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 = EA cosq , where q is the angle the electric fi eld makes with the line normal to the plane of A. If A lies in the xy-plane and E is in the z-direction, then q = 0° and Φ E = EA = 5.00 N C ()4.00 m 2 () = 20.0 N ⋅ m 2 C. 8. Choice (b). If q = 60° in Quick Quiz 15.7 above, then Φ E = EA cosq which yields Φ E = 5.00 N C ()4.00 m 2 () cos 60° ()= 10.0 N ⋅ m 2 C. 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_ch25_p001-029.indd 1 1/7/11 2:28:37 PM Show ��ǰ������
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