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Hp 16 win

1. 1. How to Use This Presentation <ul><li>To View the presentation as a slideshow with effects </li></ul><ul><li>select “View” on the menu bar and click on “Slide Show.” </li></ul><ul><li>To advance through the presentation, click the right-arrow key or the space bar. </li></ul><ul><li>From the resources slide, click on any resource to see a presentation for that resource. </li></ul><ul><li>From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. </li></ul><ul><li>You may exit the slide show at any time by pressing the Esc key. </li></ul>
2. 2. Chapter Presentation Transparencies Sample Problems Visual Concepts Standardized Test Prep Resources
3. 3. <ul><li>Section 1 Electric Charge </li></ul><ul><li>Section 2 Electric Force </li></ul><ul><li>Section 3 The Electric Field </li></ul>Table of Contents Electric Forces and Fields Chapter 16
4. 4. Objectives <ul><li>Understand the basic properties of electric charge. </li></ul><ul><li>Differentiate between conductors and insulators. </li></ul><ul><li>Distinguish between charging by contact, charging by induction, and charging by polarization. </li></ul>Section 1 Electric Charge Chapter 16
5. 5. Properties of Electric Charge <ul><li>There are two kinds of electric charge. </li></ul><ul><ul><li>like charges repel </li></ul></ul><ul><ul><li>unlike charges attract </li></ul></ul><ul><li>Electric charge is conserved. </li></ul><ul><ul><li>Positively charged particles are called protons . </li></ul></ul><ul><ul><li>Uncharged particles are called neutrons . </li></ul></ul><ul><ul><li>Negatively charged particles are called electrons . </li></ul></ul>Chapter 16 Section 1 Electric Charge
6. 6. Electric Charge Chapter 16 Section 1 Electric Charge
7. 7. Properties of Electric Charge, continued <ul><li>Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge . </li></ul><ul><li>Charge is measured in coulombs (C). </li></ul><ul><li>The fundamental unit of charge, e , is the magnitude of the charge of a single electron or proton. </li></ul><ul><ul><li>e = 1.602 176 x 10 –19 C </li></ul></ul>Chapter 16 Section 1 Electric Charge
8. 8. The Milikan Experiment Chapter 16 Section 1 Electric Charge
9. 9. Milikan’s Oil Drop Experiment Chapter 16 Section 1 Electric Charge
10. 10. Transfer of Electric Charge <ul><li>An electrical conductor is a material in which charges can move freely. </li></ul><ul><li>An electrical insulator is a material in which charges cannot move freely. </li></ul>Chapter 16 Section 1 Electric Charge
11. 11. Transfer of Electric Charge, continued <ul><li>Insulators and conductors can be charged by contact. </li></ul><ul><li>Conductors can be charged by induction . </li></ul><ul><li>Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor. </li></ul>Chapter 16 Section 1 Electric Charge
12. 12. Charging by Induction Chapter 16 Section 1 Electric Charge
13. 13. Transfer of Electric Charge, continued <ul><li>A surface charge can be induced on insulators by polarization. </li></ul><ul><li>With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation. </li></ul>Chapter 16 Section 1 Electric Charge
14. 14. Objectives <ul><li>Calculate electric force using Coulomb’s law. </li></ul><ul><li>Compare electric force with gravitational force. </li></ul><ul><li>Apply the superposition principle to find the resultant force on a charge and to find the position at which the net force on a charge is zero. </li></ul>Section 2 Electric Force Chapter 16
15. 15. Coulomb’s Law <ul><li>Two charges near one another exert a force on one another called the electric force. </li></ul><ul><li>Coulomb’s law states that the electric force is propor-tional to the magnitude of each charge and inversely proportional to the square of the distance between them. </li></ul>Chapter 16 Section 2 Electric Force
16. 16. Coulomb’s Law, continued <ul><li>The resultant force on a charge is the vector sum of the individual forces on that charge. </li></ul><ul><li>Adding forces this way is an example of the principle of superposition. </li></ul><ul><li>When a body is in equilibrium, the net external force acting on that body is zero. </li></ul>Chapter 16 Section 2 Electric Force
