Answers to Quick Quizzes,Odd-Numbered Conceptual Questionsand ProblemsChapter 15QUICK QUIZZES1. (b)2. (b)3. (c)4. (a)5. (c) and (d)6. (a)7. (c)8. (b)9. (d)10. (b) and (d)CONCEPTUAL QUESTIONS1. Electrons have been removed from the object.3. The configuration shown is inherently unstable. The negativecharges repel each other. If there is any slight rotationof one of the rods, the repulsion can result in furtherrotation away from this configuration. There are threeconceivable final configurations shown below. Configuration(a) is stable: If the positive upper ends are pushedtowards each other, their mutual repulsion will move thesystem back to the original configuration. Configuration(b) is an equilibrium configuration, but it is unstable: Ifthe lower ends are moved towards each other, theirmutual attraction will be larger than that of the upperends, and the configuration will shift to (c), another possiblestable configuration.+–(a)+–+ –(c)+–(b)+ –Figure Q15.35. Move an object A with a net positive charge so it is near,but not touching, a neutral metallic object B that is insulatedfrom the ground. The presence of A will polarize B,causing an excess negative charge to exist on the sidenearest A and an excess positive charge of equal magnitudeto exist on the side farthest from A. While A is stillnear B, touch B with your hand. Additional electrons willthen flow from ground, through your body and onto B.With A continuing to be near but not in contact with B,remove your hand from B, thus trapping the excess electronson B. When A is now removed, B is left with excesselectrons, or a net negative charge. By means of mutualrepulsion, this negative charge will now spread uniformlyover the entire surface of B.–+7. An object’s mass decreases very slightly (immeasurably)when it is given a positive charge, because it loses electrons.When the object is given a negative charge, its massincreases slightly because it gains electrons.9. Electric field lines start on positive charges and end onnegative charges. Thus, if the fair-weather field is directedinto the ground, the ground must have a negative charge.11. The two charged plates create a region with a uniformelectric field between them, directed from the positive towardthe negative plate. Once the ball is disturbed so as totouch one plate (say, the negative one), some negativecharge will be transferred to the ball and it will be actedupon by an electric force that will accelerate it to the positiveplate. Once the ball touches the positive plate, it willrelease its negative charge, acquire a positive charge, andaccelerate back to the negative plate. The ball will continueto move back and forth between the plates until ithas transferred all their net charge, thereby making bothplates neutral.13. The electric shielding effect of conductors depends onthe fact that there are two kinds of charge: positive andnegative. As a result, charges can move within the conductorso that the combination of positive and negativecharges establishes an electric field that exactly cancelsthe external field within the conductor and any cavitiesinside the conductor. There is only one type of gravitationcharge, however, because there is no negative mass. As aresult, gravitational shielding is not possible.15. The electric field patterns of each of these three configurationsdo not have sufficient symmetry to make the calculationspractical. Gauss’s law is useful only for calculatingthe electric fields of highly symmetric charge distributions,such as uniformly charged spheres, cylinders, andsheets.17. No, the wall is not positively charged. The balloon inducesa charge of opposite sign in the wall, causing theballoon and the wall to be attracted to each other. Theballoon eventually falls because its charge slowly diminishesafter leaking to ground. Some of the balloon’scharge could also be lost due to positive ions in the surroundingatmosphere, which would tend to neutralize thenegative charges on the balloon.19. When the comb is nearby, charges separate on the paper,and the paper is attracted to the comb. After contact,charges from the comb are transferred to the paper, sothat it has the same type of charge as the comb. The paperis thus repelled.21. The attraction between the ball and the object could bean attraction of unlike charges, or it could be an attractionbetween a charged object and a neutral object as a resultof polarization of the molecules of the neutral object.Two additional experiments could help us determinewhether the object is charged. First, a known neutral ballcould be brought near the object, and if there is anattraction, the object is negatively charged. Another possibilityis to bring a known negatively charged ball near theA.17
