Physics 6th Edition Solutions Serway Chapter 15 Solutions

Physics 6th Edition Solutions Serway Chapter 15 Solutions -...

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Chapter 15 Electric Forces and Electric Fields Quick Quizzes 1. (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. (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. (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. 5. (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. 6. (a). 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. 7. (c). When a plane area A is in a uniform electric field E , the flux through that area is cos E EA θ Φ = where θ is the angle the electric field 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 0 θ = ° and ( ) ( ) 2 2 N C .00 m 20.0 N m C = 5.00 E EA Φ = = 4 . 8. (b). If 60 θ = in Quick Quiz 15.7 above, then ° ( ) ( ) ( ) 2 2 cos 5.00 N C 4.00 m cos 60 10.0 N m C EA θ = ° = E Φ = 9. (d). Gauss’s law states that the electric flux 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 ( ) 0 0 6 C E Q Φ = ∈ = − . 1
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2 CHAPTER 15 10. (b) and (d). Since the net flux 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 flux through the surface if more lines entered the surface than left it (or vise-versa).
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Electric Forces and Electric Fields 3 Answers to Even Numbered Conceptual Questions 2. 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 floor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.
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