* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 1 - Sumner
Survey
Document related concepts
Superconductivity wikipedia , lookup
Electron mobility wikipedia , lookup
Magnetic monopole wikipedia , lookup
Fundamental interaction wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Speed of gravity wikipedia , lookup
Anti-gravity wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Mathematical formulation of the Standard Model wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Electromagnetism wikipedia , lookup
Maxwell's equations wikipedia , lookup
Field (physics) wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
CHAPTER 15 ELECTRIC CHARGE, FORCES, AND FIELDS Multiple Choice Questions: 1. (c). 2. (a). 3. (c). 4. (a). The fur is positively charged (see Exercise 15.5). When the positively charged fur is brought near an electroscope, the leaves are charged by polarization. So the charges on the leaves are positive. 5. (d). Water will be deflected towards the object regardless of its being positively or negatively charged. The water is still neutral but polarized. 6. (d). The balloon clings to the wall regardless of its being positively or negatively charged. The wall is still neutral but polarized. 7. (a). 8. (d). 9. (c). The electrical force between two point charges is inversely proportional to the square of the distance between them. So if this distance is tripled, the force decreases by a factor of 1/3 2 = 1/9. 10. (a). The electrical force between two charged particles is directly proportional to the product of the charges. To increase the force by a factor of 9 (back to its original value), each charge should be 3 times its original value because 3 3 = 9. 11. (c). 12. (b). 13. (a), because the electron has a negative charge. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 259 14. Chapter 15 Electric Charge, Forces, and Fields (b). At the given point, the electric fields of the electrons point toward the electrons, so their y-components cancel, leaving only the x-components pointing in the –x (left) direction. 15. (d). At the given point, both fields point downward (–y direction) so the net field is also downward. 16. (b), since there is no electric field inside a conductor in electrostatic equilibrium. 17. (a), since there is no electric field inside a conductor in electrostatic equilibrium. 18. (b). 19. (b), since electric field lines point toward negative charge. 20. (c). 21. (c). Conceptual Questions: 1. 2. 3. There would be no effect since it is simply a sign convention. No . Charges are simply moved from the object to another object. If an object is positively charged, its mass decreases, because it loses electrons. If an object is negatively charged, its mass increases, because it gains electrons. 4. Since the metal chain is a conductor, it allows any excess electrical charge that may have been induced on the truck (by friction of rubber on the road, for example) to leak off to ground instead of building up on the truck. If the excess charge is not removed, it could result in a spark, causing a gasoline explosion. 5. 6. No . The charges simply change location. There is no gain or loss of electrons. Starting with an uncharged electroscope, touch the bulb with your finger, which grounds the electroscope by providing a path by which electrons could escape from the bulb. Then bring a negatively charged rod near the bulb without touching it. This repels electrons from the rod. Removing your finger while keeping the negatively charged rod close would leave the electroscope with a net positive charge. If you bring a charged object near the top and the Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual 260 leaves collapse, the object must be negatively charged in order to push electrons to the leaf area and at least partially cancel the positive charge there. 