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HW4Sol - Platt David Homework 4 Due 4:00 am Inst Ken Shih...

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Platt, David – Homework 4 – Due: Sep 14 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Again! For 005, you need to answer in unit of uC (miu Coulomb,*E-6C). 001 (part 1 of 3) 10 points An electric field of magnitude 3450 N / C is applied along the x axis. Calculate the electric flux through a rect- angular plane 0 . 214 m wide and 0 . 41 m long if the plane is parallel to the yz plane. Correct answer: 302 . 703 N · m 2 / C. Explanation: Let : E = 3450 N / C , θ = 0 , L = 0 . 41 m , and W = 0 . 214 m . The area of the rectangle is A = L W = (0 . 41 m) (0 . 214 m) = 0 . 08774 m 2 . Since the field is uniform,the flux is φ = E · A = E A cos θ = (3450 N / C) (0 . 08774 m 2 ) (cos 0 ) = 302 . 703 N · m 2 / C . 002 (part 2 of 3) 10 points Calculate the electric flux through the same rectangle, if it is parallel to the xy plane. Correct answer: 0 N · m 2 / C. Explanation: The angle is θ = 90 ,cos θ = 0 , and φ = E · A = E A cos θ = (3450 N / C) (0 . 08774 m 2 ) (0) = 0 N · m 2 / C . 003 (part 3 of 3) 10 points Calculate the electric flux through the same rectangle, but now the rectangle contains the y axis and its normal makes an angle of 26 with the x axis. Correct answer: 272 . 068 N · m 2 / C. Explanation: Let θ = 26 φ = E · A = E A cos θ = (3450 N / C) (0 . 08774 m 2 ) (cos 26 ) = 272 . 068 N · m 2 / C . 004 (part 1 of 1) 10 points A (9 . 31 m by 9 . 31 m) square base pyramid with height of 3 . 32 m is placed in a vertical electric field of 59 . 9 N / C. 9 . 31 m 3 . 32 m 59 . 9 N / C Calculate the total electric flux which goes out through the pyramid’s four slanted sur- faces. Correct answer: 5191 . 9 N m 2 / C. Explanation: Let : s = 9 . 31 m , h = 3 . 32 m , and E = 59 . 9 N / C . By Gauss’ law, Φ = ~ E · ~ 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.
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