Physics 1302W.400 Lecture 2 Introductory Physics for Scientists and Engineering II

Transcription

1 Physics 1302W.400 Lecture 2 Introductory Physics for Scientists and Engineering II In today s lecture, we will start to cover Chapter 23, with the goal to develop the force field concept for the long-range nature of the electrostatic interaction. Slide 23-1

2 Miscellaneous Class web site (announcements, lecture slides, etc.) your responsibility to be up to date Syllabus your responsibility to know it Homework typically assigned/due on Wednesdays Mastering Physics code: PH S2018 Make-up lecture this Friday, January 26 Update: Office Hours on Tu 2:30pm and We 11:15am or by appointment (Room: PAN 216) policy (see also syllabus) Slide 23-2

3 Course Philosophy Read first (textbook) Listen, think & discuss (lectures, check points, self quizzes) Play (labs) Work out (homework) Work out more (discussion; practice problems) Ask for help, if needed (office hours) Slide 23-3

4 Tips Tip 1: Read ahead Tip 2: Do self quizzes; solve example problems Tip 3: Form study groups (3-4 students) Tip 4: Maintain a homework notebook Tip 5: Ask questions! J Slide 23-4

5 iclicker 2 Make sure to bring your clicker to class As of tomorrow, your clicker answers will count For each question, you will receive one point for participation You will receive one extra point for selecting the correct answer You are strongly encouraged to discuss the clicker questions with those seated near you! You can change your answer until the time is up Slide 23-5

6 Review: Electric interactions An electric interaction is a long-range interaction between objects that carry a surplus electrical charge. The electric force is sometimes called the electrostatic force because interactions between charged objects become more complicated for objects that are not at rest. There are only two types of charge: positive charge and negative charge. Objects that carry like charges repel each other; objects that carry opposite charges attract each other. Any microscopic object that carries electrical charge, such as an electron or ion, is called a charge carrier. Slide 23-6

7 Review: Electric interactions The SI unit of charge is the coulomb (C), which is the magnitude of the charge carried by about electrons. It is also equal to the amount of charge transported in 1 s by a steady current of 1 ampere. Coulomb s law:! F 12 E = k q 1 q 2 r 12 2 In SI units, the proportionality constant k is: ˆr 12 k = N m 2 /C 2 The charge on any object is quantized it comes in only whole-number multiples of e: q = ne, n = 0, ± 1, ± 2, ± 3,. Slide 23-7

8 Review: Movement of charge carriers Charge carriers can flow through an electrical conductor but cannot readily flow through an electrical insulator. Charge carriers can be transferred from one nonconductor to another only by bringing the nonconductors into contact with each other, and the carriers remain near the spot at which they were deposited. We ground an object by connecting it electrically to Earth. Grounding permits the exchange of charge carriers with Earth, a huge reservoir of charge carriers. A charged, conducting object that is grounded will retain no surplus of either type of charge, assuming no other nearby electrical influences. Slide 23-8

9 Review: Movement of charge carriers An ion is an atom or molecule that has lost or gained one or more electrons. According to the principle of conservation of charge, electrical charge can be created or destroyed only in identical positive-negative pairs, with the result that the charge of a closed system always remains constant. Polarization, which is the separation of charge carriers in an object, allows neutral objects to interact electrically. Charging by induction is a method of using a charged object to charge a neutral object without the two objects touching each other. Slide 23-9

10 Electrostatic Ping Pong Description: A Wimshurst machine is connected to parallel aluminum plates. A ping pong ball coated with conducting paint is suspended between the plates. The plates are charged up (with opposite charge). The ball is initially attracted to one plate by induction, but is repelled when it comes in contact. When the charge on the ball switches sign, it is attracted to the other plate. The ball will continue to oscillate for some time after the production of static charge is stopped. The Wimshurst machine is an electrostatic generator, a machine for generating high voltages developed between 1880 and 1883 by British inventor James Wimshurst ( ). For details regarding its operation, please see: Slide 23-10

