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Unformatted text preview: marout (ssm979) – 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 2 . 9 cm with a uni formly distributed charge of +5 . 6 μ 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: 1 . 05754 × 10 8 N / C. Explanation: Let : R = 2 . 9 cm = 0 . 029 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 5 . 6 μ C = 5 . 6 × 10 − 6 C , and x = 3 . 4 mm = 0 . 0034 m . The surface charge density is σ = Q π R 2 = 5 . 6 × 10 − 6 C π (0 . 029 m) 2 = 0 . 00211954 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 − . 0034 m radicalbig (0 . 0034 m) 2 + (0 . 029 m) 2 = 0 . 883556 , E = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00211954 C / m 2 ) × (0 . 883556) = 1 . 05754 × 10 8 N / C so bardbl vector E bardbl = 1 . 05754 × 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 . 19692 × 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 . 00211954 C / m 2 ) = 1 . 19692 × 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 48 cm from the center of the disk....
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 Fall '10
 TURNER,J
 Electrostatics, Electric charge, Erod

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