(Solution Manual) Serway And Vuille's College Physics (Volume 2)

Serway, Vuille, 9

Physics

Solution Manual

9th Edition

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INSTRUCTOR ’S SOLUTIONS MANUAL FOR SERWAY AND VUILLE’S COLLEGE PHYSICS NINTH EDITION , VOLUME 2 INSTRUCTOR ’S SOLUTIONS MANUAL FOR SERWAY AND VUILLE ’S COLLEGE PHYSICS NINTH EDITION , VOLUME 2 Charles Teague Emeritus, Eastern Kentucky University Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States 68719_00_fm_pi-viii.indd i 68719_00_fm_pi-viii.indd i 1/7/11 3:07:36 PM 1/7/11 3:07:36 PM© 2012 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmit-ted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, with-out the prior written permission of the publisher. For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionrequest@cengage.comISBN-13: 978 0 8400 6871 9 ISBN-10: 0 8400 6871 9 Brooks/Cole 20 Channel Center StreetBoston, MA 02210USA Cengage Learning is a leading provider of customized learning solutions with offi ce locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local offi ce at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below. Cengage Learning has established these use limita-tions in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements. Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions. The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third party. You may not sell, license, auction, or otherwise redistribute the Supplement in any form. We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution. Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement. If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt. All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors. The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied. This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s confl ict of law rules. Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement. We trust you fi nd the Supplement a useful teaching tool.READ IMPORTANT LICENSE INFORMATION Printed in the United States of America 1 2 3 4 5 6 7 14 13 12 11 10NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN. 68719_00_fm_pi-viii.indd ii 68719_00_fm_pi-viii.indd ii 1/7/11 3:07:37 PM 1/7/11 3:07:37 PMTABLE OF CONTENTS Acknowledgements v Preface vii Part 4 – Electricity and Magnetism Chapter 15 – Electric Forces and Electric Fields 1 Chapter 16 – Electrical Energy and Capacitance 30 Chapter 17 – Current and Resistance 62 Chapter 18 – Direct-Current Circuits 83 Chapter 19 – Magnetism 120 Chapter 20 – Induced V oltages and Inductance 148 Chapter 21 – Alternating Current Circuits and Electromagnetic Waves 171 Part 5 – Light and Optics Chapter 22 – Refl ection and Refraction of Light 198 Chapter 23 – Mirrors and Lenses 223 Chapter 24 – Wave Optics 256 Chapter 25 – Optical Instruments 282 Part 6 – Modern Physics Chapter 26 – Relativity 305 Chapter 27 – Quantum Physics 324 Chapter 28 – Atomic Physics 341 Chapter 29 – Nuclear Physics 360 Chapter 30 – Nuclear Physics and Elementary Particles 377 iii 68719_00_fm_pi-viii.indd iii 68719_00_fm_pi-viii.indd iii 1/7/11 3:07:38 PM 1/7/11 3:07:38 PM68719_00_fm_pi-viii.indd iv 68719_00_fm_pi-viii.indd iv 1/7/11 3:07:38 PM 1/7/11 3:07:38 PMvACKNOWLEDGEMENTS The author would like to thank everyone who has contributed to this work. In particular, thanks go to the support staff at Cengage Learning for their excellent guidance and support in all phases of this project. Special mention goes to Physics Publisher, Charles Hartford; Development Editor, Ed Dodd; Associate Content Project Manager, Holly Schaff ; Associate Development Editor, Brandi Kirksey; and Editorial Assistant, Brendan Killion. Susan English of Durham Technical Community College served as accuracy reviewer for this manual. Her contributions are deeply appreciated. Any remaining errors in this work are the responsibility of the author alone. I would like to acknowledge the staff of MPS Limited, a Macmillan Company for their excellent work in assembling and typing this manual and preparing diagrams and page layouts. Finally, the author would like to thank his wife, Carol, for her patience, understanding, and great support during this eff ort. v 68719_00_fm_pi-viii.indd v 68719_00_fm_pi-viii.indd v 1/7/11 3:07:38 PM 1/7/11 3:07:38 PM68719_00_fm_pi-viii.indd vi 68719_00_fm_pi-viii.indd vi 1/7/11 3:07:38 PM 1/7/11 3:07:38 PMPREFACE This manual is written to accompany