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Unformatted text preview: segovia (cs39966) Homework 2 Spurlock (14441) 1 This printout should have 58 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A charge of 5 . 6 C is at the geometric center of a cube. What is the electric flux through one of the faces? Correct answer: 1 . 05412 10 5 N m 2 / C. Explanation: Let : q = 5 . 6 C = 5 . 6 10 6 C . By Gauss law, = contintegraldisplay vector E d vector A = q The total flux through the cube is given by tot = q = 5 . 6 10 6 C 8 . 85419 10 12 C 2 / N m 2 = 6 . 32469 10 5 N m 2 / C . The flux through one side of the cube is then = 1 6 tot = 1 . 05412 10 5 N m 2 / C . 002 10.0 points A closed surface with dimensions a = b = . 269 m and c = 0 . 2421 m is located as in the figure. The electric field throughout the region is nonuniform and given by vector E = ( + x 2 ) where x is in meters, = 4 N / C, and = 6 N / (C m 2 ). E y x z a c b a What is the magnitude of the net charge enclosed by the surface? Correct answer: 7 . 26022 10 13 C. Explanation: Let : a = b = 0 . 269 m , c = 0 . 2421 m , = 4 N / C , and = 6 N / (C m 2 ) . The electric field throughout the region is directed along the xaxis and the direction of d vector A is perpendicular to its surface. Therefore, vector E is parallel to d vector A over the four faces of the surface which are perpendicular to the yz plane, and vector E is perpendicular to d vector A over the two faces which are parallel to the yz plane. That is, only the left and right sides of the right rectangular parallel piped which encloses the charge will contribute to the flux. The net electric flux through the cube is = integraldisplay right side E x dA  integraldisplay left side E x dA = a b bracketleftbig + ( a + c ) 2  a 2 bracketrightbig = a b (2 a c + c 2 ) = a b c (2 a + c ) = (0 . 269 m) (0 . 269 m) (0 . 2421 m) [6 N / (C m 2 )] [2 (0 . 269 m) + 0 . 2421 m] = 0 . 0819975 N m 2 / C , so the enclosed charge is q = = [8 . 85419 10 12 C 2 / (N m 2 )] segovia (cs39966) Homework 2 Spurlock (14441) 2 (0 . 0819975 N m 2 / C) = 7 . 26022 10 13 C . 003 10.0 points Consider a long, uniformly charged, cylindri cal insulator of radius R with charge density 1 . 6 C / m 3 . (The volume of a cylinder with radius r and length is V = r 2 .) The value of the Permittivity of free space is 8 . 85419 10 12 C 2 / N m 2 R 2 . 8 cm What is the magnitude of the electric field inside the insulator at a distance 2 . 8 cm from the axis (2 . 8 cm < R )?...
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 Fall '11
 HOFFMAN
 Charge, Work

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