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Unformatted text preview: lim (kl9356) Homework 5 Weathers (22202) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points On a clear, sunny day, there is a vertical electrical field of about 54 N / C pointing down over flat ground or water. What is the magnitude of the surface charge density on the ground for these conditions? Correct answer: 4 . 78126 10 10 C / m 2 . Explanation: Let : E = 54 N / C . By Gauss law, contintegraldisplay vector E d vector A = q . If we consider the electric field as being im mediately above the ground or water, then we can think of the ground or water as an infi nite sheet of charge. The question is, should the electric field beneath the surface be the same as the field above? Should it be zero? The real answer is quite complicated. In or der to work this problem, we assume the field is zero beneath the surface. This makes sense: since water is a conductor, the field inside the water should be zero. Using Gauss Law, E A = A , we have = E = ( 54 N / C) ( 8 . 85419 10 12 C 2 / N m 2 ) = 4 . 78126 10 10 C / m 2 , with a magnitude of 4 . 78126 10 10 C / m 2 . 002 (part 1 of 2) 10.0 points A conducting spherical shell having an inner radius of 2 . 6 cm and outer radius of 3 . 5 cm carries a net charge of 13 . 4 C. A conducting sphere is placed at the center of this shell having a radius of 1 . 3 cm carring a net charge of 2 . 8 C. 1 . 3 cm 2 . 6 cm , Q 2 inside 3 . 5 cm Q 2 outside 2 . 8 C 13 . 4 C P Determine the surface charge density on the inner surface of the shell. Correct answer: . 000329611 C / m 2 . Explanation: Let : r = 1 . 3 cm , not required a = 2 . 6 cm = 0 . 026 m , b = 3 . 5 cm = 0 . 035 m , and q = 2 . 8 C = 2 . 8 10 6 C . Basic Concept: E = 0 inside a Conductor. Gauss Law, contintegraldisplay vector E d vector A = Q . Solution: Since the electric field is zero inside any conductor in electrostatic equi librium, the net charge is zero inside any spherical Gaussian surface of radius r , where a < r < b . Thus the charge on the inner sur face of the sphere must be q . If we call the charge density on the inner surface in , then q = 4 a 2 in in = q 4 a 2 = 2 . 8 10 6 C 4 (0 . 026 m) 2 = . 000329611 C / m 2 . 003 (part 2 of 2) 10.0 points Determine the surface charge density on the outer surface of the shell. Correct answer: 0 . 00105237 C / m 2 . lim (kl9356) Homework 5 Weathers (22202) 2 Explanation: Let : Q = 13 . 4 C = 1 . 34 10 5 C ....
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 Spring '00
 Littler
 Magnetism, Work

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