College Physics , Ninth Edition, by Raymond A. Serway and Chris Vuille. For each chapter in that text, the manual includes solutions to all end-of- chapter problems, more detailed answers to Quick Quizzes and Multiple Choice Questions than available in the main text, and answers to the even-numbered Conceptual Questions. Considerable eff ort has been made to ensure that the solutions and answers given in this manual comply with the rules on signifi cant fi gures and rounding given in the Chapter 1 of the textbook. This means that intermediate answers are rounded to the proper number of signifi cant fi gures when written, and that rounded value is used in all subsequent calculations. Users should not be concerned if their answers diff er slightly in the last digit from the answers given here. Most often, this will be caused by choosing to round intermediate answers at diff erent stages of the solution. You are encouraged to keep this manual out of the hands of students as instructors in many col-leges throughout the country use this textbook, and many of them use graded problem assignments as part of the fi nal course grade. Additionally, even when the problems are not used in such a direct fashion, it is advantageous for students to struggle with some problems in order to improve their problem-solving skills. Feel free to post answers and solutions to selected questions and problems but please preserve the manual as a whole. You may also encourage students to purchase a copy of the Student Solutions Manual & Study Guide, which provides chapter summaries as well as detailed solutions to selected problems in the main text. Attempting to keep the manual of manageable size, and recognizing that the primary users will be instructors well versed in the fi eld, answers and solutions are kept fairly brief. Answers to conceptual questions have been shortened by not off ering detailed arguments that lead to the answer. Problem solutions often omit commentary, intermediate steps, as well as initial steps that could be necessary for clear understanding by students. On occasions where selected problem solutions are to be shared with students, you may wish to supply intermediate steps and additional comments as needed. An electronic version of this manual can be obtained by requesting the Instructor’s Power Lecture CD from your local Cengage Learning Sales Representative. Contact information for your sales representative is available under the “Find Your Rep” tab found at the bottom of the page at www. academic.cengage.com. We welcome your comments on the accuracy of the solutions as presented here, as well as sugges-tions for alternative approaches. Charles Teague vii 68719_00_fm_pi-viii.indd vii 68719_00_fm_pi-viii.indd vii 1/7/11 3:07:38 PM 1/7/11 3:07:38 PM68719_00_fm_pi-viii.indd viii 68719_00_fm_pi-viii.indd viii 1/7/11 3:07:38 PM 1/7/11 3:07:38 PM15 Electric Forces and Electric Fields QUICK QUIZZES 1. Choice (b). Object A must have a net charge because two neutral objects do not attract each other. Since object A is attracted to positively-charged object B, the net charge on A must be negative. 2. Choice (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other. 3. Choice (c). The electric fi eld at point P is due to charges other than the test charge. Thus, it is unchanged when the test charge is altered. However, the direction of the force this fi eld exerts on the test change is reversed when the sign of the test charge is changed. 4. Choice (a). If a test charge is at the center of the ring, the force exerted on the test charge by charge on any small segment of the ring will be balanced by the force exerted by charge on the diametrically opposite segment of the ring. The net force on the test charge, and hence the electric fi eld at this location, must then be zero. 5. Choices (c) and (d). The electron and the proton have equal magnitude charges of opposite signs. The forces exerted on these particles by the electric fi eld have equal magnitude and opposite directions. The electron experiences an acceleration of greater magnitude than does the proton because the electron’s mass is much smaller than that of the proton. 