11 The field model How do objects separated in space influence each other? What is the fundamental mechanism for action at a distance? Physicists use the interaction field model of longrange interactions to explain forces such as gravity and electricity. In the field model, an interacting object fills the space around itself with a field. When another object is placed in the field of the first it feels the presence of the first. Slide 23-11

12 The field model In the case of gravity: At any given location in the space surrounding a source object S, the magnitude of the gravitational field created by S is the magnitude of the gravitational force exerted on an object B placed at that location divided by the mass of B. Slide 23-12

13 The field model Note that the gravitational field is a vector field: The magnitude and direction of the field can be determined by using a test particle (an idealized particle whose mass is small enough that its presence does not perturb the object whose gravitational field we are measuring). Measure, at each location, the gravitational force exerted by Earth on the test particle, and then divide that force by the mass of the test particle to obtain the direction and the magnitude of the gravitational field at that location. Slide 23-13

14 The field model As shown in the figure below, two objects of mass m 1 and m 2 are placed at the same height above the Earth s surface. We can observe that The two objects are subject to different gravitational forces m 1 g and m 2 g. However, the magnitude of the gravitational field g = F G E /m is the same for any object. Slide 23-14

15 The field model Unlike the temperature field which is a scalar field (see figure bottom left), the gravitational field is a vector field. The figure on the right shows a vector field diagram representing the gravitational field near Earth. Slide 23-15

16 Clicker Question 1 When two charged Ping-Pong balls, A and B, are held a small distance apart, which ball is the source of the electric field that acts on ball B? 1. Ball A 2. Ball B 3. Both balls 4. Either ball 5. Not enough information to tell Slide 23-16

17 Electric field diagrams Let s apply the field concept to the electric interaction. In analogy with the gravitational case: At any given location in the space surrounding a source object S, the electric field created by S is the electric force exerted on a charged test particle placed at that location divided by the charge of the test particle: E S = F E St / q t The direction of the electric field at a given location is the same as the direction of the electric force exerted on a positively charged object at that location. Slide 23-17

18 Electric field diagrams The figure shows the pattern of the electric field around single isolated positive and negative charges. Notice that at each point the direction of the field arrow is the same as would be the direction of the electric force on a positive charge under the influence of the source charge. Slide 23-18

19 Fuzzy Fur Field Tank Description: A Plexiglas tank contains 'fur' suspended in oil which covers the bottom. Various shaped electrodes can be placed in the oil. These electrodes are connected to a Wimshurst machine. When the machine is cranked, the fur aligns itself with the electric field. Slide 23-19

20 Physics 1302W.400 Lecture 3 Introductory Physics for Scientists and Engineering II In today s lecture, we will continue to cover electric fields & forces, and consider electric dipoles. Slide 23-20

21 Clicker Question 1 An electron initially moving horizontally near Earth s surface enters a uniform electric field and is deflected upward. What can you say about the direction of the electric field (assuming no other interaction such as gravity)? 1. The electric field points upward. 2. The electric field points downward. 3. The electric field has an upward component. 4. The electric field has a downward component. 5. There is not enough information to tell. Slide 23-21

22 Electric field of a charged particle Let s derive the equation for the electric field produced by a single charged object from Coulomb s law. The figure shows the geometry of the situation. The electric field at a certain point P in space is the electric force experienced at P by a test particle carrying a charge q t divided by the charge of the test particle: E F t E q t Slide 23-22

23 Electric field of a charged particle If we place a test particle carrying a charge q t at P, Coulomb s law tells us that the force exerted on the test particle is F E st = k q q s t If we divide the electric force exerted by the source particle on the test particle by the charge q t of the test particle, we obtain an expression for the electric field created by the source charge at P: E s = F st E q t r st 2 ˆr st = k q s r st 2 ˆr st Slide 23-23