17. 17. Superposition Principle Chapter 16 Section 2 Electric Force
18. 18. Sample Problem <ul><li>The Superposition Principle </li></ul><ul><li>Consider three point charges at the corners of a triangle, as shown at right, where q 1 = 6.00  10 –9 C, q 2 = –2.00  10 –9 C, and q 3 = 5.00  10 –9 C. Find the magnitude and direction of the resultant force on q 3 . </li></ul>Chapter 16 Section 2 Electric Force
19. 19. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>1. Define the problem, and identify the known variables. </li></ul><ul><li>Given: </li></ul><ul><ul><li>q 1 = +6.00  10 –9 C r 2,1 = 3.00 m </li></ul></ul><ul><ul><li>q 2 = –2.00  10 –9 C r 3,2 = 4.00 m </li></ul></ul><ul><ul><li>q 3 = +5.00  10 –9 C r 3,1 = 5.00 m </li></ul></ul><ul><ul><li> = 37.0º </li></ul></ul><ul><li>Unknown: F 3,tot = ? Diagram: </li></ul>Chapter 16 Section 2 Electric Force
20. 20. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>Tip: According to the superposition principle, the resultant force on the charge q 3 is the vector sum of the forces exerted by q 1 and q 2 on q 3 . First, find the force exerted on q 3 by each, and then add these two forces together vectorially to get the resultant force on q 3 . </li></ul><ul><li>2. Determine the direction of the forces by analyzing the charges. </li></ul><ul><ul><li>The force F 3,1 is repulsive because q 1 and q 3 have the same sign. </li></ul></ul><ul><ul><li>The force F 3,2 is attractive because q 2 and q 3 have opposite signs. </li></ul></ul>Chapter 16 Section 2 Electric Force
21. 21. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>3. Calculate the magnitudes of the forces with Coulomb’s law. </li></ul>Chapter 16 Section 2 Electric Force
22. 22. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>4. Find the x and y components of each force. </li></ul><ul><ul><li>At this point, the direction each component must be taken into account. </li></ul></ul><ul><li>F 3,1 : F x = (F 3,1 )(cos 37.0º) = (1.08  10 –8 N)(cos 37.0º) </li></ul><ul><ul><li>F x = 8.63  10 –9 N </li></ul></ul><ul><ul><li>F y = (F 3,1 )(sin 37.0º) = (1.08  10 –8 N)(sin 37.0º) </li></ul></ul><ul><ul><li>F y = 6.50  10 –9 N </li></ul></ul><ul><li>F 3,2 : F x = – F 3,2 = –5.62  10 –9 N </li></ul><ul><ul><li>F y = 0 N </li></ul></ul>Chapter 16 Section 2 Electric Force
23. 23. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>5. Calculate the magnitude of the total force acting in both directions. </li></ul><ul><li>F x,tot = 8.63  10 –9 N – 5.62  10 –9 N = 3.01  10 –9 N </li></ul><ul><li>F y,tot = 6.50  10 –9 N + 0 N = 6.50  10 –9 N </li></ul>Chapter 16 Section 2 Electric Force
24. 24. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>6. Use the Pythagorean theorem to find the magni-tude of the resultant force. </li></ul>Chapter 16 Section 2 Electric Force
25. 25. Sample Problem, continued <ul><li>The Superposition Principle </li></ul><ul><li>7. Use a suitable trigonometric function to find the direction of the resultant force. </li></ul><ul><ul><li>In this case, you can use the inverse tangent function: </li></ul></ul>Chapter 16 Section 2 Electric Force
26. 26. Coulomb’s Law, continued <ul><li>The Coulomb force is a field force . </li></ul><ul><li>A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects. </li></ul>Chapter 16 Section 2 Electric Force
27. 27. Objectives <ul><li>Calculate electric field strength. </li></ul><ul><li>Draw and interpret electric field lines. </li></ul><ul><li>Identify the four properties associated with a conductor in electrostatic equilibrium. </li></ul>Section 3 The Electric Field Chapter 16
28. 28. Electric Field Strength <ul><li>An electric field is a region where an electric force on a test charge can be detected. </li></ul><ul><li>The SI units of the electric field, E, are newtons per coulomb (N/C). </li></ul><ul><li>The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge. </li></ul>Chapter 16 Section 3 The Electric Field