6. Choice (a). The fi eld is greatest at point A because this is where the fi eld lines are closest together. The absence of lines at point C indicates that the electric fi eld there is zero. 7. Choice (c). When a plane area A is in a uniform electric fi eld E, the fl ux through that area is Φ E=EAcosq, where q is the angle the electric fi eld makes with the line nor- mal to the plane of A. If A lies in the xy-plane and E is in the z-direction, then q=0° and ΦE=EA=5.00 N C() 4.00 m2() =20.0 N ⋅m2C. 8. Choice (b). If q=60° in Quick Quiz 15.7 above, then ΦE=EAcosq which yields ΦE=5.00 N C() 4.00 m2() cos 60°() =10.0 N ⋅m2C. 9. Choice (d). Gauss’s law states that the electric fl ux through any closed surface is equal to the net enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28, the net enclosed charge is Q=−6 C, which gives Φ E=Q∈0=−6 C() ∈0. 10. Choices (b) and (d). Since the net fl ux through the surface is zero, Gauss’s law says that the net change enclosed by that surface must be zero as stated in (b). Statement (d) must be true because there would be a net fl ux through the surface if more lines entered the surface than left it (or vise-versa). 1 68719_15_ch15_p001-029.indd 1 1/7/11 2:28:37 PM2 Chapter 15 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. To balance the weight of the ball, the magnitude of the upward electric force must equal the mag- nitude of the downward gravitation force, or qE=mg, which gives E=mg q=5.0×10−3 kg () 9.80 m s2() 4.0×10−6 C=1.2×104 NC and the correct choice is (b). 2. The magnitude of the electric fi eld at distance r from a point charge q is E=keqr2, so E=8.99 ×109 N⋅m2C2() 1.60 ×10−19 C () 5.29 ×10−11 m ()2 =5.14 ×1011 NC∼1012 NC making (e) the best choice for this question. 3. The magnitude of the electric force between two protons separated by distance r is F=k ee2r2, so the distance of separation must be r=kee2 F=8.99 ×109 N⋅m2C2() 1.60 ×10−19 C ()2 2.3×10−26 N=0.10 m and (a) is the correct choice. 4. The ball is made of a metal, so free charges within the ball will very quickly rearrange themselves to produce electrostatic equilibrium at all points within the ball. As soon as electrostatic equilib- rium exists inside the ball, the electric fi eld is zero at all points within the ball. Thus, the correct choice is (c). 5. Choosing the surface of the box as the closed surface of interest and applying Gauss’s law, the net electric fl ux through the surface of the box is found to be ΦE=Qinside ∈0=3.0−2.0−7.0+1.0 () ×10−9 C 8.85 ×10−12 C2N⋅m2=−5.6×102 N⋅m2C meaning that (b) is the correct choice. 6. From Newton’s second law, the acceleration of the electron will be ax=Fx m=qEx m=−1.60 ×10−19 C () 1.00 ×103 NC () 9.11×10−31 kg=−1.76 ×1014 ms2 The kinematics equation vx2=v0x2+2axΔx() , with vx=0, gives the stopping distance as Δx=−v0x2 2ax=−3.00 ×106 ms ()2 2−1.76 ×1014 ms2()=2.56 ×10−2 m=2.56 cm so (a) is the correct response for this question. 68719_15_ch15_p001-029.indd 2 1/7/11 2:28:38 PMElectric Forces and Electric Fields 3 7. The displacement from the −4.00 nC charge at point 0, 1.00() m to the point 4.00, −2.00 () m has components rx=xf−xi () =+4.00 m and ry=yf−yi () =−3.00 m , so the magnitude of this displacement is r=rx2+ry2=5.00 m and its direction is q=tan−1ryrx() =−36.9° . The x-component of the electric fi eld at point 4.00, −2.00 () m is then Ex=Ecosq=keq r2cosq=8.99 ×109 N⋅m2C2() −4.00 ×10−9 C () 5.00 m()2 cos−36.9°() =−1.15 N C and the correct response is (d). 8. The magnitude of the electric force between charges Q1i and Q2i, separated by distance ri, is Fi=keQ1iQ2iri2. If changes are made so Q1f=Q1i, Q2f=Q2i3, and rf=2ri, the magnitude of the new force will be Ff=keQ1fQ2f rf2=keQ1iQ2i3() 2ri()2 =1 32()2keQ1iQ2i ri2⎛ ⎝⎜⎞ ⎠⎟=1 12Fi so choice (a) is the correct answer for this question. 9. Each of the situations described in choices (a) through (d) displays a high degree of symmetry, and as such, readily lends itself to the use of Gauss’s law to determine the electric fi elds gener- ated. Thus, the best answer for this question is choice (e), stating that Gauss’s law can be readily applied to fi nd the electric fi eld in all of these contexts. 10. When a charged insulator is brought near a metallic object, free charges within the metal move around, causing the metallic object to become polarized. Within the metallic object, the center of charge for the type of charge opposite to that on the insulator will be located closer to the charged insulator than will the center of charge for the same type as that on the insulator. This causes the attractive force between the charged insulator and the opposite type of charge in the metal to exceed the magnitude of the repulsive force between the insulator and the same type of charge in the metal. Thus, the net electric force between the insulator and the metallic object is one of attraction, and choice (b) is the correct answer. 