24 Superposition of electric fields The principle of superposition allows us to calculate the net electric influence produced by the individual charges: The combined electric field created by a collection of charged objects is equal to the vector sum of the electric fields created by the individual objects. It points in the same direction as the vector sum of the forces. E = E 1 + E 2 + = k q i ˆr ip r ip 2 Slide 23-24

25 Superposition of electric fields The diagram shown is obtained by vectorially adding the electric field vectors of the two individual particles. The length of the arrows is drawn to be inversely proportional to the distance from the charge, consistent with Coulomb s law Slide 23-25

26 Clicker Question 2 Consider the four field patterns shown. Assuming there are no charges in the regions shown, which of the patterns represent(s) a possible electrostatic field? 1. (a) 2. (b) 3. (b) and (d) 4. (a) and 5. (b) and (c) Slide 23-26

27 Clicker Question 3 A delicate instrument is two meters away from a highly charged metallic sphere. If you want to reduce the magnitude of the electric field at the instrument to 1% of its present value, how many meters away from the charge must you move the instrument? meters meters meters meters 5. Some other number Slide 23-27

28 Van De Graaf Generator A simple Van de Graaff-generator consists of a belt of silk, or a similar flexible dielectric material, running over two metal pulleys, one of which is surrounded by a hollow metal sphere. The fundamental idea for the friction machine as high-voltage supply, using electrostatic influence to charge rotating disk or belt, can be traced back to the 17th century or even before. The Van de Graaff generator was developed, starting in 1929, by American physicist Robert J. Van de Graaff. Slide 23-28

29 Van De Graaf Streamers Styrofoam Peanuts Description: Nylon strings are placed on the generator. When the generator is switched on, the strings align with the field produced. Description: A bucket of styrofoam packing peanuts is placed on the dome of a Van de Graaff generator. When the generator is switched on, the peanuts fly out of the bucket, following the electric field lines produced. Slide 23-29

30 Electric fields and forces What are the forces exerted by an electric field on charged or objects? For the case of a uniform electric field, the figure shows a charged particle placed in a uniform electric field undergoing constant acceleration. [Insert Fig ] Slide 23-30

31 Electric fields and forces For the case of a nonuniform electric field: A positively charged particle placed in a nonuniform electric field has an acceleration in the same direction as the electric field A negatively charged particle placed in a nonuniform electric field has an acceleration in the opposite direction Slide 23-31

32 Electric fields and forces The electric object in this figure is called a dipole. A dipole consists of equal amounts of positive and negative charge separated by a small distance. The orientation of an electric dipole can be characterized by a vector, the dipole moment, that, by definition, points from the center of negative charge to the center of positive charge. Slide 23-32

33 Electric fields and forces A permanent electric dipole placed in an electric field is subject to a torque that tends to align the dipole moment with the direction of the electric field. If the field is uniform, the dipole has zero acceleration. If the electric field is nonuniform, the dipole has a nonzero acceleration. Slide 23-33

34 Volta s Hailstorm Description: A hollow cylinder contains vermiculite. The ends of the cylinder are capped with aluminum plates. When the apparatus is placed in the field of a van de Graaf generator, the particle begin to bounce around. Slide 23-34

35 Dipole field Let s calculate the electric field due to a permanent dipole Notice in the diagram that two chosen observation points along the axis of the dipole and along the perpendicular bisector have been chosen. For the case with the observation point on the bisector of the dipole, symmetry demands that the magnitude of the electric fields produced by each charge be the same. It also demands that the horizontal components of the individual electric fields sum to zero. The resultant field then points in the negative y direction. Slide 23-35

36 Dipole field The two contributions and are the same: " = 2 $ k # E + y E y E y = E + y + E y = (E + + E )cosθ = 2E + cosθ q % p " d 2 % q x 2 + (d 2) 2 ' $ ' = k p d &#[x 2 + (d 2) 2 ] 1 2 & [x 2 + (d 2) 2 ] 3 2 The dipole moment (vector p, from to + ) is defined as p q rp p Taking the limit as x >> d/2, which implies that the observation point is far away compared with the separation of the charges (i.e, compared to the size of the dipole), gives E y k p x 3 (far from dipole along the positive x axis) Slide 23-36