29. 29. Electric Fields and Test Charges Chapter 16 Section 3 The Electric Field
30. 30. Electric Field Strength, continued <ul><li>Electric field strength depends on charge and distance. An electric field exists in the region around a charged object. </li></ul><ul><li>Electric Field Strength Due to a Point Charge </li></ul>Chapter 16 Section 3 The Electric Field
31. 31. Calculating Net Electric Field Chapter 16 Section 3 The Electric Field
32. 32. Sample Problem <ul><li>Electric Field Strength </li></ul><ul><li>A charge q 1 = +7.00 µC is at the origin, and a charge q 2 = –5.00 µC is on the x-axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin. </li></ul>Chapter 16 Section 3 The Electric Field
33. 33. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>1. Define the problem, and identify the known variables. </li></ul><ul><li>Given: </li></ul><ul><ul><li>q 1 = +7.00 µC = 7.00  10 –6 C r 1 = 0.400 m </li></ul></ul><ul><ul><li>q 2 = –5.00 µC = –5.00  10 –6 C r 2 = 0.500 m </li></ul></ul><ul><ul><li>= 53.1º </li></ul></ul><ul><li>Unknown: </li></ul><ul><ul><li>E at P ( y = 0.400 m) </li></ul></ul>Chapter 16 Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors. Section 3 The Electric Field
34. 34. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge. </li></ul>Chapter 16 Section 3 The Electric Field
35. 35. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>3. Analyze the signs of the charges. </li></ul><ul><li>The field vector E 1 at P due to q 1 is directed vertically upward, as shown in the figure, because q 1 is positive. Likewise, the field vector E 2 at P due to q 2 is directed toward q 2 because q 2 is negative. </li></ul>Chapter 16 Section 3 The Electric Field
36. 36. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>4. Find the x and y components of each electric field vector. </li></ul><ul><li>For E 1 : E x,1 = 0 N/C </li></ul><ul><ul><li> E y,1 = 3.93  10 5 N/C </li></ul></ul><ul><li>For E 2 : E x,2 = (1.80  10 5 N/C)(cos 53.1º) = 1.08  10 5 N/C </li></ul><ul><ul><li> E y,1 = (1.80  10 5 N/C)(sin 53.1º)= –1.44  10 5 N/C </li></ul></ul>Chapter 16 Section 3 The Electric Field
37. 37. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>5. Calculate the total electric field strength in both directions. </li></ul><ul><li>E x,tot = E x,1 + E x,2 = 0 N/C + 1.08  10 5 N/C </li></ul><ul><ul><li> = 1.08  10 5 N/C </li></ul></ul><ul><li>E y,tot = E y,1 + E y,2 = 3.93  10 5 N/C – 1.44  10 5 N/C </li></ul><ul><ul><li> = 2.49  10 5 N/C </li></ul></ul>Chapter 16 Section 3 The Electric Field
38. 38. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector. </li></ul>Chapter 16 Section 3 The Electric Field
39. 39. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector. </li></ul><ul><li>In this case, you can use the inverse tangent function: </li></ul>Chapter 16 Section 3 The Electric Field
40. 40. Sample Problem, continued <ul><li>Electric Field Strength </li></ul><ul><li>8. Evaluate your answer. </li></ul><ul><li>The electric field at point P is pointing away from the charge q 1 , as expected, because q 1 is a positive charge and is larger than the negative charge q 2 . </li></ul>Chapter 16 Section 3 The Electric Field
41. 41. Electric Field Lines <ul><li>The number of electric field lines is proportional to the electric field strength. </li></ul><ul><li>Electric field lines are tangent to the electric field vector at any point. </li></ul>Chapter 16 Section 3 The Electric Field
42. 42. Rules for Drawing Electric Field Lines Chapter 16 Section 3 The Electric Field
43. 43. Rules for Sketching Fields Created by Several Charges Chapter 16 Section 3 The Electric Field
44. 44. Conductors in Electrostatic Equilibrium <ul><li>The electric field is zero everywhere inside the conductor. </li></ul><ul><li>Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface. </li></ul><ul><li>The electric field just outside a charged conductor is perpendicular to the conductor’s surface. </li></ul><ul><li>On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points. </li></ul>Chapter 16 Section 3 The Electric Field
