17.59. Set Up: The method of Example 17.10 shows that the electric field outside the sphere is the same as for a point charge of the same charge located at the center of the sphere. The charge of an electron has magnitude Solve: (a) For so The number of excess electrons is (b) 17.60. Set Up: The charge distribution has spherical symmetry, so the electric field is radial and depends only on the distance from the center of the charged sphere. The surface area of a sphere is Solve: (a) Apply Gauss’s law to a spherical surface of radius concentric with the sphere of charge. The elec-tric field is constant over the Gaussian surface and perpendicular to it, so The charge enclosed is Q, so Gauss’s law gives and This is the same E as for a point charge Q at the center of the sphere. (b) No direction is preferred, so 17.61. Set Up: The charge distribution has spherical symmetry, so the electric field, if nonzero, is radial and depends only on the distance from the center of the shell. Solve: (a)
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.