37 Dipole field The axis of the dipole is a symmetry axis (azimuthal symmetry). Along this axis the problem is only one-dimensional. The x- component of the field is zero. For the y-component, we have ( y > +d /2) ( E y = k q p y d 2 " % $ ' q 2 p y + d 2 " % + * $ ' - ) * # & # 2 &, - = k q ( p 1 d 2 " % $ ' 1+ d 2 " % + * $ ' - y 2 ) *# 2y & # 2y &, - Taking the limit as y >> d/2, which implies that the observation point is far from the dipole, gives E y k q )" p $ 1+ 2 d % " ' 1 2 d %, + $ '. y 2 *# 2y & # 2y &- = k q p y 2 ) + * 2d y,. = 2k q d p - y = 2k p ( y >> d /2) 3 y 3 Slide 23-37

38 Dipoles in electric fields The forces exerted by a uniform electric field on the charged ends of the dipole are equal in magnitude but opposite in direction, and so the vector sum of the forces exerted on the dipole is zero. Consequently the acceleration of the center of mass of the dipole is zero. Approximately, this is also the case if the dipole is tiny and/or the electric field varies only gradually in space Slide 23-38

39 Dipoles in electric fields The figure shows a dipole consisting of two particles that carry charges of equal magnitude, but opposite sign connected by a rod of length d. The dipole moment is: p q rp p The dipole makes an angle θ with a uniform electric field E created by some unseen distant source. The charge of the positively charged pole is called dipole charge. The distance d is called the dipole separation. Slide 23-39

40 Dipoles in electric fields Because the forces are exerted on opposite ends of the dipole, however, they create torques that cause the dipole to rotate counterclockwise about its center of mass. Half of the torque results from the force on the positive charge: Slide 23-40

41 Dipoles in electric fields Note that r = d/2 and p = qd. The total torque can be expressed by: τ ϑ = 2( 1 d sinθ )( q 2 p E) = ( q p d) E sinθ pe sinθ Or more compactly in vector form: τ = τ + + τ = p E Slide 23-41

42 Torque on Electric Dipole Description: A rod consists of half acrylic, half PVC. When each half is charged with charges of the opposite sign, a dipole is formed. The dipole is placed between two conducting plates, which are charged with a Wimshurst machine. The rod will align itself with the electric field between the plates. Slide 23-42

43 Dipoles in electric fields Note that r = d/2 and p = qd. The total torque can be expressed by: τ ϑ = 2( 1 d sinθ )( q 2 p E) = ( q p d) E sinθ pe sinθ Or more compactly in vector form: τ = τ + + τ = p E The torque on the dipole is maximum when the dipole moment is perpendicular to the electric field and zero when it is parallel or antiparallel to the electric field. The dipole tends to align itself with the field. Slide 23-43

44 Dipoles in electric fields As we saw in Chapter 22, electrically neutral objects interact with a charged object because they become polarized in the presence of the charged object. Consider an isolated neutral atom (i.e., q = 0): The centers of the atom s positive and negative charge distributions coincide (d = 0), so the atom s dipole moment is zero: p = 0. Slide 23-44

45 Dipoles in electric fields The presence of an external electric field that is, an electric field created by some other charged object causes a separation between the positive and negative charge centers and so induces a dipole moment. To understand the interaction between charged objects and neutral ones, we must therefore study the interaction between a charged particle and what is called an induced dipole. In contrast, the water molecule constitutes a permanent dipole. Slide 23-45

46 Clicker Question 1 How does the force on the charge q and the induced dipole moment depend on the distance y between them (assume that forces are not too large)? 1. F ~ 1/y 2. F ~ 1/y 2 3. F ~ 1/y 3 4. F ~ 1/y 4 5. F ~ 1/y 5 Slide 23-46