7. The water would bend the same way as with the balloon. In this case, the positive charges in the rod would attract the negative ends of the water molecules, causing the stream to bend toward the rod. 8. The electrical force between two point charges is directly proportional to the product of the charges. If one charge is doubled and the other tripled, the new force is proportional to (2q1)(3q2) = 6q1q2, which makes the force 6 times as strong, or 6F . 9. A positive charge can be placed midway between the two electrons so that each will experience a repulsive force from the other and an attractive force from the positive charge. With the proper amount of positive charge, the two forces on each electron will cancel. 10. In the original case, Fe = kq1q2/d2, and in the new case Fe = k(27q1)(q2/3)/r2. Since the forces are equal, we can equate them and solve for r, giving kq1q2/d2 = k(27q1)(q2/3)/r2, which gives r = 3d . 11. (a) The object must be positively charged because the downward repulsion of the nearby positive charge of the dipole is greater than the upward attraction of the more distant negative charge of the dipole. (b) The dipole would accelerate downward because the downward attraction on the nearby negative end would be greater than the upward repulsion of the more distant positive end. 12. It is determined from the relative lengths of the electric field vectors. 13. It is determined by the relative density or spacing of the field lines. The closer the lines are together, the greater the field magnitude. 14. The electric field is the force per unit charge at any location. The force on a charge can only point in one direction and so therefore must the electric field. Thus the field lines never cross. If they did, that would indicate two different force directions at the crossing point, something that is physically unreasonable since a charge placed and released there will accelerate only in one direction, not two. 15. If a positive charge is at center of the spherical shell, the electric field is not zero inside. The field lines run radially outward to the inside surface of the shell where they stop at the induced negative charges on this surface. The field lines reappear on the outside shell surface (which is positively charged) and continue radially outward as if Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 261 Chapter 15 Electric Charge, Forces, and Fields emanating from the point charge at the center. If the charge were negative, the field lines would point in the reverse directions. 16. The Earth’s surface must carry a negative charge, because the electric fields terminate at negative charges. 17. (a) Yes, this is possible when the electric fields are equal in magnitude and opposite in direction at some location. For example, at the midway point in between and along a line joining two charges of the same type and magnitude, the electric field is zero. (b) No, this is not possible . Between the charges, the fields are in the same direction, so they could not cancel. At other points, the fields could not cancel because they would either have different magnitudes (for points along the line connecting the two charges) or would not have opposite directions (for all other points). 