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Unformatted text preview: song (shs546) Homework 5 weathers (22202) 1 This print-out should have 12 questions. Multiple-choice 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 elec- trical field of about 165 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: 1 . 46094 10 9 C / m 2 . Explanation: Let : E =- 165 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 = (- 165 N / C) ( 8 . 85419 10 12 C 2 / N m 2 ) =- 1 . 46094 10 9 C / m 2 , with a magnitude of 1 . 46094 10 9 C / m 2 . 002 (part 1 of 2) 10.0 points A conducting spherical shell having an inner radius of 3 . 6 cm and outer radius of 4 . 1 cm carries a net charge of 4 . 1 C. A conducting sphere is placed at the center of this shell having a radius of 0 . 7 cm carring a net charge of 2 . 5 C. . 7 cm 3 . 6 cm , Q 2 inside 4 . 1 cm Q 2 outside 2 . 5 C 4 . 1 C P Determine the surface charge density on the inner surface of the shell. Correct answer:- . 000153506 C / m 2 . Explanation: Let : r = 0 . 7 cm , not required a = 3 . 6 cm = 0 . 036 m , b = 4 . 1 cm = 0 . 041 m , and q = 2 . 5 C = 2 . 5 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 . 5 10 6 C 4 (0 . 036 m) 2 =- . 000153506 C / m 2 . 003 (part 2 of 2) 10.0 points song (shs546) Homework 5 weathers (22202) 2 Determine the surface charge density on the outer surface of the shell. Correct answer: 0 . 00031244 C / m 2 ....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
- Spring '00