47 Dipoles in electric fields If the electric forces on the charges of a neutral atom are not too large, the induced dipole separation d is proportional to the external electric field (Hooke s law: force proportional to separation). So the magnitude of the induced dipole moment is proportional to the external electric field: p ind = αe α is called the polarizability of the atom For a single external charge q, the electric field falls off as the inverse square of the distance, so we have: p ind = αe = α k q y 2 In contrast, the dipole moment of a permanent dipole is constant. Slide 23-47

48 Dipoles in electric fields We can now substitute the induced-dipole result into the equation for the electric field along the dipole axis E y 2k p y 3 ( y >> d /2) to determine the force exerted by a charged particle on an induced dipole: F E pd = 2k p q ind = α 2k 2 q 2 y 3 y 5 This result shows that the interaction between a charged particle and a polarized object depends much more strongly on the distance between them (1/y 5 ) than does the interaction between two charged objects (1/y 2 ). Slide 23-48

49 Physics 1302W.400 Lecture 5 Introductory Physics for Scientists and Engineering II In today s lecture, we will review some properties of electric dipoles and discuss continuous charge distributions. Slide 23-49

50 Dipoles in electric fields If the electric forces on the charges of a neutral atom are not too large, the induced dipole separation d is proportional to the external electric field (Hooke s law: force proportional to separation). So the magnitude of the induced dipole moment is proportional to the external electric field: p ind = αe α is called the polarizability of the atom For a single external charge q, the electric field falls off as the inverse square of the distance, so we have: p ind = αe = α k q y 2 In contrast, the dipole moment of a permanent dipole is constant. Slide 23-50

51 Dipoles in electric fields We can now substitute the induced-dipole result into the equation for the electric field along the dipole axis E y 2k p y 3 ( y >> d /2) to determine the force exerted by a charged particle on an induced dipole: F E pd = 2k p q ind = α 2k 2 q 2 y 3 y 5 This result shows that the interaction between a charged particle and a polarized object depends much more strongly on the distance between them (1/y 5 ) than does the interaction between two charged objects (1/y 2 ). Slide 23-51

52 Clicker Question 1 An electrically neutral dipole is placed in an external field. In which situation(s) is the net force on the dipole zero? 1. (a) 2. (c) 3. (a) and (c) 4. (c) and (d) 5. Some other combination Slide 23-52

53 Discussion 2 Group Problem II Slide 23-53

54 Discussion 2 Group Problem II Slide 23-54

55 Discussion 2 Group Problem II Slide 23-55

56 Electric fields of continuous charge distributions Most charged objects of interest are extended objects: Extended objects can be modeled as being comprised of an infinite number of point charges. The finite sum contained in the superposition principle can then be extended to a continuous summation using integration: E = E 1 + E 2 + = k q ˆr i ip = de 2 s = k r ip dq s r sp 2 ˆr sp Slide 23-56

57 Electric fields of continuous charge distributions For continuous distributions of charge it is useful to express the charge on the object in terms of the charge density, i.e., the electric charge per unit length, area, or volume. For a one-dimensional object, such as a thin charged wire of length l carrying a charge q uniformly distributed along the wire, the linear charge density the amount of charge per unit of length (in coulombs per meter) is given by λ q (uniform charge distribution) Slide 23-57

58 Electric fields of continuous charge distributions For uniformly charged two-dimensional objects, we use the surface charge density the amount of charge per unit of area (in coulombs per square meter). For example, the surface charge density of a flat plate of area A carrying a uniformly distributed charge q is σ q A (uniform charge distribution) Slide 23-58

59 Electric fields of continuous charge distributions For a uniformly charged three-dimensional object, we use the volume charge density which gives the amount of charge per cubic meter: ρ q V (uniform charge distribution) Slide 23-59

60 Electric fields of continuous charge distributions Example 23.5 Electric field created by a uniformly charged thin ring (Be sure to work through all related examples) A thin ring of radius R carries a uniformly distributed charge q. What is the electric field at point P along an axis that is perpendicular to the plane of the ring and passes through its center? Slide 23-60