What is the electric potential at point A? B) V = 0 C) V < 0 A B ConcepTest 23.3a Electric Potential I A) V > 0 What is the electric potential at point A? B) V = 0 C) V < 0 Since Q2 (which is positive) is closer to point A than Q1 (which is negative) and since the total potential is equal to V1 + V2, the total potential is positive. A B ConcepTest 23.3b Electric Potential II A) V > 0 What is the electric potential at point B? B) V = 0 C) V < 0 A B ConcepTest 23.3b Electric Potential II A) V > 0 What is the electric potential at point B? B) V = 0 C) V < 0 Since Q2 and Q1 are equidistant from point B, and since they have equal and opposite charges, the total potential is zero. Follow-up: What is the potential at the origin of the x y axes? A B ConcepTest 23.4 Hollywood Square Four point charges are arranged at the corners of a square. Find the electric field E and the potential V at the center of the square. A) E = 0 V=0 B) E = 0 V0 C) E 0 V0 D) E 0 V=0 E) E = V regardless of the value -Q +Q -Q +Q ConcepTest 23.4 Hollywood Square Four point charges are arranged at the corners of a square. Find the electric field E and the potential V at the center of the square. A) E = 0 V=0 B) E = 0 V0 C) E 0 V0 D) E 0 V=0 E) E = V regardless of the value The potential is zero: the scalar contributions from the two positive charges cancel the two minus charges. However, the contributions from the electric field add up as vectors, and they do not cancel (so it is non-zero). Follow-up: What is the direction of the electric field at the center? -Q +Q -Q +Q ConcepTest 23.5a Equipotential Surfaces I 1 E) all of them At which point does V = 0? 2 +Q 3 4 –Q ConcepTest 23.5a Equipotential Surfaces I 1 E) all of them At which point does V = 0? 2 +Q 3 –Q 4 All of the points are equidistant from both charges. Since the charges are equal and opposite, their contributions to the potential cancel out everywhere along the mid-plane between the charges. Follow-up: What is the direction of the electric field at all 4 points? ConcepTest 23.5b Equipotential Surfaces II Which of these configurations gives V = 0 at all points on the x axis? +2mC +1mC +2mC +1mC x -1mC -2mC +2mC -2mC x -2mC -1mC A) B) D) all of the above x +1mC -1mC C) E) none of the above ConcepTest 23.5b Equipotential Surfaces II Which of these configurations gives V = 0 at all points on the x axis? +2mC +1mC +2mC +1mC x -1mC -2mC +2mC -2mC x -2mC -1mC A) x B) D) all of the above +1mC -1mC C) E) none of the above Only in case (A), 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. ConcepTest 23.5c Equipotential Surfaces III Which of these configurations gives V = 0 at all points on the y axis? +2mC +1mC +2mC +1mC x -1mC -2mC +2mC -2mC x -2mC -1mC A) B) D) all of the above x +1mC -1mC C) E) none of the above ConcepTest 23.5c Equipotential Surfaces III Which of these configurations gives V = 0 at all points on the y axis? +2mC +1mC +2mC +1mC x -1mC -2mC +2mC -2mC x -2mC -1mC A) x B) D) all of the above +1mC -1mC C) E) none of the above Only in case (C), where opposite charges lie directly across the y axis from each other, do the potentials from the two charges above the y axis cancel the ones below the y axis. Follow-up: Where is V = 0 for configuration #2? ConcepTest 23.6 Equipotential of Point Charge A) A and C Which two points have the same potential? B) B and E C) B and D D) C and E E) no pair A C B E Q D ConcepTest 23.6 Equipotential of Point Charge A) A and C Which two points have the same potential? B) B and E C) B and D D) C and E E) no pair Since the potential of a point charge is: A Q V k r only points that are at the same distance from charge Q are at the same potential. This is true for points C and E. C B They lie on an equipotential surface. Follow-up: Which point has the smallest potential? E Q D ConcepTest 23.7a Work and Electric Potential I A) P 1 Which requires the most work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. B) P 2 C) P 3 D) P 4 E) all require the same amount of work 3 2 1 P E 4 ConcepTest 23.7a Work and Electric Potential I Which requires the most work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. For path #1, you have to push the positive charge against the E field, which is hard to do. By contrast, path #4 is the easiest, since the field does all the work. A) P 1 B) P 2 C) P 3 D) P 4 E) all require the same amount of work 3 2 1 P E 4 ConcepTest 23.7b Work and Electric Potential II A) P 1 Which requires zero work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. B) P 2 C) P 3 D) P 4 E) all require the same amount of work 3 2 1 P E 4 ConcepTest 23.7b Work and Electric Potential II Which requires zero work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. A) P 1 B) P 2 C) P 3 D) P 4 E) all require the same amount of work For path #3, you are moving in a direction perpendicular to the field lines. This means you are moving along an equipotential, which requires no work (by definition). Follow-up: Which path requires the least work? 3 2 1 P E 4 Ch. 23 Assignments • Group problems #’s 4, 10, 12, 30, 34, 36, 42 • Homework problems #’s 1, 9, 11, 13, 15, 27, 29, 33, 49, 51, 57 • Finish reading Ch. 23 and begin looking over Ch. 24. Copyright © 2009 Pearson Education, Inc.