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Unformatted text preview: Platt, David Homework 5 Due: Sep 16 2005, 4:00 am Inst: Ken Shih 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Note: 1,Always give your answer using SI unit system unless specifically pointed out. Consider it as a rule. 2,For problems 001 and 002, we assume that the filament is long enough compared to the cylinder that field lines emerge from the filament only radially and the field doesnt change appreciably over the surface of the cylinder. 001 (part 1 of 2) 10 points A uniformly charged, straight filament 7 m in length has a total positive charge of 2 C. An uncharged cardboard cylinder 1 cm in length and 5 cm in radius surrounds the filament using the filament as its axis of symmetry, with the filament as the central axis of the cylinder. Find the total electric flux through the cylinder. Correct answer: 322 . 695 N m 2 / C. Explanation: Let : L = 7 m , r = 5 cm = 0 . 05 m , l = 1 cm = 0 . 01 m , and Q = 2 C = 2 10 6 C . Calculate the flux through the cylinder us ing Gauss law. The flux through a closed surface is = q enclosed . Our Gaussian surface will be the cardboard tube, but we will close off the ends of the cylinder for our imaginary surface. We will assume that the filament is long enough (com pared tp the cylinder) that field lines emerge from the filament only radially they do not penetrate the caps on our Gaussian surface. The charge which the cylinder encloses is equal to the charge density of the filament times the length of the cylinder l q = l = Q L l . where Q is the total charge on the filament and L is the length of the filament. Sub stituting this result into Gauss law, we see that = Ql L = (2 10 6 C)(0 . 01 m) (8 . 854 10 12 C 2 / N m 2 )(7 m) = 322 . 695 N m 2 / C . 002 (part 2 of 2) 10 points What is the electric field at the surface of the cylinder? Correct answer: 102717 V / m. Explanation: Gausss law can also tell us the field at the surface of the cylinder. We know that = I ~ E d ~ A where the integral is over the area through which the field penetrates. We will assume that the field does not change appreciably...
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This note was uploaded on 04/10/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Work

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