18. (a) Directly upward because the electric field due to a positive charge points away from the charge. (b) Away from the plate in the same horizontal plane as the plate because the field points away from the positive plate. (c) The field would have an upward component and a horizontal component horizontal component away from the plate. 19. At very large distances, the object looks very small—like a point. So the electric field pattern looks like the field due to a point charge located at the object. 20. Yes , because the car (a metal frame) keeps the electric field inside it zero. 21. The surface must be spherical . 22. This is because charges accumulate at sharp points , which makes the electric field strongest there. So lightning hits the tall sharp rods first. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual 23. 262 Note that electric field is zero inside the metallic slab. Since charges are mobile, the negative charges are attracted toward the upper portion of the slab, while the positive charges move toward the lower portion. The amount of charge induced on each side of the slab is the same in magnitude as that on each of the plates. + - 24. The field lines start out perpendicular to the plates and end up curving so they are perpendicular to the surface of the metal sphere. + F2 q1 Fnet 60 20 cm 25. The net charges are equalF3 in magnitude but opposite in sign . q2 26. 20 cm q3 No . All it means is that there is more positive charge than negative charge. In other words, there is a net positive charge. Exercises: 1. q = ne = (106)(1.6 1019 C) = 1.6 1013 C . 2. q = ne, n= q –50 106 C = = 3.1 1014 electrons . e 1.60 1019 C Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 263 3. Chapter 15 Electric Charge, Forces, and Fields (a) q = 2(2e) = 4e = 4(1.60 1019 C) = 6.40 10 19 C (b) An alpha particle has two protons, so it needs two electrons to make it a neutral atom. 4. (a) The charge on the silk must be (3) negative because of the conservation of charge. When one object becomes positively charged, it loses electrons. These same electrons must be gained by another object, and therefore it is negatively charged. (b) 8.0 1010 C according to charge conservation. q = ne, n= 8.0 1010 C q = = 5.0 109 electrons . e 1.6 1019 C (c) The mass is m = (9.11 1031 kg/electron)(5.0 109 electrons) = 4.6 1021 kg . 5. (a) The charge on the fur must be (1) positive because of the conservation of charge. When one object becomes negatively charged, it gains electrons. These same electrons must be lost by another object, and therefore it is positively charged. (b) +4.8 109 C according to charge conservation. From q = +ne, n= q 4.8 109 C = = 3.0 1010 electrons. e 1.6 1019 C The mass is (3.0 1010)(9.11 1031 kg) = 2.7 1020 kg . (c) The electrons moved from fur to the rubber rod, so the mass is still 2.7 1020 kg . 6. (2.50 1012 C)[(1 electron)/(1.60 1019 C)] = 1.56 107 electrons 7. (3.22 108 electrons)[(1.60 1019 C)/(1 electron)] = 5.15 10 11 C = 51.5 pC 8. (a) The answer is (4) 1/4 . Since Fe = kq1q2 , F is inversely proportional to the square of the distance, r. If r r2 doubles, Fe becomes 1/4 times the original force. (b) 9. Fe ro2 12 = 2 = = 9. Feo r (1/3)2 (a) Since Fe = So Fe = 9F . kq1q2 q1q2, Fe is 2 12 = 1 , i.e., the same. r2 (b) Fe = 12 12 = 1/4 . (c) Fe = 12 1 = 1/2 . Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual 264 kq1q2 k(6e)(6e) (9.0 109 N·m2/C2)(6)2 (1.6 1019 C)2 = = = 1.3 107 N . 2 2 r r (0.25 109 m)2 10. Fe = 11. (a) Fe = kq1q2 (9.0 109 N·m2/C2)(1.6 1019 C)2 = = 5.8 1011 N . 2 r (2.0 109 m)2 (b) Zero , because they are internal forces of the system. 