A.18 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problemsobject. In that case, if there is a repulsion, then the objectis negatively charged. If there is an attraction, then theobject is neutral.PROBLEMS1. 1.1 10 8 N (attractive)3. 91 N (repulsion)5. (a) 36.8 N (b) 5.54 10 27 m/s 27. 5.12 10 5 N9. (a) 2.2 10 5 N (attraction)(b) 9.0 10 7 N (repulsion)11. 1.38 10 5 N at 77.5° below the negative x -axis13. 0.872 N at 30.0° below the positive x -axis15. 7.2 nC17. 1.5 10 3 C19. 7.20 10 5 N/C (downward)21. 1.2 10 4 N/C23. (a) 6.12 10 10 m/s 2 (b) 19.6 s (c) 11.8 m(d) 1.20 10 15 J25. zero27. 1.8 m to the left of the 2.5-C charge33. (a) 0 (b) 5 C inside, 5 C outside (c) 0 inside, 5 C outside (d) 0 inside, 5 C outside35. 1.3 10 3 C37. (a) 4.8 10 15 N (b) 2.9 10 12 m/s 239. (a) 858 N m 2 /C (b) 0 (c) 657 N m 2 /C41. 4.1 10 6 N/C43. (a) 0 (b) k e q/r 2 outward47. 57.5 N49. 24 N/C in the positive x -direction51. (a) E 2k e qb (a 2 b 2 ) 3/2 in the positive x-direction(b) E k e Qb(a 2 b 2 ) 3/2 in the positive x-direction53. (a) 0 (b) 7.99 10 7 N/C (outward)(c) 0 (d) 7.34 10 6 N/C (outward)55. 3.55 10 5 N m 2 /C57. 4.4 10 5 N/C59. (a) 10.9 nC (b) 5.44 10 3 N61. 10 7 C63. (a) 1.00 10 3 N/C (b) 3.37 10 8 s (c) accelerate at1.76 10 14 m/s 2 in the direction opposite that of theelectric fieldChapter 16QUICK QUIZZES1. (b)2. (b), (d)3. (d)4. (c)5. (a)6. (c)7. (a) C decreases. (b) Q stays the same. (c) E staysthe same. (d) V increases. (e) The energy storedincreases.8. (a) C increases. (b) Q increases. (c) E stays the same.(d) V remains the same. (e) The energy stored increases.9. (a)CONCEPTUAL QUESTIONS1. (a) The proton moves in a straight line with constant accelerationin the direction of the electric field. (b) As itsvelocity increases, its kinetic energy increases and the electricpotential energy associated with the proton decreases.3. The work done in pulling the capacitor plates fartherapart is transferred into additional electric energy storedin the capacitor. The charge is constant and the capacitancedecreases, but the potential difference between theplates increases, which results in an increase in the storedelectric energy.5. If the power line makes electrical contact with the metalof the car, it will raise the potential of the car to 20 kV. Itwill also raise the potential of your body to 20 kV, becauseyou are in contact with the car. In itself, this is not a problem.If you step out of the car, however, your body at20 kV will make contact with the ground, which is at zerovolts. As a result, a current will pass through your bodyand you will likely be injured. Thus, it is best to stay in thecar until help arrives.7. If two points on a conducting object were at different potentials,then free charges in the object would move andwe would not have static conditions, in contradiction tothe initial assumption. (Free positive charges would migratefrom locations of higher to locations of lower potential.Free electrons would rapidly move from locations oflower to locations of higher potential.) All of the chargeswould continue to move until the potential became equaleverywhere in the conductor.9. The capacitor often remains charged long after the voltagesource is disconnected. This residual charge can belethal. The capacitor can be safely handled after dischargingthe plates by short-circuiting the device with a conductor,such as a screwdriver with an insulating handle.11. Field lines represent the direction of the electric force ona positive test charge. If electric field lines were to cross,then, at the point of crossing, there would be an ambiguityregarding the direction of the force on the test charge,because there would be two possible forces there. Thus,electric field lines cannot cross. It is possible for equipotentialsurfaces to cross. (However, equipotential surfacesat different potentials cannot intersect.) For example,suppose two identical positive charges are at diagonallyopposite corners of a square and two negative charges ofequal magnitude are at the other two corners. Then theplanes perpendicular to the sides of the square at theirmidpoints are equipotential surfaces. These two planescross each other at the line perpendicular to the square atits center.13. You should use a dielectric-filled capacitor whose dielectricconstant is very large. Further, you should make thedielectric as thin as possible, keeping in mind that dielectricbreakdown must also be considered.15. (a) ii (b) i17. It would make no difference at all. An electron volt is thekinetic energy gained by an electron in being acceleratedthrough a potential difference of 1 V. A proton acceleratedthrough 1 V would have the same kinetic energy, becauseit carries the same charge as the electron (exceptfor the sign). The proton would be moving in the oppositedirection and more slowly after accelerating through1 V, due to its opposite charge and its larger mass, but itwould still gain 1 electron volt, or 1 proton volt, of kineticenergy.PROBLEMS1. (a) 6.40 10 19 J (b) 6.40 10 19 J (c) 4.00 V3. 1.4 10 20 J
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