11. The outer regions of the atoms in your body and the atoms making up the ground both contain negatively charged electrons. When your body is in close proximity to the ground, these negatively charged regions exert repulsive forces on each other. Since the atoms in the solid ground are rig-idly locked in position and cannot move away from your body, this repulsive force prevents your body from penetrating the ground. The best response for this question is choice (e). 12. The positive charge +2Q makes a contribution to the electric fi eld at the upper right corner that is directed away from this charge in the direction of the arrow labeled (a). The magnitude of this contribu-tion is E +=ke(2Q)/2s2, where s is the length of a side of the square. Each of the negative charges makes a contribution of magnitude E−Q=keQs2 directed back toward that charge. The vector sum of these two contributions due to negative charges has magnitude E−=2E−Qcos 45°= 2keQs2 and is directed along the diagonal of the square in the direction of the arrow labeled (d). Since E−>E+, the resultant electric fi eld at the upper right corner of the square is in the direction of arrow (d) and has magnitude E=E−−E+=(2−1)keQs2. The correct answer to the question is choice (d). FIGURE MCQ15.12 68719_15_ch15_p001-029.indd 3 1/7/11 2:28:40 PM4 Chapter 15 13. If the positive charge +2Q at the lower left corner of the square in the above fi gure were removed, the fi eld contribution E+ discussed above would be eliminated. This would leave only E−=2keQs2 as the resultant fi eld at the upper right corner. This has a larger magnitude than the resultant fi eld E found above, making choice (a) the correct answer. 14. Metal objects normally contain equal amounts of positive and negative charge and are electri- cally neutral. The positive charges in both metals and nonmetals are bound up in the nuclei of the atoms and cannot move about or be easily removed. However, in metals, some of the negative charges (the outer or valence electrons in the atoms) are quite loosely bound, can move about rather freely, and are easily removed from the metal. When a metal object is given a positive charge, this is accomplished by removing loosely bound electrons from the metal rather than by adding positive charge to it. Taking away the electrons to leave a net positive charge behind very slightly decreases the mass of the coin. Thus, choice (d) is the best choice for this question. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. Electrons are more mobile than protons and are more easily freed from atoms than the protons which are tightly bound within the nuclei of the atoms. 4. Conducting shoes are worn to avoid the build up of a static charge on them as the wearer walks. Rubber-soled shoes acquire a charge by friction with the fl oor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen. 6. No. Object A might have a charge opposite in sign to that of B, but it also might be neutral. In this latter case, object B causes object A to be polarized, pulling charge of the sign opposite the change on B toward the near face of A and pushing an equal amount of charge of the same sign as that on B toward the far face. Then, due to difference in distances, the force of attraction exerted by B on the induced charge of opposite sign is slightly larger than the repulsive force exerted by B on the induced charge of like sign. Therefore, the net force on A is toward B . 8. (a) Yes. The positive charges create electric fi elds that extend in all directions from those charges. The total fi eld at point A is the vector sum of the individual fi elds produced by the charges at that point. (b) No, because there are no fi eld lines emanating from or converging on point A. (c) No. There must be a charged object present to experience a force.10. Electric fi eld lines start on positive charges and end on negative charges. Thus, if the fair-weather fi eld is directed into the ground, the ground must have a negative charge. 