12. kq1q2 , r is proportional to 1/ Fe . If Fe decreases by a factor of 10, r will increase but not by a r2 (a) Since Fe = factor of 10. So the answer is (1) less than 10 times the original distance. (b) Fe = kq1q2 , r2 F2 r12 = 2, F1 r2 F2 r12 = 2, F1 r2 so F1 r = F2 1 r2 = 13. Fe = kq1q2 , r2 14. Fe = kq1q2 (9.00 109 Nm2/C2)(1.60 1019 C)2 = = 2.90 109 N . 2 r (2.82 1010 m)2 15. (a) If q1 and q2 are like charges, the third charge must be placed between q1 and q2 for it to be in electrostatic so r1 = F2 r = F1 2 10 (30 cm) = 95 cm . 5 (100 cm) = 2.24 m . equilibrium. Also, q1 and q2 have the same magnitude; it must be at the center or at x = 0.25 m from symmetry. (b) If q1 and q2 are unlike charges, the third charge could be placed outside q1 and q3 (x < 0 and x > 0.50 m). However, since the magnitudes +q3 F1 q1 q2 F2 d of q1 and q2 are equal, there will be nowhere for it to happen according –q3 F2 0 0.50 m F1 to Coulomb’s law. (c) Since q1 = +3.0 C and q2 = 7.0 C, a third charge of either type can be placed outside q1 (x < 0) for it to be in electrostatic equilibrium. It cannot be placed to the right of q2 (x > 0.50 m), since the force by q2 will always be larger due to the closer distance from it. Assume q3 is placed at d from q1 (or x = d). From Coulomb’s law, we have kq1q3 k|q2|q3 = , 2 2 r13 r23 Taking the square root on both sides gives Solving, 16. d = 0.94 m. Thus q1 |q2| = . d2 (d + 0.50 m)2 3.0 7.0 = , d d + 0.50 so 1.73(d + 0.50) = 2.65d. x = 0.94 m for q3 . (a) Calling q the smaller charge, Coulomb’s law gives Fe = kq1q2/r2 = kq(3q)/r2 = 3kq2/r2 Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 265 Chapter 15 Electric Charge, Forces, and Fields 3.15 10–6 N = (9.00 109 Nm2 /C2 )(3q2)/(0.10 m)2 q = 1.08 00–9 C = 1.08 nC (b) Now each point charge has 2q of charge. Fe = k(2q)(2q)/r2 = 4(kq2/r2) = 4/3 of the original force Fe = (4/3)(3.15 10–6 N) = 4.20 10–6 N = 4.20 µN 17. (a) The forces on the electron add numerically because they are in the same direction. Putting the negative charge on the left and the positive charge on the right, we have Fnet 1 1 1 1 kqe kqe 2 kqe 2 2 6.48 10 21 2 2 2 r1 r2 r1 r2 r1 r2 1 1 At 5.0 cm: Fnet 6.48 10 21 2 (0.05 m) (0.25 m)2 18 2.7 10 N Likewise, at the other locations we get the forces At 10.0 cm: 8.1 10 19 N At 15.0 cm: 5.8 1019 N At 20.0 cm: 8.1 10 19 N At 25.0 cm: 2.7 1018 N (b) According to the graph, the electron feels the least force midway between the two charges. Fnet | x 15 cm 18. F2 = F3 = kq1q2 (9.0 109 N·m2/C2)(4.0 106 C)2 = = 3.6 N. 2 r (0.20 m)2 According to symmetry, the net force on q1 points in the +x direction. Fnet = Fx = F2 cos 60 + F3 cos 60 = 2(3.6 N) cos 60 = 3.6 N in the positive x direction . 19. (a) From Coulomb’s law: F1 = So kq1q2 (9.0 109 N·m2/C2)(10 106 C)(10 106 C) = = 90 N. 2 r (0.10 m)2 F 1 = (90 N) ^x. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual F3 = 266 (9.0 109 N·m2/C2)(5.0 106 C)(10 106 C) = 45 N. (0.10 m)2 q1 F 3 = (45 N) ŷ . So F1 F4 (9.0 109 N·m2/C2)(5.0 106 C)(10 106 C) F4 = = 22.5 N. (0.10 m)2 + (0.10 m)2 0.10 m F 4 = F4 (cos 45) ^x + (sin 45) ŷ = (15.9 N) (^x + ŷ ). So q2 F3 q4 0.10 m q3 Therefore the net force is F 2 = F 1 + F 3 + F 4 = (74.1 N) x + (60.9 N) ŷ . Thus (74.1 N)2 + (60.9 N)2 = 96 N , F2 = . 609 = tan1 = 39 below positive x-axis . 741 . (b) F1 = (9.0 109 N·m2/C2)(10 106 C)(5.0 106 C) = 45 N. (0.10 m)2 So F 1 = (45 N) ŷ . q1 q2 0.10 m F1 F2 (9.0 109 N·m2/C2)(10 106 C)(5.0 106 C) F2 = = 22.5 N. (0.10 m)2 + (0.10 m)2 F3 q4 0.10 m q3 F 2 = F2 (cos 45) ^x + (sin 45) ŷ = (15.9 N) (^x + ŷ ). So F3 = (9.0 109 N·m2/C2)(5.0 106 C)(5.0 106 C) = 22.5 N. (0.10 m)2 So F3 = (22.5 N) ^x. Therefore the net force is F 4 = F 1 + F 2 + F 3 = (6.6 N) ^x + (60.9 N) ŷ . Thus 20. Fe = 60.9 = tan1 = 84 above negative x-axis . 