Physics 1302W.500 Lecture 9 Introductory Physics for Scientists and Engineering II

Physics 1302W.500 Lecture 9 Introductory Physics for Scientists and Engineering II In today s lecture, we will finish our discussion of Gauss law. Slide 25-1

Applying Gauss s law Procedure: Calculating the electric field using Gauss s Law Gauss s law allows you to calculate the electric field for charge distributions that exhibit spherical, cylindrical, or planar symmetry without having to carry out any major integrations. Slide 25-2

Applying Gauss s law Procedure: Calculating the electric field using Gauss s Law 1. Identify the symmetry of the charge distribution. This symmetry determines the general pattern of the electric field and the type of Gaussian surface you should use. Slide 25-3

Applying Gauss s law Procedure: Calculating the electric field using Gauss s Law 2. Sketch the charge distribution and the electric field by drawing a number of field lines, remembering that the field lines start on positively charged objects and end on negatively charged ones. A two-dimensional drawing should suffice. Slide 25-4

Applying Gauss s law Procedure: Calculating the electric field using Gauss s Law 3. Draw a Gaussian surface such that the electric field is either parallel or perpendicular (and constant) to each face of the surface. If the charge distribution divides space into distinct regions, draw a Gaussian surface in each region where you wish to calculate the electric field. Slide 25-5

Applying Gauss s law Procedure: Calculating the electric field using Gauss s Law 4. For each Gaussian surface determine the charge q enc enclosed by the surface. 5. For each Gaussian surface calculate the electric flux Φ E through the surface. Express the electric flux in terms of the unknown electric field E. 6. Use Gauss s law to relate q enc and Φ E and solve for E. Slide 25-6

Applying Gauss s law Procedure: Calculating the electric field using Gauss s Law You can use the same general approach to determine the charge carried by a charge distribution given the electric field of a charge distribution exhibiting one of the three symmetries. Follow the same procedure, but in steps 4 6, express q enc in terms of the unknown charge q and solve for q. Slide 25-7

Applying Gauss s law Electric field of infinite charged sheet (Ex. 24.8) What is the electric field a distance d from a thin, infinite nonconducting sheet with a uniform positive surface charge density σ? q enc = σa Φ E = 2EA 2EA = σ A 0 E = σ 2 0 (constant) Same result (E = 2πkσ) obtained by direct integration in Example 23.6 for a uniformly charged disk (in limit very close to the disk) Slide 25-8

Problem Consider a system that consists of an infinitely long non-conducting charged rod centered in a thick, conducting cylindrical shell of inner radius R and outer radius 2R. The rod has a uniform linear charge density. For each finite length l of the system there is a charge -q on the shell s inner surface. For that length l, determine the charge (a) on the rod and (b) on the shell s outer surface. (c) In length l, what is the charge on the shell? (d) what is the ratio of the surface charge density on the shell s inner surface to that on its outer surface? Slide 25-9

Solution Slide 25-10

Problem Can the radial electric flux dependence shown in the figure be produced by a system of charged particles? If not, explain why. If so, describe the system that contains the minimum number of charged particles. Slide 25-11

Clicker Question A positively charged non-conducting infinite sheet with uniform surface charge density σ lies in the xy plane. A circular region of radius R has been cut out of the sheet. What is the electric field magnitude at the center of the circle? σ σ 1. E = 2. E = 3. E = 0 2 0 2 0 kσ kσ 4. E = 5. E = 2 2 Slide 25-12

Clicker Question 1. 2. 3. 4. 5. Slide 25-13

Physics 1302W.500 Lecture 10 Introductory Physics for Scientists and Engineering II In today s lecture, we will begin to develop tools for analyzing electrostatic properties of charge distributions using the concepts of work and energy. Slide 25-14