12. To some extent, a television antenna will act as a lightning rod on the house. If the antenna is connected to the Earth by a heavy wire, a lightning discharge striking the house may pass through the metal support rod and be safely carried to the Earth by the ground wire. 68719_15_ch15_p001-029.indd 4 1/7/11 2:28:42 PMElectric Forces and Electric Fields 5 14. (a) If the charge is tripled, the fl ux through the surface is also tripled because the net fl ux is proportional to the charge inside the surface. (b) The fl ux remains constant when the volume changes because the surface surrounds the same amount of charge, regardless of its volume. (c) The fl ux does not change when the shape of the closed surface changes. (d) The fl ux through the closed surface remains unchanged as the charge inside the surface is moved to another location inside that surface. (e) The fl ux is zero because the charge inside the surface is zero. All of these conclusions are arrived at through an understanding of Gauss’s law. 16. All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a result—it is diffi cult to accumulate large amounts of excess charge on an object in a humid climate. During a North American winter, the cold, dry air allows accumulation of signifi cant excess charge, giving the possibility for shocks caused by static electricity sparks. ANSWERS TO EVEN NUMBERED PROBLEMS 2. 1.57 N directed to the left 4. (a) 0.115 N (b) 1.25 cm 6. 2.25 ×10−9 Nm 8. 4.33keq2a2 to the right and 45° above the horizontal 10. F6mC=46.7 N to the left; F1.5mC=157 N to the right; F−2 mC=111 N to the left 12. 5.15 ×103 Nm 14. 16.7 m C 16. (a) 0 (b) 30.0 N (c) 21.6 N (d) 17.3 N (e) –13.0 N (f) 17.3 N (g) 17.0 N (h) I F R=24.3 N at 44.5° above the +x-axis 18. (a) 2.00 ×107 NC to the right (b) 40.0 N to the left 20. (a) 5.27 ×1013 ms2 (b) 5.27 ×105 ms 22. (a) See Solution. (b) 1.4×103 NC (c) 7.5×10−2 N 24. (a) 2.19 ×105 NC at 85.2° below the +x-axis (b) The electric fi eld would be unchanged, but the force would double in magnitude. 68719_15_ch15_p001-029.indd 5 1/7/11 2:28:43 PM6 Chapter 15 26. (a) 1.80 ×104 NC to the right (b) 9.00 ×10−5 N to the left 28. 1.88 ×103 NC at 4.40° below the +x-axis 30. (a) q1q2=−13 (b) q1<0, q2>0 32. (a), (b), and (c) See Solution for Sketches. 34. (a) See Solution. (b) I E=0 at the center of the triangle. (c) 1.73keqa2 (b) positive y-direction 36. (a) zero (b) 1.8×106 Nm (c) 1.1×105 Nm 38. r∼10−6 m 40. 1.38 ×103 N⋅m2C 42. (a) 3.38 nC (b) The sphere contains a net positive charge in a spherically symmetric distribution. 44. (a) 6.55 ×105 N⋅m2C (b) 1.64 ×105 N⋅m2C 46. (a) 1.92 ×107 N⋅m2C (b) 3.20 ×106 N⋅m2C (c) The answer to (b) would change because of the loss of symmetry; however, the answer to (a) would be unchanged since Qinside is unaltered. 48. The electric fi eld is everywhere parallel to the z-axis with the following magnitudes: (a) Ez=+s2∈0 (b) Ez=+3s2∈0 (c) Ez=−s2∈0 50. (a) 2.94 ×105 NC (b) The magnitude of the fi eld is independent of distance from the sheet, provided that distance is small compared to the dimensions of the sheet. 52. +5.25 m C 54. E=K ed in the direction of the electron’s motion. 56. (a) 0.307 s (b) Yes, neglecting gravity causes a 2.28% error. 58. (a) For 0<x<1.00 m , the fi eld contributions made by the two charges are in the same direc- tion and cannot add to zero. (b) Any point for x<0 is nearer the larger charge, so the fi eld contributions made by the two charges cannot have equal magnitudes and add to zero. (c) Points for x>1.00 m are nearer the smaller charge and the fi eld contributions by the two charges are in opposite directions. Thus, it is possible for them to add to zero. (d) E=0 at x =+9.47 m 68719_15_ch15_p001-029.indd 6 1/7/11 2:28:44 PMElectric Forces and Electric Fields 7 60. 0.951 m 62. 1.98 m C 64. (a) 37.0° or 53.0° (b) 1.67 ×10−7s for q =37.0° ; 2.21 ×10−7s for q =53.0° PROBLEM SOLUTIONS 15.1 (a) From Coulomb’s law, F=keQ1Q2 r2, we have F=8.99 ×109 N⋅m2C2()7.50 ×10−9 C () 4.20 ×10−9 C () 1.80 m()2 =8.74 ×10−8 N (b) Since these are like charges (both positive), the force is repulsive . 15.2 Particle A exerts a force toward the right on particle B. By Newton’s third law, particle B will then exert a of equal magnitude force toward the left on particle A. The ratio of the fi nal magnitude of the force to the original magnitude of the force is Ff Fi=keq1q2rf2 keq1q2ri2=ri rf⎛ ⎝⎜⎞ ⎠⎟2 so Ff=Firi rf⎛ ⎝⎜⎞ ⎠⎟2 =2.62 N()13.7 mm 17.7 mm⎛ ⎝⎜⎞ ⎠⎟2 =1.57 N The fi nal vector force that B exerts on A is 1.57 N directed to the left . 