6.6 (6.6 N)2 + (60.9 N)2 = 61 N , F2 = kq1q2 kq2 = 2 . 2 r r sin = In the vertical direction, 9.0 cm , 30 cm = 17.5. T cos = mg, T= F2 mg . cos q1 Fnet 60 In the horizontal direction, Therefore Thus 21. kq2 = mg tan, r2 q= (a) Since E = F = T sin = q= mg sin = mg tan. cos mg tan r. k 20 cm F3 q2 20 cm q3 (0.10 103 kg)(9.80 m/s2) tan 17.5 (0.18 m) = 3.3 108 C . 9.0 109 N·m2/C2 kq , E is inversely proportional to the square of the distance. r2 So if r doubled, E (2) decreased . (b) ro2 E 12 = 2 = 2 = 1/4. Eo r 2 Therefore E= Eo 1.0 104 N/C = = 2.5 105 N/C . 4 4 Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 267 Chapter 15 Electric Charge, Forces, and Fields F 3.2 1014 N = = 2.0 105 N/C . q 1.6 1019 C 22. E= 23. Use E = F / q to find the components of the electric field. Note that the direction of the field is opposite to the direction of the force since the electron is negative. Vertical component: Ey = Fy/q = (2.7 1014 N)/(1.60 1019 C) = 1.69 105 N/C downward (y) Horizontal component: Ex = Fx/q = (3.8 1014 N)/(1.60 1019 C) = 2.38 105 N/C left (x) The magnitude of the field is E = (Ex2 + Ey2)1/2 = [(1.69 105 N/C)2 + (2.38 105 N/C)2]1/2 = 2.9 105 N/C kq (9.0 109 N·m2/C2)(2.0 1012 C) = 3.2 102 N/C away from the charge . 2 = r (0.75 102 m)2 24. E= 25. Using E = r= 26. kq . r2 kq = E (9.00 109 Nm2/C2)(1.60 1019 C ) = 1.2 107 m away from the charge . 1.0 105 N/C (a) The net field is (3) directed toward the 5.0 C charge . The electric fields by the two charges are opposite in direction, but the one by the –5.0-C charge has a greater magnitude. (b) E4.0 = E5.0 = kq (9.0 109 N·m2/C2)(4.0 106 C) = 3.6 106 N/C, 2 = r (0.10 m)2 (9.0 109 N·m2/C2)(5.0 106 C) = 4.5 106 N/C. (0.10 m)2 So the net field is E = E5.0 E4.0 = 9.0 105 N/C . 27. Electric force balances the weight of the particle, F = mg. Proton: E = F mg (1.67 1027 kg )(9.80 m/s2) = = = 1.0 107 N/C upward . q q 1.6 1019 C Electron: E = 28. (9.11 1031 kg )(9.80 m/s2) = 5.6 1011 N/C downward . 1.6 1019 C (a) Since both charges are negative, a point between the two charges could have zero electric field. The location is at (1) left of the origin , because the charge closer to the point has the –3.0 C E3.0 –0.50 m 0 E4.0 –4.0 C 0.50 m x d Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual 268 smaller magnitude. The distance from the smaller charge has to be smaller to match the field by the bigger charge. (b) Assume the point is d from the –3.0-C charge. For the electric field to be zero, E3.0 = E4.0. So E3.0 = kq k(3.0 C) k(4.0 C) = = E4.0 = . r2 d2 (1.0 d)2 Taking the square root on both sides gives Solving for 29. d = 0.464 m. 3.0 4.0 = . d 1.0 d Or Therefore the coordinates of the point are (0.036 m‚ 0) . y = tan1 x . The electric field by the +2.5-C charge is r= x2 + y2 , E2.5 = kq (9.0 109 N·m2/C2)(2.5 106 C) = 3.6 105 N/C, 2 = r (0.20 m)2 + (0.15 m)2 and 1.73(1.0 d) = 2.0d. +2.5 C (–0.20 m, 0.15 m) 2.5 = 36.9, so E 2.5 = (3.6 105 N/C)[(cos36.9) ^x (sin 36.9) ŷ ] –6.3 C = (2.88 105 N/C) ^x + (2.16 105 N/C) ŷ . –4.8 C (–0.42 m, –0.32 m) (0.50 m, –0.35 m) The electric field by the 4.8-C charge is E4.8 = (9.0 109 N·m2/C2)(4.8 106 C) = 1.16 105 N/C, (0.50 m)2 + (0.35 m)2 4.8 = 35.0, so E 4.8 = (1.16 105 N/C)[(cos 35.0) ^x (sin 35.0) ŷ ] = (9.50 104 N/C) ^x + (6.65 104 N/C) ŷ . The electric field by the –6.3-C charge is E6.3 = c (9.0 109 N·m2/C2)(6.3 106 C) = 2.03 105 N/C, (0.42 m)2 + (0.32 m)2 6.3 = 37.3, so E 6.3 = [(2.03 105 N/C)(cos 37.3) ^x (sin 37.3) ŷ ] = [(1.61 105 N/C) ŷ + (1.23 105 N/C) ŷ ]. ˆ . Therefore the net electric field is E = E 2.5 + E 4.8 + E 6.3 = (2.2 105 N/C)xˆ + ( 4.1 105 N/C) y. 30. The point has to be between the two charges. Assume it is d from the +4.0 C charge. For the electric field to be zero, kq1 kq2 = 2, r12 r2 Taking the square root on both sides gives Solving, 31. r= 4.0 C 9.0 C = . d2 (0.30 m d)2 2 3 = . d 0.30 d Therefore 3d = 2(0.30 d). d = 0.12 m = 12 cm from the charge of 4.0 C (between the charges) . 