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Unformatted text preview: linares (jl36797) homework 04 Turner (56705) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points Consider a disk of radius 4 . 5 cm with a uni formly distributed charge of +4 . 3 C. Compute the magnitude of the electric field at a point on the axis and 3 . 4 mm from the center. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . Correct answer: 3 . 52936 10 7 N / C. Explanation: Let : R = 4 . 5 cm = 0 . 045 m , k e = 8 . 98755 10 9 N m 2 / C 2 , Q = 4 . 3 C = 4 . 3 10 6 C , and x = 3 . 4 mm = 0 . 0034 m . The surface charge density is = Q R 2 = 4 . 3 10 6 C (0 . 045 m) 2 = 0 . 000675917 C / m 2 . The field at the distance x along the axis of a disk with radius R is E = 2 k e parenleftbigg 1 x x 2 + R 2 parenrightbigg , Since 1 x x 2 + R 2 = 1 . 0034 m radicalbig (0 . 0034 m) 2 + (0 . 045 m) 2 = 0 . 924659 , E = 2 (8 . 98755 10 9 N m 2 / C 2 ) (0 . 000675917 C / m 2 ) (0 . 924659) = 3 . 52936 10 7 N / C so bardbl vector E bardbl = 3 . 52936 10 7 N / C , 002 (part 2 of 4) 10.0 points Compute the field from the nearfield ap proximation x R . Correct answer: 3 . 81694 10 7 N / C. Explanation: x R , so the second term in the parenthe sis can be neglected and E approx = 2 k e = 2 ( 8 . 98755 10 9 N m 2 / C 2 ) (0 . 000675917 C / m 2 ) = 3 . 81694 10 7 N / C close to the disk. 003 (part 3 of 4) 10.0 points Compute the electric field at a point on the axis and 26 cm from the center of the disk....
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This note was uploaded on 09/07/2010 for the course PHY 57970 taught by Professor Antoniewicz during the Spring '10 term at University of Texas at Austin.
 Spring '10
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