45. 45. Multiple Choice <ul><li>1. In which way is the electric force similar to the gravitational force? </li></ul><ul><li>A. Electric force is proportional to the mass of the object. </li></ul><ul><li>B. Electric force is similar in strength to gravitational force. </li></ul><ul><li>C. Electric force is both attractive and repulsive. </li></ul><ul><li>D. Electric force decreases in strength as the distance between the charges increases. </li></ul>Standardized Test Prep Chapter 16
46. 46. Multiple Choice, continued <ul><li>1. In which way is the electric force similar to the gravitational force? </li></ul><ul><li>A. Electric force is proportional to the mass of the object. </li></ul><ul><li>B. Electric force is similar in strength to gravitational force. </li></ul><ul><li>C. Electric force is both attractive and repulsive. </li></ul><ul><li>D. Electric force decreases in strength as the distance between the charges increases. </li></ul>Standardized Test Prep Chapter 16
47. 47. Multiple Choice, continued <ul><li>2. What must the charges be for A and B in the figure so that they produce the electric field lines shown? </li></ul><ul><li>F. A and B must both be positive. </li></ul><ul><li>G. A and B must both be negative. </li></ul><ul><li>H. A must be negative, and B must be positive. </li></ul><ul><li>J. A must be positive, and B must be negative. </li></ul>Standardized Test Prep Chapter 16
48. 48. Multiple Choice, continued <ul><li>2. What must the charges be for A and B in the figure so that they produce the electric field lines shown? </li></ul><ul><li>F. A and B must both be positive. </li></ul><ul><li>G. A and B must both be negative. </li></ul><ul><li>H. A must be negative, and B must be positive. </li></ul><ul><li>J. A must be positive, and B must be negative. </li></ul>Standardized Test Prep Chapter 16
49. 49. Multiple Choice, continued <ul><li>3. Which activity does not produce the same results as the other three? </li></ul><ul><li>A. sliding over a plastic-covered automobile seat </li></ul><ul><li>B. walking across a woolen carpet </li></ul><ul><li>C. scraping food from a metal bowl with a metal spoon </li></ul><ul><li>D. brushing dry hair with a plastic comb </li></ul>Standardized Test Prep Chapter 16
50. 50. Multiple Choice, continued <ul><li>3. Which activity does not produce the same results as the other three? </li></ul><ul><li>A. sliding over a plastic-covered automobile seat </li></ul><ul><li>B. walking across a woolen carpet </li></ul><ul><li>C. scraping food from a metal bowl with a metal spoon </li></ul><ul><li>D. brushing dry hair with a plastic comb </li></ul>Standardized Test Prep Chapter 16
51. 51. Multiple Choice, continued <ul><li>4. By how much does the electric force between two charges change when the distance between them is doubled? </li></ul>Standardized Test Prep Chapter 16
52. 52. Multiple Choice, continued <ul><li>4. By how much does the electric force between two charges change when the distance between them is doubled? </li></ul>Standardized Test Prep Chapter 16
53. 53. Multiple Choice, continued <ul><li>Use the passage below to answer questions 5–6. </li></ul><ul><li>A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded. </li></ul><ul><li>5. What is this process of charging called? </li></ul><ul><li>A. charging by contact </li></ul><ul><li>B. charging by induction </li></ul><ul><li>C. charging by conduction </li></ul><ul><li>D. charging by polarization </li></ul>Standardized Test Prep Chapter 16
54. 54. Multiple Choice, continued <ul><li>Use the passage below to answer questions 5–6. </li></ul><ul><li>A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded. </li></ul><ul><li>5. What is this process of charging called? </li></ul><ul><li>A. charging by contact </li></ul><ul><li>B. charging by induction </li></ul><ul><li>C. charging by conduction </li></ul><ul><li>D. charging by polarization </li></ul>Standardized Test Prep Chapter 16