A few comments regarding Quiz 1 The Group Quiz on Thursday will consist of one problem worth 25 points The Quiz on Friday will cover Chapters 22, 23 and 24 (homework 1, 2 and 3) and be worth 75 points All Friday quizzes will be in Willey 125 We will provide an equation sheet (posted on the course site) On Friday, there will be will be 2 long problems In addition, there will be 5 multiple-choice questions No textbooks or notes are allowed A non-programmable calculator is allowed (you are not allowed to store formulae or specific problems on your calculator) Remember to write your name, your ID number, your TA s name, and the problem number on each sheet Slide 25-15

Electric Wind Cotrell Precipitator Description: A pointed electrode and plane electrode are connected to a Wimshurst machine. A lit candle is placed between the electrodes. When the Wimshurst is cranked, the flame is blown away from the point and towards the plate. The candle flame is rich in positive ions, and the positive terminal of the electrostatic machine must be used for the pointed electrode. There is an electrostatic force of repulsion acting on the flame, pushing it away from the sharp point. Description: A glass tube has two electrodes at each end. These electrodes are connected to a Wimshurst machine. A lit piece of smoke paper is placed in the tube. When the tube fills with smoke, the Wimshurst is cranked. The smoke dissipates. [ An electrostatic precipitator (ESP) is a filtration device that removes fine particles, like dust and smoke, from a flowing gas using the force of an induced electrostatic charge minimally impeding the flow of gases through the unit. (wikipedia)] Slide 25-16

Electric potential energy Consider a system of electric charges. When the charges move in the vicinity of each other, the force of interaction causes the system energy to change. This energy change is analogous to an inertial object in free fall near Earth s surface. The electrostatic interaction is associated with stored energy called electric potential energy. Slide 25-17

Electric potential energy The figure shows the situation for both the attractive and repulsive cases: The movement of an electric charge near a sheet of charge causes the electrostatic potential energy of the system to change. Slide 25-18

Electric potential energy The total energy of the system is conserved. Notice that the charges move on their own to a more stable situation. The change in electrostatic potential energy is associated with a change in configuration of the charged objects. Slide 25-19

Clicker Question You release three balls simultaneously from the same height above the floor. The balls (A, B, C) all carry the same quantity of surplus positive charge, but they have different masses: A) 1 kg, B) 2 kg, C) 3 kg. In addition to the gravitational field due to earth, there is a uniform electric field directed downward. Assume that the balls have negligible influence on each other and that air resistance can be ignored. Which ball has the greatest speed when it reaches the floor? 1. A 2. B 3. C 4. They ll all have the same speed. Slide 25-20

Clicker Question How does the electric potential energy of a dipole placed near a stationary charged object depend on the orientation of the dipole relative to the charged object? 1. It is constant, independent of the orientation of the dipole with the direction of the electric field created by the charge. 2. It is largest if the dipole moment points along the electric field direction. 3. It is smallest if the dipole moment points along the electric field direction. Slide 25-21

Dipole in uniform electric field As the dipole in the figure continues to rotate, it reaches the point where its axis is aligned with the electric field of the large object. (a) What happens to the electric potential energy as the dipole moves beyond that point? (b) Describe the motion of the dipole beyond that point. (c) How would the motion of the dipole change if it were released with a different orientation from the one shown in the figure? Slide 25-22

Electrostatic work Work done by an electrostatic field is referred to as electrostatic work. Electrostatic work obeys this general principle: The electrostatic work done on a charged particle as it moves from one point to another is independent of the path taken by the particle and depends on only the positions of the endpoints of the path of motion. Slide 25-23

Clicker Question When a charged particle moves in an electrostatic field, on which aspects of its path does the electrostatic work done on the particle depend? 1. The starting point, the ending point, and the path 2. The path only 3. The starting point only 4. The ending point only 5. Only the starting and ending points Slide 25-24

Calculating work and energy in electrostatics The figure shows the geometry we will use to quantify the electrostatic work. Consider moving an electric charge q 2 in the electric field of charge q 1 from points A to B along the path shown. The work done by particle 1 on particle 2 is: W 12 (A B) = Paths AB and ACB are equivalent. For CB: F 12 E d = k q 1 q 2 r 2 B A dr F 12 E d Slide 25-25