15.3 (a) When the balls are an equilibrium distance apart, the tension in the string equals the magni- tude of the repulsive electric force between the balls. Thus, F=keq2q() r2=2.50 N ⇒ q2=2.50 N() r2 2ke or q=2.50 N() 2.00 m()2 2 8.99 ×109 N⋅m2C2()=2.36 ×10−5 C=23.6 m C (b) The charges induce opposite charges in the bulkheads, but the induced charge in the bulk- head near ball B is greater, due to B’s greater charge. Therefore, the system moves slowly towards the bulkhead closer to ball B . 15.4 (a) The gravitational force exerted on the upper sphere by the lower one is neg- ligible in comparison to the gravitational force exerted by the Earth and the downward electrical force exerted by the lower sphere. Therefore, ΣFy=0 ⇒ T−mg−Fe=0 or T=mg+keq1q2 d2 T=7.50 ×10−3 kg () 9.80 m s2⎛ ⎝⎜⎞ ⎠⎟+8.99 ×109 N⋅m2C2() 32.0 ×10−9 C () 58.0 ×10−9 C () 2.00 ×10−2 m ()2 giving T=0.115 N continued on next pageT m, q1 m, q2Femgd 68719_15_ch15_p001-029.indd 7 1/7/11 2:28:47 PM8 Chapter 15 (b) ΣFy=0 → Fe=keq1q2 d2=T−mg, and d=keq1q2 T−mg Thus, if T=0.180 N , d=8.99 ×109 N⋅m2C2() 32.0 ×10−9 C () 58.0 ×10−9 C () 0.180 N −7.50 ×10−3 kg () 9.80 m s2⎛ ⎝⎜⎞ ⎠⎟=1.25 ×10−2 m=1.25 cm 15.5 (a) F=ke2e()2 r2=8.99 ×109 N⋅m2C2() 4 1.60 ×10−19 C ()2⎡ ⎣⎤⎦ 5.00 ×10−15 m ()2 =36.8 N (b) The mass of an alpha particle is m=4.002 6 u , where 1 u=1.66 ×10−27 kg is the unifi ed mass unit. The acceleration of either alpha particle is then a=F m=36.8 N 4.002 6 1.66 ×10−27 kg ()=5.54 ×1027 ms2 15.6 The attractive force between the charged ends tends to compress the molecule. Its magnitude is F=ke1e()2 r2=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟1.60 ×10−19 C ()2 2.17 ×10−6 m ()2=4.89 ×10−17 N The compression of the “spring” is x=0.010 0() r=0.010 0() 2.17 ×10−6 m () =2.17 ×10−8 m, so the spring constant is k=F x=4.89 ×10−17 N 2.17 ×10−8 m=2.25 ×10−9 Nm . 15.7 In the new equilibrium position, ΣFy=Fs−Fe=kd−Fe=0. Thus, k=Fs d=keq1q2r2 d=keq1q2 d⋅r2 or k=8.99 ×109 N⋅m2C2() 0.800 ×10−6 C () −0.600 ×10−6 C () 3.50 ×10−2 m () 5.00 ×10−2 m ()2 k=49.3 N m 15.8 See the sketch at the right. The magnitudes of the forces are F1=F2=keq2q() a2=2keq2 a2 and F3=keq3q() a2()2=1.50keq2 a2 The components of the resultant force on charge q are Fx=F1+F3cos 45°= 2+1.50 cos 45°() keq2 a2=3.06keq2 a2 Fs = kd q1 q2Fe r continued on next page 68719_15_ch15_p001-029.indd 8 1/7/11 2:28:49 PMElectric Forces and Electric Fields 9 and Fy=F1+F3sin 45 °=2+1.50 sin 45°() keq2 a2=3.06keq2 a2 The magnitude of the resultant force is Fe=Fx2+Fy2=2 3.06 keq2 a2⎛ ⎝⎜⎞ ⎠⎟=4.33keq2 a2 and it is directed at q=tan−1Fy Fx⎛ ⎝⎜⎞ ⎠⎟=tan−11.00() =45° above the horizontal . 15.9 (a) The spherically symmetric charge distributions behave as if all charge was located at the centers of the spheres. Therefore, the magnitude of the attractive force is F=keq1q2 r2=8.99 ×109 N⋅m2C2()12×10−9 C () 18×10−9 C () 0.30 m()2 =2.2×10−5 N (b) When the spheres are connected by a conducting wire, the net charge qnet=q1+q2=−6.0×10−9 C will divide equally between the two identical spheres. Thus, the force is now F=keqnet2()2 r2=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟−6.0×10−9 C ()2 4 0.30 m()2 or F=9.0×10−7 N (repulsion) 15.10 The forces are as shown in the sketch below. F1=keq1q2 r122=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟6.00 ×10−6 C () 1.50 ×10−6 C () 3.00 ×10−2 m ()2 =89.9 N F2=keq1q3 r132=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟6.00 ×10−6 C () 2.00 ×10−6 C () 5.00 ×10−2 m ()2 =43.2 N F3=keq2q3 r232=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟1.50 ×10−6 C () 2.00 ×10−6 C () 2.00 ×10−2 m ()2 =67.4 N The net force on the 6mC charge is F6mC=F1−F2=46.7 N to the left . The net force on the 1.5mC charge is F1.5mC=F1+F3=157 N to the right . The net force on the −2mC charge is F−2mC=F2+F3=111 N to the left . 