Chapter 15

# Chapter 15 - CHAPTER 15 Quick Quizzes 1(d Object A could...

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C HAPTER 15 Quick Quizzes 1. (d). Object A could possess a net charge whose sign is opposite that of the excess charge on B. If object A is neutral, B would also attract it by creating an induced charge on the surface of A. This situation is illustrated in Figure 15.5 of the textbook. 2. (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other. 3. (c). The electric field 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 field exerts on the test change is reversed when the sign of the test charge is changed. 4. (b). The magnitude of the upward electrical force must equal the weight of the ball. That is: , so qE m g = ( ) ( ) 32 4 6 5. 01 0 kg 9.80 m s 1. 21 0 NC 4. 0 C mg E q × == × 5. (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 field at this location, must then be zero. 6. (c) and (d) The electron and the proton have equal magnitude charges of opposite signs. The forces exerted on these particles by the electric field 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. 7. A, B, and C . The field is greatest at point A because this is where the field lines are closest together. The absence of lines at point C indicates that the electric field there is zero. 8. Statements (b) and (d) are true and follow from Gauss’s law. Statement (a) is not necessarily true because Gauss’s law says that the net flux through any closed surface equals the net charge inside the surface divided by 0 . For example, a positive and a negative charge could be inside the surface. Statement (c) is not necessarily true. Although the net flux through the surface is zero, the electric field in that region may not be zero. 3

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CHAPTER 15 Problem Solutions 15.1 Since the charges have opposite signs, the force is one of attraction . Its magnitude is ( )( ) () 99 2 12 9 2 22 4. 51 0 C2 . 81 0 C Nm 8.99 10 C 3.2 m e kqq F r −− ×× ⎛⎞ == × = 8 1. 11 0 N × ⎜⎟ ⎝⎠ 15.2 (a) The force is ( ) ( ) 2 14 2 2 9 2 6 1.60 10 C C 2. 0 m e kq F r × == × = ( × ) 7 3. 71 0 N rep u lsion × (b) With r two times larger, is four times larger and the force is 2 r red u ced to one fourth of its previous value . 15.3 ( ) 2 27 9 e ke e F r = ( ) ( ) 2 19 2 9 2 2 158 1.60 C C 01 m × = × ( ) 91 N rep u lsion 15.4 The electrical force is 2 2 e e ke F r = , and the gravitational force is 2 2 g Gm F r = . Thus, if ge FF = , the mass would have to be ( ) 92 2 2 2 10 N m C C 6.67 N m k e k me Gg ×⋅ ==× = 9 1.86 kg × This result is times greater than the actual mass of a proton.
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Chapter 15 - CHAPTER 15 Quick Quizzes 1(d Object A could...

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