SM_chapter24

SM_chapter24 - 24 Gausss Law CHAPTER OUTLINE 24.1 24.2 24.3...

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24 Gauss’s Law CHAPTER OUTLINE 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions 24.4 Conductors in Electrostatic Equilibrium ANSWERS TO QUESTIONS Q24.1 The luminous fl ux on a given area is less when the sun is low in the sky, because the angle between the rays of the sun and the local area vector, d r A , is greater than zero. The cosine of this angle is reduced. The decreased fl ux results, on the average, in colder weather. Q24.2 The surface must enclose a positive total charge. Q24.3 The net fl ux through any gaussian surface is zero. We can argue it two ways. Any surface contains zero charge, so Gauss’s law says the total fl ux is zero. The F eld is uniform, so the F eld lines entering one side of the closed surface come out the other side and the net fl ux is zero. *Q24.4 (i) Equal amounts of fl ux pass through each of the six faces of the cube. Answer (e). (ii) Move the charge to very close below the center of one face, through which the fl ux is then q /2 0 . Answer (c). (iii) Move the charge onto one of the cube faces. Then the F eld has no component perpendicular to this face and the fl ux is zero. Answer (a). *Q24.5 (i) Answer (a). (ii) the fl ux is zero through the two faces pierced by the F lament. Answer (b). *Q24.6 (i) Answer (a). (ii) The fl ux is nonzero through the top and bottom faces, and zero through the other four faces. Answer (c). *Q24.7 (i) Both spheres create equal F elds at exterior points, like particles at the centers of the spheres. Answer (c). (ii) The F eld within the conductor is zero. The F eld within the insulator is 4/5 of its surface value. Answer (f). Q24.8 Gauss’s law cannot tell the different values of the electric F eld at different points on the surface. When E is an unknown number, then we can say Ed A E d A cos cos θθ ∫∫ = . When Exyz ,, ( ) is an unknown function, then there is no such simpliF cation. Q24.9 The electric fl ux through a sphere around a point charge is independent of the size of the sphere. A sphere of larger radius has a larger area, but a smaller F eld at its surface, so that the product of F eld strength and area is independent of radius. If the surface is not spherical, some parts are closer to the charge than others. In this case as well, smaller projected areas go with stronger F elds, so that the net fl ux is unaffected. 27
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28 Chapter 24 Q24.10 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface of the conductor. *Q24.11 (a) Let q represent the charge of the insulating sphere. The F eld at A is (4/5) 3 q /[4 p (4 cm) 2 0 ]. The F eld at B is q /[4 p (8 cm) 2 0 ]. The F eld at C is zero. The F eld at D is q /[4 p (16 cm) 2 0 ]. The ranking is A > B > D > C.
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This note was uploaded on 01/19/2012 for the course PHY 232 taught by Professor Williams,frank during the Spring '11 term at Ohio State.

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SM_chapter24 - 24 Gausss Law CHAPTER OUTLINE 24.1 24.2 24.3...

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