68719_15_ch15_p001-029.indd 9 1/7/11 2:28:51 PM10 Chapter 15 15.11 In the sketch at the right, FR is the resultant of the forces F6 and F3 that are exerted on the charge at the origin by the 6.00 nC and the –3.00 nC charges, respectively. F6=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟6.00 ×10−9 C () 5.00 ×10−9 C () 0.300 m()2 =3.00 ×10−6 N F3=8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟3.00 ×10−9 C () 5.00 ×10−9 C () 0.100 m()2 =1.35 ×10−5 N The resultant is FR=F6()2+F3()2=1.38 ×10−5 N at q=tan−1F3 F6⎛ ⎝⎜⎞ ⎠⎟=77.5° or I FR=1.38 ×10−5 N at 77.5° below the −x-axis 15.12 At equilibrium, ΣFx=Fe−Fs=0 or Fs=Fe Thus, kx=keq1q2 d2 and the force constant of the spring is k=keq1q2 xd2=8.99 ×109 N⋅m2C2() 2.70 ×10−6 C () 8.60 ×10−6 C () 5.00 ×10−3 m () 9.00 ×10−2 m ()2 k=5.15 ×103 Nm 15.13 Please see the sketch at the right. F1=8.99 ×109 N⋅m2C2() 2.00 ×10−6 C () 4.00 ×10−6 C () 0.500 m()2 or F1=0.288 N F2=8.99 ×109 N⋅m2C2() 2.00 ×10−6 C () 7.00 ×10−6 C () 0.500 m()2 or F2=0.503 N The components of the resultant force acting on the 2.00 m C charge are: Fx=F1−F2cos60.0°= 0.288 N −0.503 N() cos60.0°= 3.65 ×10−2 N and Fy=−F2sin 60.0°=− 0.503 N() sin 60.0°=− 0.436 N The magnitude and direction of this resultant force are continued on next pageFs = kxq2q1 dFe 68719_15_ch15_p001-029.indd 10 1/7/11 2:28:53 PMElectric Forces and Electric Fields 11 F=Fx2+Fy2=0.036 5 N()2+0.436 N()2=0.438 N at q=tan−1Fy Fx⎛ ⎝⎜⎞ ⎠⎟=tan−1−0.436 N 0.036 5 N⎛ ⎝⎜⎞ ⎠⎟=−85.2° or 85.2° below the +x-axis 15.14 At equilibrium, Fe=Fs or keQ2 L2=kx=kL−Li () Thus, Q=kL−Li () L2 ke=100 N m() 0.500 m −0.400 m () 0.500 m()2 8.99 ×109 N⋅m2C2 Q=1.67 ×10−5 C=16.7 m C 15.15 Consider the free-body diagram of one of the spheres given at the right. Here, T is the tension in the string and Fe is the repulsive electrical force exerted by the other sphere. ΣFy=0 ⇒ Tcos5.0°= mg, or T=mg cos5.0° ΣFx=0 ⇒ Fe=Tsin 5.0°= mgtan 5.0° At equilibrium, the distance separating the two spheres is r=2Lsin 5.0° . Thus, Fe=mgtan 5.0° becomes keq2 2Lsin 5.0°()2=mgtan 5.0° and yields q=2Lsin 5.0°()mgtan 5.0° ke =2 0.300 m() sin 5.0° []0.20 ×10−3 kg () 9.80 m s2() tan 5.0° 8.99 ×109 N⋅m2C2=7.2 nC 15.16 In the sketch at the right, rBC=4.00 m()2+3.00 m()2=5.00 m and q=tan−13.00 m 4.00 m⎛ ⎝⎜⎞ ⎠⎟=36.9° (a) FAC()x=0 (b) FAC()y=keqAqC rAC2=8.99 ×109 N⋅m2C2()3.00 ×10−4 C () 1.00 ×10−4 C () 3.00 m()2 =30.0 NQ QFs Fe L FAC FBCrBC(0, 3.00 m) (0, 0) (4.00 m, 0) 3.00 × 10 −4 C1.00 × 10−4 C −6.00 × 10−4 Cqq B AC continued on next page 68719_15_ch15_p001-029.indd 11 1/7/11 2:28:55 PM12 Chapter 15 (c) FBC=keqBqC rBC2=8.99 ×109 N⋅m2C2()6.00 ×10−4 C () 1.00 ×10−4 C () 5.00 m()2 =21.6 N (d) FBC()x=FBCcosq=21.6 N() cos 36.9°() =17.3 N (e) FBC()y=−FBCsinq=−21.6 N() sin 36.9°() =−13.0 N (f) FR()x=FAC()x+FBC()x=0+17.3 N =17.3 N (g) FR()y=FAC()y+FBC()y=30.0 −13.0 N =17.0 N (h) FR=FR()x2+FR()y2=17.3 N()2+17.0 N()2=24.3 N and j=tan−1FR()y FR()x⎡ ⎣⎢ ⎢⎤ ⎦⎥ ⎥=tan−117.0 N 17.3 N⎛ ⎝⎜⎞ ⎠⎟=44.5° or FL I R=24.3 N at 44.5° above the +x-axis 15.17 In order to suspend the object in the electric fi eld, the electric force exerted on the object by the fi eld must be directed upward and have a magnitude equal to the weight of the object. Thus, Fe=qE=mg, and the magnitude of the electric fi eld must be E=mg q=3.80 g() 9.80 m s2() 18×10−6 C1 kg 103 g⎛ ⎝⎜⎞ ⎠⎟=2.07 ×103 NC The electric force on a negatively charged object is in the direction opposite to that of the electric fi eld. Since the electric force must be directed upward, the electric fi eld must be directed downward. 15.18 (a) Taking to the right as positive, the resultant electric fi eld at point P is given by ER=E1+E3−E2 =keq1 r12+keq3 r32−keq2 r22 =8.99 ×109 N⋅m2 C2⎛ ⎝⎜⎞ ⎠⎟6.00 ×10−6 C 0.020 0 m()2+2.00 ×10−6 C 0.030 0 m()2−1.50 ×10−6 C 0.010 0 m()2⎡ ⎣⎢ ⎢⎤ ⎦⎥ ⎥ This gives ER=+2.00 ×107 NC or I ER=2.00 ×107 NC to the right (b) I F=qI ER=−2.00 ×10−6 C () 2.00 ×107 NC () =−40.0 N or I F=40.0 N to the left 68719_15_ch15_p001-029.indd 12 1/7/11 2:28:58 PMElectric Forces and Electric Fields 13 15.19 The force on a negative charge is opposite to the direction of the electric fi eld and has magnitude F=qE. Thus, F=−6.00 ×10−6 C5.25 ×105 NC () =3.15 N and I F=3.15 N due north 15.20 (a) The magnitude of the force on the electron is F=qE=eE, and the acceleration is a=F me=eE me=1.60 ×10−19 C () 300 N C() 9.11×10−31 kg=5.27 ×1013 ms2 (b) v=v0+at=0+5.27 ×1013 ms2() 1.00 ×10−8 s () =5.27 ×105 ms 15.21 (a) The electrical force must be directed up the incline and have a magnitude equal to the tangential component of the gravita- tional force. ΣFx=0 ⇒ Fe−mgsinq=0 or Fe=QE=mgsinq and E=mgsinqQ (b) E=mgsinq Q=5.40 ×10−3 kg () 9.80 m s2() sin 25.0° 7.00 ×10−6 C=3.19 ×103 NC Since Fe must be directed up the incline and the electrical force on a negative charge is directed opposite to the fi eld, it is necessary to have the electric fi eld directed down the incline. Thus, I E=3.19 ×103 NC down the incline 15.22 (a) Fe T mgq (b) ΣFy=0 ⇒ Fesinq−mg=0 or Fe=mg sinq Since Fe=qE, this gives E=mg qsinq=5.8×10−3 kg () 9.8 m s2() 68×10−6 C () sin37 °=1.4×103 NC (c) ΣFx=0 ⇒ Fecosq−T=0 or T=Fecosq=mg sinq⎛ ⎝⎜⎞ ⎠⎟cosq=mg tanq and T=5.8×10−3 kg () 9.8 m s2() tan 37 °=7.5×10−2 Nq q mgFe Q, mnxy 68719_15_ch15_p001-029.indd 13 1/7/11 2:29:00 PM14 Chapter 15 15.23 (a) a=F m=qE mp=1.60 ×10−19 C () 640 N C() 1.673 ×10−27 kg=6.12 ×1010 ms2 (b) t=Δv a=1.20 ×106 ms 6.12 ×1010 ms2=1.96 ×10−5 s=19.6 m s (c) Δx=vf2−v02 2a=1.20 ×106 ms ()2−0 2 6.12 ×1010 ms2()=11.8 m (d) KEf=1 2mpvf2=1 21.673 ×10−27 kg () 1.20 ×106 ms ()2=1.20 ×10−15 J 15.24 (a) Please refer to the solution of Problem 15.13 earlier in this chapter. There it is shown that the resultant electric force experienced by the 2.00 m C located at the origin is I F=0.438 N at 85.2° below the +x-axis . Since the electric fi eld at a location is defi ned as the force per unit charge experienced by a test charge placed in that location, the electric fi eld at the origin in the charge confi guration of Figure P15.13 is I E=I F q0=0.438 N 2.00 ×10−6 C at −85.2° =2.19 ×105 N C at 85.2° below the +x-axis (b) The electric force experienced by the charge at the origin is directly proportional to the magnitude of that charge. Thus, doubling the magnitude of this charge would double the magnitude of the electric force. However, the electric fi eld is the force per unit charge and the field would be unchanged if the charge was doubled. This is easily seen in the calculation of part (a) above. Doubling the magnitude of the charge at the origin would double both the numerator and the denominator of the ratio I Fq0, but the value of the ratio (i.e., the electric fi eld) would be unchanged. 15.25 From the fi gure at the right, observe that 2rcos 45.0°+ d=2d. Thus, r=d 2 cos 45.0°=d 2 22()=d 2 Ex=E2−E1cos 45.0°=keQ d2−keQ r22 2⎛ ⎝⎜⎞ ⎠⎟=keQ d2−keQ d22()2 2⎛ ⎝⎜⎞ ⎠⎟=1−2()keQ d2 Ey=E1sin 45.0°+ 0=keQ r22 2⎛ ⎝⎜⎞ ⎠⎟=keQ d22()2 2⎛ ⎝⎜⎞ ⎠⎟=2keQ d2 15.26 (a) Observe the fi gure at the right: E1=E2=keQ r2 and E3=keq r2 Ey=E1sin 30.0°− E2sin 30.0°= 0 Ex=E1cos30.0°+ E2cos30.0°− E3 continued on next page45.0°+Q +QE2E1 45.0°rP45.0° d 2d 30.0°30.0° Q = +3.00 nCE2y x rPE1r = 4.00 cmQ = +3.00 nC q = −2.00 nCE3 68719_15_ch15_p001-029.indd 14 1/7/11 2:29:02 PMElectric Forces and Electric Fields 15 Ex=2keQ r2cos30.0°−keq r2=ke r22Qcos30.0°− q () =8.99 ×109 N⋅m2C2() 4.00 ×10−2 m ()2 2 3.00 ×10−9 C () cos30.0°− 2.00 ×10−9 C ⎡⎣⎤⎦ E x=+1.80 ×104 NC Then, E=Ex2+Ey2=1.80 ×104 NC and I E=1.80 ×104 NC to the right (b) Fe=′qE=−5.00 ×10−9 C () 1.80 ×104 NC () =−9.00 ×10−5 N or I Fe=9.00 ×10−5 N to the left 15.27 If the resultant fi eld is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x-axis, with the origin at the –2.5 µC charge. Then, the two contributions will have opposite directions only in the regions x<0 and x>1.0 m. For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of zero resultant fi eld is on the x-axis at x<0. Requiring equal magnitudes gives keq1 r12=keq2 r22 or 2.5 m C d2=6.0 m C 1.0 m +d ()2 Thus, 1.0 m +d ()2.5 6.0=d Solving for d yields d=1.8 m , or 1.8 m to the left of the −2.5 m C charge 15.28 The altitude of the triangle is h=0.500 m() sin 60.0°= 0.433 m and the magnitudes of the fi elds due to each of the charges are E1=keq1 h2=8.99 ×109 N⋅m2C2() 3.00 ×10−9 C () 0.433 m()2 =144 N C E2=keq2 r22=8.99 ×109 N⋅m2C2() 8.00 ×10−9 C () 0.250 m()2 =1.15 ×103 NC continued on next page 68719_15_ch15_p001-029.indd 15 1/7/11 2:29:04 PM16 Chapter 15 and E3=keq3 r32=8.99 ×109 N⋅m2C2() 5.00 ×10−9 C () 0.250 m()2 =719 N C Thus, ΣEx=E2+E3=1.87 ×103 NC and ΣEy=−E1=−144 N C giving ER=Σ Ex()2+ΣEy()2=1.88 ×103 NC and q=tan−1ΣEyΣEx () =tan−1−0.0769() =−4.40° Hence I ER=1.88 ×103 NC at 4.40° below the + x-axis 15.29 From the symmetry of the charge distribution, students should recognize that the resultant electric fi eld at the center is I ER=0 If one does not recognize this intuitively, consider: I ER=I E1+I E2+I E3 so Ex=E1x−E2x=keq r2cos30°−keq r2cos30°= 0 and Ey=E1y+E2y−E3=keq r2sin 30 °+keq r2sin 30 °−keq r2=0 Thus, ER=Ex2+Ey2=0 15.30 The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines emerge from q2 as enter q1. q2=3q1 (a) Then, q1q2=−13 (b) q2>0 because lines emerge from it, and q1<0 because lines terminate on it. 15.31 Note in the sketches at the right that electric fi eld lines originate on positive charges and terminate on negative charges. The density of lines is twice as great for the −2q charge in (b) as it is for the 1q charge in (a). 68719_15_ch15_p001-029.indd 16 1/7/11 2:29:06 PM