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# HW 6 - terry(ect328 homework 06 Turner(59130 This print-out...

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terry (ect328) – homework 06 – Turner – (59130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 106 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 7 . 9 × 10 5 N · m 2 / C. What is the electric field strength? Correct answer: 8 . 95211 × 10 5 N / C. Explanation: Let : r = 53 cm = 0 . 53 m and Φ = 7 . 9 × 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 = 7 . 9 × 10 5 N · m 2 / C π (0 . 53 m) 2 = 8 . 95211 × 10 5 N / C . 002 10.0 points A (8 . 8 m by 8 . 8 m) square base pyramid with height of 6 . 35 m is placed in a vertical electric field of 53 N / C. 8 . 8 m 6 . 35 m 53 N / C Calculate the total electric flux which goes out through the pyramid’s four slanted sur- faces. Correct answer: 4104 . 32 N m 2 / C. Explanation: Let : s = 8 . 8 m , h = 6 . 35 m , and E = 53 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 = (53 N / C) (8 . 8 m) 2 = 4104 . 32 N m 2 / C . 003 10.0 points A spherical shell of radius 7 . 5 m is placed in a uniform electric field with magnitude 1070 N / C. Determine the total electric flux through the shell. Correct answer: 0 N · m 2 / C. Explanation: Let : r = 7 . 5 m and E = 1070 N / C .

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