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hw4solution_pdf - klarin(sjk772 homework 04 Turner(56705...

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klarin (sjk772) – homework 04 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice 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 . 9 cm with a uni- formly distributed charge of +5 μ C. Compute the magnitude of the electric field at a point on the axis and 3 . 2 mm from the center. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 3 . 49932 × 10 7 N / C. Explanation: Let : R = 4 . 9 cm = 0 . 049 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 5 μ C = 5 × 10 6 C , and x = 3 . 2 mm = 0 . 0032 m . The surface charge density is σ = Q π R 2 = 5 × 10 6 C π (0 . 049 m) 2 = 0 . 000662869 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 0 . 0032 m radicalbig (0 . 0032 m) 2 + (0 . 049 m) 2 = 0 . 934833 , E = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 000662869 C / m 2 ) × (0 . 934833) = 3 . 49932 × 10 7 N / C so bardbl vector E bardbl = 3 . 49932 × 10 7 N / C , 002 (part 2 of 4) 10.0 points Compute the field from the near-field ap- proximation x R . Correct answer: 3 . 74325 × 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 . 000662869 C / m 2 ) = 3 . 74325 × 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 30 cm from the center of the disk. Correct answer: 4 . 89535 × 10 5 N / C. Explanation: Let : x = 30 cm . E = 2 π k e σ = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 000662869 C / m 2 ) × parenleftBigg 1 0 . 3 m radicalbig (0 . 3 m) 2 + (0 . 049 m) 2 parenrightBigg = 4 . 89535 × 10 5 N / C .
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