0.10 m = 0.115 m. cos 30 q1 kq (9.0 109 N·m2/C2)(4.0 106 C) E1 = E2 = E3 = 2 = = 2.72 106 N/C. r (0.115 m)2 E1 E3 E2 r Due to symmetry, the resultant of E 1 and E 2 will point toward q3. 30 The net field E = E3 + E1 cos 60 + E2 cos 60 q2 20 cm q3 Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 269 Chapter 15 Electric Charge, Forces, and Fields = 2.72 106 N/C + 2(2.72 106 N/C) cos 60 = 5.4 106 N/C toward the charge of –4.0 C . q1 32. Due to symmetry, E1 and E2 cancel out. The net field is the field by q3. 30 kq3 (9.0 109 N·m2/C2)(4.0 106 C) E = E3 = 2 = r [(0.20 m) cos30]2 r E2 E3 = 1.2 106 N/C toward the charge of –4.0 C . E1 q2 33. E1 = E2 = kq (9.0 109 N·m2/C2)(10 106 C) = 1.8 107 N/C. 2 = r (0.05 m)2 + (0.05 m)2 E3 = E4 = (9.0 109 N·m2/C2)(5.0 106 C) = 9.0 106 N/C. (0.05 m)2 + (0.05 m)2 20 cm q1 q2 E1 0.10 m Due to symmetry, the net electric field will be upward. It is equal to q3 E2 E3 E4 r q4 0.10 m q3 E = (E1 + E3) cos 45 + (E2 + E3) cos 45 = 2(1.8 107 N/C + 9.0 106 N/C) cos 45 = 3.8 107 N/C in the +y direction . 34. (a) First find the acceleration. a= F qE (2.0 106 C)(12 N/C) = = = 1.2 m/s2. m m 2.0 105 kg x = vot + 12 at2 = 12 (1.2 m/s2)(0.50 s)2 = 0.15 m . (b) v = vo + at = (1.2 m/s2)(0.50 s) = 0.60 m/s . (c) E = 4 kQ , A so Q= EA (12 N/C)(0.050 m)2 13 = C. 9 2 2 = 2.7 10 4 k 4 (9.00 10 Nm /C ) The plates carry opposite charge, so they are 0.27 pC . 35. (a) The field between two large parallel plates is E = 4πkQ/A, which gives Q/A = E/4πk 2 2 Q/A = (1.7 106 N/C)/[4π(9.00 109 Nm /C )] = 1.5 105 C/m 2 = 15 µC/m 2 (b) Q = A (charge density) = (0.150 m)2(15 µC/m2) = 3.4 107 C = 0.34 µC 36. (a) The field between two large parallel plates is E = 4πkQ/A 2 2 E = 4π(9.00 Nm /C )(4.0 109 C)/(0.20 m)2 = 1.1 104 N/C (b) F = qE = (1.60 1019 C)(1.13 104 N/C) = 1.8 10 15 N toward the positive plate. (c) a = F/m = (1.18 1015 N)/(9.11 1031 kg) = 2.0 1015 m/s2 37. E1 = 4.0 kq (9.0 109 N·m2/C2)(10 106 C) = 7.76 106 N/C, 1 = tan1 10 = 21.8. 2 = 2 2 r (0.10 m) + (0.040 m) Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 270 College Physics Seventh Edition: Instructor Solutions Manual E 1 = (7.76 106 N/C)[(cos 21.8) ^x + (sin 21.8) ŷ ] q1 = (7.20 106 N/C) ^x + (2.88 106 N/C) ŷ ; 1 0.10 m (9.0 109 N·m2/C2)(10 106 C) E2 = = 5.63 107 N/C, (0.040 m)2 E2 q2 4.0 cm E1 E3 4 So E 2 = (5.63 107 N/C) ^y; E3 = (9.0 109 N·m2/C2)(5.0 106 C) = 1.25 107 N/C, (0.060 m)2 so E4 = (9.0 109 N·m2/C2)(5.0 106 C) = 3.31 106 N/C, (0.10 m)2 + (0.060 m)2 6.0 4 = tan1 10 = 31.0, q4 E4 0.10 m 6.0 m q3 E 3 = (1.25 107 N/C) ŷ ; Thus E 4 = (3.31 106 N/C)[(cos 31.0) ^x + (sin 31.0) ŷ ] = (2.84 106 N/C) ^x + (1.70 106 N/C) ŷ . E = E 1 + E 2 + E 3 + E 4 = (4.4 106 N/C)xˆ + (7.3 107 N/C) yˆ . Therefore 38. (a) From symmetry, the net electric field is in the –y direction. (b) r = E– = E+ = (d/2)2 + x2 , sin = d/2 = r +q d/2 . (d/2)2 + x2 r d/2 x kq kq = . r2 (d/2)2 + x2 d/2 E = Ey = E– sin + E+ sin = 2E+ sin 2kq = (d/2)2 + x2 E– –q E+ E d/2 kqd = . 