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Unformatted text preview: Reilly (mar3978) – homework 04 – Turner – (56725) 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 . 2 cm with a uni formly distributed charge of +6 . 8 μ C. Compute the magnitude of the electric field at a point on the axis and 3 . 7 mm from the center. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 05656 × 10 8 N / C. Explanation: Let : R = 3 . 2 cm = 0 . 032 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 6 . 8 μ C = 6 . 8 × 10 − 6 C , and x = 3 . 7 mm = 0 . 0037 m . The surface charge density is σ = Q π R 2 = 6 . 8 × 10 − 6 C π (0 . 032 m) 2 = 0 . 00211378 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 parenleftbigg 1 − x √ x 2 + R 2 parenrightbigg , Since 1 − x √ x 2 + R 2 = 1 − . 0037 m radicalbig (0 . 0037 m) 2 + (0 . 032 m) 2 = 0 . 88514 , E = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00211378 C / m 2 ) × (0 . 88514) = 1 . 05656 × 10 8 N / C so bardbl vector E bardbl = 1 . 05656 × 10 8 N / C , 002 (part 2 of 4) 10.0 points Compute the field from the nearfield ap proximation x ≪ R . Correct answer: 1 . 19366 × 10 8 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 . 00211378 C / m 2 ) = 1 . 19366 × 10 8 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 33 cm from the center of the disk....
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This note was uploaded on 11/30/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas.
 Spring '10
 TURNER,J

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