Calculating work and energy in electrostatics Consequently, inserting Coulomb s law into the definition of work yields: W 12 = q q 1 2 1 1 4π 0 r 12,i r 12,f Note: i and f stand for initial and final Notice that the expression is independent of the path taken. It does not require particle 1 to be at the origin. It is the same if we move particle 1 instead. Slide 25-26

Calculating work and energy in electrostatics Now, the change in electrostatic potential energy, ΔU E, is the negative of the work done on particle 2 by the system comprised of particle 1: ΔU E = W 12 = q q 1 2 1 1 4π 0 r 12,f If we establish as a reference the configuration where the force between the interacting charges is zero that is to say, for infinite separation then U E = 0 for r 12 =. Thus, the electric potential energy for two particles carrying charges q 1 and q 2 and separated by distance r 12 is given by: U E = q 1 q 2 4π 0 1 r 12 r 12,i (U E zero at infinite separation) Slide 25-27

Calculating work and energy in electrostatics Notice that electric potential energy of a pair of charges is positive if they have the same sign and negative for opposite signs. This pattern is consistent with Coulomb s law: opposite sign charges attract and like sign charges repel. Slide 25-28

Physics 1302W.500 Lecture 11 Introductory Physics for Scientists and Engineering II In today s lecture, we will continue to develop tools for analyzing electrostatic properties of charge distributions using the concepts of work and energy. Slide 25-29

Calculating work and energy in electrostatics W 12 (A B) = B A F 12 E d W 12 = q q 1 2 1 1 4π 0 r 12,i r 12,f Slide 25-31

Calculating work and energy in electrostatics The change in electrostatic potential energy, ΔU E, is the negative of the work done on particle 2 by the system comprised of particle 1: ΔU E = W 12 = q q 1 2 1 1 4π 0 r 12,f If we establish as a reference the configuration where the force between the interacting charges is zero that is to say, for infinite separation then U E = 0 for r 12 =. Thus, the electric potential energy for two particles carrying charges q 1 and q 2 and separated by distance r 12 is given by: U E = q 1 q 2 4π 0 1 r 12 r 12,i (U E zero at infinite separation) Slide 25-32

Calculating work and energy in electrostatics The superposition principle allows us to generalize this analysis to systems of many electric charges. For a system of three charges, q 1, q 2, and q 3, we have: W = q 1 q 2 4π 0 1 r 12 q 1 q 3 4π 0 1 r 13 q 2 q 3 4π 0 1 r 23 U E = q 1 q 2 4π 0 1 r 12 + q 1 q 3 4π 0 1 r 13 + q 2 q 3 4π 0 1 r 23 (U E zero at infinite separation) Slide 25-33

Clicker Question Two test charges are brought separately into the vicinity of a charge +Q. First, test charge +q is brought to a point a distance r from +Q. Then this charge is removed and test charge q is brought to the same point. The electrostatic potential energy of which test charge is greater? 1. +q 2. q 3. It is the same for both. Slide 25-34

Electrostatic work The force field representation of the electrostatic interaction allows for an important insight when considering moving an electric charge in the electric field produced by other charges: The electrostatic work done on a charged particle is proportional to the charge carried by that particle Slide 25-35

Electrostatic work This observation allows us to define a new quantity called the electrostatic potential difference (or simply potential difference): The potential difference between point A and point B in an electrostatic field is equal to the negative of the electrostatic work per unit charge done on a charged particle as it moves from A to B. V AB V B V A W (A B) q q Slide 25-36

Potential difference V AB V B V A W (A B) q q = q 1 1 1 4π 0 r B r A Potential difference is a scalar (field), and the SI units of potential difference are joules per coulomb (J/C). In honor of Alessandro Volta (1745 1827), who developed the first battery, this derived unit is given the name volt (V); 1 V = 1 J/C. Minus sign: we consider work done on the particle, not by the particle Slide 25-37