55. 55. Multiple Choice, continued <ul><li>Use the passage below to answer questions 5–6. </li></ul><ul><li>A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded. </li></ul><ul><li>6. What kind of charge is left on the conductor’s surface? </li></ul><ul><li>F. neutral </li></ul><ul><li>G. negative </li></ul><ul><li>H. positive </li></ul><ul><li>J. both positive and negative </li></ul>Standardized Test Prep Chapter 16
56. 56. Multiple Choice, continued <ul><li>Use the passage below to answer questions 5–6. </li></ul><ul><li>A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded. </li></ul><ul><li>6. What kind of charge is left on the conductor’s surface? </li></ul><ul><li>F. neutral </li></ul><ul><li>G. negative </li></ul><ul><li>H. positive </li></ul><ul><li>J. both positive and negative </li></ul>Standardized Test Prep Chapter 16
57. 57. Multiple Choice, continued <ul><li>7. What is the electric field strength 2.0 m from the center of the conducting sphere? </li></ul><ul><li>A. 0 N/C </li></ul><ul><li>B. 5.0  10 2 N/C </li></ul><ul><li>C. 5.0  10 3 N/C </li></ul><ul><li>D. 7.2  10 3 N/C </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
58. 58. Multiple Choice, continued <ul><li>7. What is the electric field strength 2.0 m from the center of the conducting sphere? </li></ul><ul><li>A. 0 N/C </li></ul><ul><li>B. 5.0  10 2 N/C </li></ul><ul><li>C. 5.0  10 3 N/C </li></ul><ul><li>D. 7.2  10 3 N/C </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
59. 59. Multiple Choice, continued <ul><li>8. What is the strength of the electric field at the surface of the conducting sphere? </li></ul><ul><li>F. 0 N/C </li></ul><ul><li>G. 1.5  10 2 N/C </li></ul><ul><li>H. 2.0  10 2 N/C </li></ul><ul><li>J. 7.2  10 3 N/C </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
60. 60. Multiple Choice, continued <ul><li>8. What is the strength of the electric field at the surface of the conducting sphere? </li></ul><ul><li>F. 0 N/C </li></ul><ul><li>G. 1.5  10 2 N/C </li></ul><ul><li>H. 2.0  10 2 N/C </li></ul><ul><li>J. 7.2  10 3 N/C </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
61. 61. Multiple Choice, continued <ul><li>9. What is the strength of the electric field inside the conducting sphere? </li></ul><ul><li>A. 0 N/C </li></ul><ul><li>B. 1.5  10 2 N/C </li></ul><ul><li>C. 2.0  10 2 N/C </li></ul><ul><li>D. 7.2  10 3 N/C </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
62. 62. Multiple Choice, continued <ul><li>9. What is the strength of the electric field inside the conducting sphere? </li></ul><ul><li>A. 0 N/C </li></ul><ul><li>B. 1.5  10 2 N/C </li></ul><ul><li>C. 2.0  10 2 N/C </li></ul><ul><li>D. 7.2  10 3 N/C </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
63. 63. Multiple Choice, continued <ul><li>10. What is the radius of the conducting sphere? </li></ul><ul><li>F. 0.5 m </li></ul><ul><li>G. 1.0 m </li></ul><ul><li>H. 1.5 m </li></ul><ul><li>J. 2.0 m </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
64. 64. Multiple Choice, continued <ul><li>10. What is the radius of the conducting sphere? </li></ul><ul><li>F. 0.5 m </li></ul><ul><li>G. 1.0 m </li></ul><ul><li>H. 1.5 m </li></ul><ul><li>J. 2.0 m </li></ul><ul><li>Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. </li></ul>Standardized Test Prep Chapter 16
65. 65. Short Response <ul><li>11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center? </li></ul>Standardized Test Prep Chapter 16
66. 66. Short Response, continued <ul><li>11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center? </li></ul><ul><li>Answer: 0.0 N/C </li></ul>Standardized Test Prep Chapter 16
67. 67. Short Response, continued <ul><li>12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer. </li></ul>Standardized Test Prep Chapter 16
68. 68. Short Response, continued <ul><li>12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer. </li></ul><ul><li>Answer: not necessarily; The suspended object might have a charge induced on it, but its overall charge could be neutral. </li></ul>Standardized Test Prep Chapter 16
