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Unformatted text preview: mcconnell (kam2342) – homework06 – Turner – (60230) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 68 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 6 . 2 × 10 5 N · m 2 / C. What is the electric field strength? Correct answer: 1 . 7072 × 10 6 N / C. Explanation: Let : r = 34 cm = 0 . 34 m and Φ = 6 . 2 × 10 5 N · m 2 / C . By Gauss’ law, Φ = contintegraldisplay vector E · d vector A The position of maximum electric flux will be that position in which the plane of the loop is perpendicular to the electric field; i.e. , when vector E · d vector A = E dA . Since the field is constant, Φ = E A = Eπ r 2 E = Φ π r 2 = 6 . 2 × 10 5 N · m 2 / C π (0 . 34 m) 2 = 1 . 7072 × 10 6 N / C . 002 10.0 points A (6 . 84 m by 6 . 84 m) square base pyramid with height of 5 . 45 m is placed in a vertical electric field of 38 . 5 N / C. b 6 . 84 m 5 . 45 m 38 . 5 N / C Calculate the total electric flux which goes out through the pyramid’s four slanted sur faces. Correct answer: 1801 . 25 N m 2 / C. Explanation: Let : s = 6 . 84 m , h = 5 . 45 m , and E = 38 . 5 N / C . By Gauss’ law, Φ = vector E · vector A Since there is no charge contained in the pyra mid, the net flux through the pyramid must be 0 N/C. Since the field is vertical, the flux through the base of the pyramid is equal and opposite to the flux through the four sides. Thus we calculate the flux through the base of the pyramid, which is Φ = E A = E s 2 = (38 . 5 N / C) (6 . 84 m) 2 = 1801 . 25 N m 2 / C . 003 10.0 points A spherical shell of radius 1 . 2 m is placed in a uniform electric field with magnitude 6020 N / C....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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