2 2 2 [(d/2) + x2]3/2 (d/2) + x (c) If x >> d, we can ignore the term (d/2)2 in (d/2)2 + x2. So E= kqd kqd kqd 2 3/2 = 3 . [(d/2)2 + x2]3/2 (x ) x (d) This is because the field is not created by a single point charge. We expect inverse square from single point charges, but vector addition can cause other distance dependences as happens here. 39. (a) The inner surface of the shell will have (1) negative charge due to induction. (b) Zero since all excess charge resides on the surface of the conductor in electrostatic equilibrium. (c) +Q since all excess charge resides on the surface of the conductor in electrostatic equilibrium. (d) Q by induction and the conservation of charge. (e) +Q by induction and the conservation of charge. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 271 40. Chapter 15 Electric Charge, Forces, and Fields (a) There is none , since the electric field is zero in the interior of the solid sphere. (b) Outward from the center of the sphere, since the charge on the surface of the sphere is positive. (c) There is none , since the electric field is zero. The field by the sphere cancels out the field by the inner surface of the shell. (d) Outward from the center of the sphere. The net excess charge is positive on the outer surface of the shell. 41. (a) It is zero , since there is no electric field inside a conductor in electrostatic equilibrium. (b) The charge on the surface of the sphere can be considered as if it were concentrated at the center. So E= k(Q) kQ = 2 , the negative sign indicates it is toward the center. r2 r (c) Again, it is zero . (d) All charges can be considered as if they were concentrated at the center. So E= k(Q) kQ k(Q) kQ + 2 + = 2 , the negative sign indicates it is toward the center. r2 r r2 r 42. More charges accumulate at sharper points, and the electric field lines are perpendicular to the surface. 43. See the diagram below. 44. (a) The electric field has moved negative charges in the slab to the top of the slab, so this field must point downward . (b) The net field in the metal must be zero. So the field due to the generated surface charges must be opposite to the external field and of the same magnitude. Hence, E = 1250 N/C , and it points upward . Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual 272 (c) The top and bottom of the slab have equal but opposite charges Q. Inside the slab, these charges generate an electric field of 1250 N/C. We can use the formula for the field between parallel plates, treating the surface charges as the parallel plates. E = 4πkQ/A Q = EA/4πk = (1250 N/C)(0.0500 m)2/[4π(9.00 109 Nm2 /C2 )] Q = 2.76 10 11 C = 27.6 pC 45. The net number of field lines is +6 2 6 = 6 lines , or net of 6 lines entering it . 46. (a) The magnitude of the unknown charge is (1) greater than 10.0 C , because the number of field lines is proportional to the magnitude of the charge inside the Gaussian surface. (b) q2 = 47. 75 (+10.0 C) = 46.9 C . 16 (a) The other end of the dipole has the same magnitude charge as the positive end, except it is negative. Therefore 10 field lines will enter the Gaussian surface. (b) The net charge of the dipole is zero, so the net number of field lines passing through the Gaussian surface would be zero . 48. (a) Fe = kq1q2 (9.0 109 N·m2/C2)(1.6 1019 C)2 = = 8.2 108 N . r2 (5.3 1011 m)2 (b) The electric force provides the required centripetal force. so v= (c) ac = Fc r = m Fe = Fc = m v2 , r (8.2 108 N)(5.3 1011 m) = 2.2 106 m/s . 