69. 69. Short Response, continued <ul><li>13. One gram of hydrogen contains 6.02  10 23 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38  10 6 m, what is the magnitude of the resulting compressional force on Earth? </li></ul>Standardized Test Prep Chapter 16
70. 70. Short Response, continued <ul><li>13. One gram of hydrogen contains 6.02  10 23 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38  10 6 m, what is the magnitude of the resulting compressional force on Earth? </li></ul><ul><li>Answer: 5.12  10 5 N </li></ul>Standardized Test Prep Chapter 16
71. 71. Short Response, continued <ul><li>14. Air becomes a conductor when the electric field strength exceeds 3.0  10 6 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius. </li></ul>Standardized Test Prep Chapter 16
72. 72. Short Response, continued <ul><li>14. Air becomes a conductor when the electric field strength exceeds 3.0  10 6 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius. </li></ul><ul><li>Answer: 1.3  10 –3 C </li></ul>Standardized Test Prep Chapter 16
73. 73. Extended Response <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>15. What is the magnitude of the acceleration of the proton? </li></ul>Standardized Test Prep Chapter 16
74. 74. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>15. What is the magnitude of the acceleration of the proton? </li></ul><ul><li>Answer: 6.1  10 10 m/s 2 </li></ul>Standardized Test Prep Chapter 16
75. 75. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>16. How long does it take the proton to reach this speed? </li></ul>Standardized Test Prep Chapter 16
76. 76. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>16. How long does it take the proton to reach this speed? </li></ul><ul><li>Answer: 2.0  10 –5 s </li></ul>Standardized Test Prep Chapter 16
77. 77. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>17. How far has it moved in this time interval? </li></ul>Standardized Test Prep Chapter 16
78. 78. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>17. How far has it moved in this time interval? </li></ul><ul><li>Answer: 12 m </li></ul>Standardized Test Prep Chapter 16
79. 79. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>18. What is its kinetic energy at the later time? </li></ul>Standardized Test Prep Chapter 16
80. 80. Extended Response, continued <ul><li>Use the information below to answer questions 15–18. </li></ul><ul><li>A proton, which has a mass of 1.673  10 –27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  10 6 m/s. </li></ul><ul><li>18. What is its kinetic energy at the later time? </li></ul><ul><li>Answer: 1.2  10 –15 J </li></ul>Standardized Test Prep Chapter 16
81. 81. Extended Response, continued <ul><li>19. A student standing on a piece of insulating material places her hand on a Van de Graaff generator. She then turns on the generator. Shortly thereafter, her hairs stand on end. Explain how charge is or is not transferred in this situation, why the student is not shocked, and what causes her hairs to stand up after the generator is started. </li></ul>Standardized Test Prep Chapter 16
82. 82. Extended Response, continued <ul><li>19. (See previous slide for question.) </li></ul><ul><li>Answer: The charge on the sphere of the Van de Graaff generator is transferred to the student by means of conduction. This charge remains on the student because she is insulated from the ground. As there is no path between the student and the generator and the student and the ground by which charge can escape, the student is not shocked. The accumulation of charges of the same sign on the strands of the student’s hair causes the strands to repel each other and so stand on end. </li></ul>Standardized Test Prep Chapter 16
83. 83. Charging By Induction Chapter 16 Section 1 Electric Charge
84. 84. Transfer of Electric Charge Chapter 16 Section 1 Electric Charge
85. 85. Electric Field Lines Chapter 16 Section 3 The Electric Field