Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces.

22-3 Applications of Gauss’s Law Conceptual Example 22-9: Conductor with charge inside a cavity. Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor? Solution: The field must be zero within the conductor, so the inner surface of the cavity must have an induced charge totaling –q (so that a gaussian surface just around the cavity encloses no charge). The charge +Q resides on the outer surface of the conductor.

22-3 Applications of Gauss’s Law Procedure for Gauss’s law problems: Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). Draw the surface. Use the symmetry to find the direction of E. Evaluate the flux by integrating. Calculate the enclosed charge. Solve for the field.

22-4 Experimental Basis of Gauss’s and Coulomb’s Laws In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero. Figure 22-22. (a) A charged conductor (metal ball) is lowered into an insulated metal can (a good conductor) carrying zero net charge. (b) The charged ball is touched to the can and all of its charge quickly flows to the outer surface of the can. (c) When the ball is then removed, it is found to carry zero net charge.

Summary of Chapter 22 Electric flux: Gauss’s law: Gauss’s law can be used to calculate the field in situations with a high degree of symmetry. Gauss’s law applies in all situations. Follows from Coulomb’s Law

Chapter 23 Electric Potential Chapter 23 opener. We are used to voltage in our lives—a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge. The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

Units of Chapter 23 Electric Potential Energy and Potential Difference Relation between Electric Potential and Electric Field Electric Potential Due to Point Charges Potential Due to Any Charge Distribution Equipotential Surfaces Electric Dipole Potential

Units of Chapter 23 E Determined from V Electrostatic Potential Energy; the Electron Volt

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: Figure 23-1. Work is done by the electric field in moving the positive charge q from position a to position b.

ConcepTest 23.1 Electric Potential Energy 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? 1) proton 2) electron 3) both feel the same acceleration 4) neither – there is no acceleration 5) they feel the same magnitude acceleration but opposite direction Electron electron - + Proton proton

ConcepTest 23.1 Electric Potential Energy 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? 1) proton 2) electron 3) both feel the same acceleration 4) neither – there is no acceleration 5) 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

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.

23-1 Electrostatic Potential Energy and Potential Difference Only changes in potential can be measured, allowing free assignment of V = 0: Sound like gravity?

23-1 Electrostatic Potential Energy and Potential Difference Analogy between gravitational and electrical potential energy: Figure 23-3. (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

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? Figure 23-2. Central part of Fig. 23–1, showing a negative point charge near the negative plate, where its potential energy (PE) is high. Example 23–1. Solution: The electron will move towards the positive plate if released, thereby increasing its kinetic energy. Its potential energy must therefore decrease. However, it is moving to a region of higher potential V; the potential is determined only by the existing charge distribution and not by the point charge. U and V have different signs due to the negative charge.

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:

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 +1 d +1 +2 d 1 3 Both took the same amount of work.

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 +1 d +1 +2 d 1 3 Both took the same amount of work. For case 1: only 1 pair The work needed to assemble a collection of charges is the same as the total PE of those charges: For case 2: there are 3 pairs added over all pairs

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: Figure 23-5. To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

23-2 Relation between Electric Potential and Electric Field The simplest case is a uniform field:

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. Solution: E = V/d = 1000 V/m.

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. Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞: a. V = Q/4πε0r. b. V = Q/4πε0r0. c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

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: Figure 23-8. (a) E versus r, and (b) V versus r, for a uniformly charged solid conducting sphere of radius r0 (the charge distributes itself on the surface); r is the distance from the center of the sphere.

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: Figure 23-9. We integrate Eq. 23–4a along the straight line (shown in black) from point a to point b. The line ab is parallel to a field line.

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: Figure 23-10. Potential V as a function of distance r from a single point charge Q when the charge is positive. Figure 23-11. Potential V as a function of distance r from a single point charge Q when the charge is negative.

23-3 Electric Potential Due to Point Charges Example: Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a proton q = 1.60×10-19 C from a great distance away (take r = ∞) to a point 1.60×10-15 m from another proton? Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

23-3 Electric Potential Due to Point Charges Example: Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a proton q = 1.60×10-19 C from a great distance away (take r = ∞) to a point 1.60×10-15 m from another proton? Wext = ke2/r - ke2/ = (9×109) (1.6×10-19)2/ (1×10-15) = 2.3×10-13 J = 2.3×10-13 J

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:

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.

Back to the two protons Wext = ke2/r - ke2/ = (9×109) (1.6×10-19)2/ (1×10-15) = 2.3×10-13 J = 2.3×10-13 J/(1.6×10-19J/eV) = 1.4×106 eV = 1.4 MeV meV/eV convenient for Atomic/Molecular Physics MeV/GeV convenient for Nuclear Physics GeV/TeV convenient of High Energy Physics A = Angstrom = 10-10 m for AMO Physics & Chem fm = Fermi = femtometer = 10-15 m for NP & HEP o

ConcepTest 23.3 Electric Potential 1) V > 0 2) V = 0 3) V < 0 What is the electric potential at point B? A B

ConcepTest 23.3 Electric Potential 1) V > 0 2) V = 0 3) V < 0 What is the electric potential at point B? A B 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?

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. Solution: The total potential is the sum of the potential due to each charge; potential is a scalar, so there is no direction involved, but we do have to keep track of the signs. a. V = 7.5 x 105 V b. V = 0 (true at any point along the perpendicular bisector)

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

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. Solution: Each point on the ring is the same distance from point P, so the potential is just that of a charge Q a distance (R2 + x2)1/2 from point P.

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. Solution: Consider the disk to be made up of infinitely thin rings, each at a radius R with a thickness dR. Each ring then carries a charge dq = 2QR dR/R02. Integrating to find V then gives the solution in the text.

ConcepTest 23.5 Equipotential Surfaces Which of these configurations gives V = 0 at all points on the x axis? 1) x +2mC -2mC +1mC -1mC 2) 3) 4) all of the above 5) none of the above

ConcepTest 23.5 Equipotential Surfaces Which of these configurations gives V = 0 at all points on the x axis? 1) x +2mC -2mC +1mC -1mC 2) 3) 4) all of the above 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.

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) Figure 23-16. Equipotential lines (the green dashed lines) between two oppositely charged parallel plates. Note that they are perpendicular to the electric field lines (solid red lines).

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). Figure 23-18. Equipotential lines (green, dashed) are always perpendicular to the electric field lines (solid red) shown here for two equal but oppositely charged particles.

23-5 Equipotential Surfaces A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude). Figure 23-19. A topographic map (here, a portion of the Sierra Nevada in California) shows continuous contour lines, each of which is at a fixed height above sea level. Here they are at 80 ft (25 m) intervals. If you walk along one contour line, you neither climb nor descend. If you cross lines, and especially if you climb perpendicular to the lines, you will be changing your gravitational potential (rapidly, if the lines are close together).

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: Figure 23-20. Electric dipole. Calculation of potential V at point P.

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:

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. Solution: a. Just do the derivatives of the result of Example 23-8; the only nonzero component is in the x direction. b. Same as (a); use the result of Example 23-9.

ConcepTest 23.7 Work and Electric Potential 5) all require the same amount of work 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. P 1 2 3 4

ConcepTest 23.7 Work and Electric Potential 5) all require the same amount of work 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. P 1 2 3 4 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.

Summary of Chapter 23 Electric potential is potential energy per unit charge: Potential difference between two points: Potential of a point charge:

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:

Questions?