9.11 1031 kg v2 F 8.2 108 N = = = 9.0 1022 m/s2 = 9.2 1021 g , where g = 9.80 r m 9.11 1031 kg m/s2. T 49. (a) Since the charge is negative, and the electric force has to be to the left to keep the ball F balanced, the direction of the electric field is to the right . (b) In the vertical direction: T cos = mg, T= mg . cos mg In the horizontal direction: F = qE = T sin . So E = T sin mg sin mg tan (6.00 106 kg)(9.80 m/s2) tan 12.3 = = = = 8.55 103 N/C . q q q cos 1.50 109 C Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 273 50. Chapter 15 Electric Charge, Forces, and Fields (a) The bottom plate must be positively charged. In this configuration, the electric force (upward) can balance the weight of the particle. 4 kQ 4 (9.00 10 Nm /C )(5.50 105 C) = = 5.14 108 N/C. A (0.110 m)2 9 (b) E = F = qE = mg, 51. m= 2 2 qE (9.35 1012 C)(5.14 108 N/C) = = 4.90 104 kg . g 9.80 m/s2 (a) The electron gains speed, so the force on it is to the right. Since electron carries negative charge, the charge is positive on right plate and negative on left plate . (b) The direction of the electric field is from right to left . (c) First calculate the acceleration of the electron. v = v + 2a(x xo), 2 E= So 52. 2 o 4 kQ F qE 4 kqQ and a = m = m = . A Am Ama (0.254 m)2 (9.11 1031 kg) (3.468 1010 m / s) 45 Q = 4 kq = = 1.13 10 C . 4 (9.00 109 N m 2 / C2 ) (1.60 1019 C) (a) t = x 0.10 m = = 1.626 109 s. vx 6.15 107 m/s d = 12 ay t2 , ay = v2 v2o (4.15 104 m/s)2 (1.63 104 m/s)2 a= = = 3.468 1010 m/s2. 2(0.0210 m) 2(x xo) F qE = , m m (b) Since E = ay = 2d 2(4.70 103 m) = 3.555 1015 m/s2. 2 = t (1.626 109 s)2 E= ay m ay m (3.555 1015 m/s2)(9.11 1031 kg) = = = 2.02 104 N/C . q q (1.60 1019 C) 4 kQ , the surface charge density is A Q E 2.024 104 N/C 7 2 = = 9 2 2 = 1.79 10 C/m . A 4 k 4 (9.00 10 Nm /C ) 53. (a) The forces F and F (see figure) create a torque on the dipole which will turn it until the dipole moment p is parallel to the electric field E . + p F+ E F- - (b) Add the torques due to the two forces, realizing that the torques are equal. Calling d the distance between the charges and the angle between p and E , we get Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. College Physics Seventh Edition: Instructor Solutions Manual 274 = 2F+(d/2) sin = qdE sin = pE sin (c) Fnet 0 because the two forces are of equal magnitude but opposite direction. (d) The torque is a maximum when |sin| = 1, which is when = 90° or 270°. When = 0° or 180° , sin = 0 for the minimum torque of zero. 54. (a) First use kinematics to find the acceleration of the proton. 0 = (3.15 105 m/s)2 + 2a(0.0525 m) a = 9.45 1011 m/s2 Newton’s second law gives F = ma = qE 3 E = ma/q = (1.67 1027 kg)(9.45 1011 m/s2)/(1.60 1019 C) = 9.86 10 N/C (b) Using kinematics, we have v v02 2a( x x0 ) 3.15 10 5 2 m/s 2 9.45 1011 m/s2 (0.0350 m 5 v = 1.82 10 m/s There are two answers because the proton first stops and then reverses direction. 55. (a) The positive charge (+q) of the dipole is to the right of the origin with the negative charge (q) to the left of the origin. The charge Q = 2.50 pC will attract the positive charge and repel the negative charge of the dipole. The ycomponents of the force will cancel, leaving only two equal x-components in the –x-direction. Calling r the distance between the upper charge and each of the charges of the dipole and the angle between the force and the x-axis, we get kqQ Fx 2 cos r2 9.00 109 N m2 /C2 4.55 1012 C 2.50 1012 C 3 Fx = 2 5 (0.0500 m)2 = 4.91 1011 N in the –x direction Newton’s second law gives ax = Fx/m = (4.91 1011 N)/(7.25 1012 kg) = 6.78 m/s 2 in the – x direction (b) Since the torques are equal, the net torque is twice the torque on each charge of the dipole. Therefore = 2(F sin d/2) = kqQd/r2 sin = (9.00 109 Nm2 /C2 )(4.55 1012 C)(2.50 1012 C)(0.0600 m)(4/5)/(0.0500 m)2 = 1.97 1012 mN in the counterclockwise direction Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.