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Unformatted text preview: howard (cah3459) – 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 3 . 3 cm with a uni formly distributed charge of +4 . 1 μ C. Compute the magnitude of the electric field at a point on the axis and 2 . 4 mm from the center. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 6 . 2766 × 10 7 N / C. Explanation: Let : R = 3 . 3 cm = 0 . 033 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 4 . 1 μ C = 4 . 1 × 10 6 C , and x = 2 . 4 mm = 0 . 0024 m . The surface charge density is σ = Q π R 2 = 4 . 1 × 10 6 C π (0 . 033 m) 2 = 0 . 00119841 C / m 2 . The field at the distance x along the axis of a disk with radius R is E = 2 π k e σ 1 x √ x 2 + R 2 , Since 1 x √ x 2 + R 2 = 1 . 0024 m (0 . 0024 m) 2 + (0 . 033 m) 2 = 0 . 927464 , E = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00119841 C / m 2 ) × (0 . 927464) = 6 . 2766 × 10 7 N / C so E = 6 . 2766 × 10 7 N / C , 002 (part 2 of 4) 10.0 points Compute the field from the nearfield ap proximation x R . Correct answer: 6 . 76749 × 10 7 N / C. Explanation: x R , so the second term in the parenthe sis can be neglected and E approx = 2 π k e σ = 2 π ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00119841 C / m 2 ) = 6 . 76749 × 10 7 N / C close to the disk....
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This note was uploaded on 09/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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