Potential difference Once we know the potential difference V AB between A and B, we can obtain the electrostatic work done on any object carrying a charge q as it is moved along any path from A to B: W q (A B) = qv AB Remember: the subscript AB means from A to B. Because B is the final position, we write V AB V B V A Note the different units for energy and potential Slide 25-38

Potential difference Just as with potential energy, only potential differences are physically relevant. If we choose a reference point, however, we can determine the value for the potential at any other point. We usually chose infinity as the reference point for the electric potential energy for charged particles because they do not interact at infinite distance U E ( ) = 0. When we deal with electrical circuits it is customary to assign zero potential to Earth (ground) because Earth is a good and very large conducting object through which the motion of charge carriers requires negligible energy. Slide 25-39

Potential difference To obtain an explicit expression for the potential difference between two points A and B in the electric field of particle 1 carrying charge q 1, we start with the expression for the electrostatic work done by particle 1 on a particle 2 carrying charge q 2 as it is moved from A to B. All we need to do is add a minus sign and divide by q 2 : Slide 25-40

Potential difference For charged particles, the choice of ground as a zero for the potential is not very meaningful. However, if we set the zero for potential at infinity and let r A be at infinity, we obtain for the potential at a distance r = r B from a single charged particle: V (r) = 1 q 1 4π 0 r (potential zero at infinity) Slide 25-41

Clicker Question A negative charge, if free, will tend to move 1. From high potential to low potential. 2. From low potential to high potential. 3. Always toward infinity. 4. Always from infinity. 5. In the direction of the electric field. Slide 25-42

Equipotentials The non-dissipative nature of the electric force results in another important feature of electrostatic work called equipotential lines (or just equipotentials): An equipotential line is a line along which the value of the electrostatic potential does not change. The electrostatic work done on a charged particle as it moves along an equipotential line is zero. Analogy: Contour lines on a topological map. Since the elevation along a contour line does not change, no gravitational work is done on you if you walk along a contour line. Slide 25-43

Equipotentials Equipotential lines are actually a cross-section of more general equipotential surfaces. Slide 25-44

Equipotentials Several important observations concerning equipotential surfaces are: 1. The equipotential surfaces of a stationary charge distribution are everywhere perpendicular to the corresponding electric field lines. 2. An electrostatic field is directed from points of higher potential to points of lower potential. 3. In an electrostatic field, positively charged particles tend to move toward regions of lower potential, whereas negatively charged particles tend to move toward regions of higher potential. Slide 25-45

Clicker Question When you hold a positively charged rod above a metallic sphere without touching it, a surplus of negative charge carriers accumulates at the top of the sphere, leaving a surplus of positive charge carriers at the bottom. The potential difference between the top and the bottom of the sphere is (see checkpoint 25.8): 1. positive 2. negative 3. zero Slide 25-46

Clicker Question In an electric field, how does the electrostatic potential vary along a given field line? 1. It is constant. 2. It increases in the direction of the electric field. 3. It decreases in the direction of the electric field. 4. It depends on the sign of the charges producing the electric field. 5. There is not enough information to determine this. Slide 25-47

Surface Charge Density Charged Ovoid Description: The pairs of spheres, each pair larger than the next, are connected to a Wimshurst machine. A spacer is used to ensure that the spheres have the same separation. The Wimshurst is cranked and the spheres are observed to see which pair sparks first. Description: A Zeppelin-shaped conductor is mounted on an insulated stand. The conductor is charged with a PVC rod. Two small conductors, one pointed and one round (to fit the ends of the large conductor), are available to transfer charge to electrosopes. The conductors are placed on the ovoid simultaneously and then on to the electroscopes. The amount of charge is observed. Note: Charge densities and electric fields are larger for surfaces with larger curvature. Slide 25-48