HW 04 - saldana(avs387 homework 04 Turner(56705 This...

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saldana (avs387) – 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 3 . 2 cm with a uni- formly distributed charge of +3 . 3 μ C. Compute the magnitude of the electric field at a point on the axis and 2 . 7 mm from the center. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 5 . 30572 × 10 7 N / C. Explanation: Let : R = 3 . 2 cm = 0 . 032 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 3 . 3 μ C = 3 . 3 × 10 6 C , and x = 2 . 7 mm = 0 . 0027 m . The surface charge density is σ = Q π R 2 = 3 . 3 × 10 6 C π (0 . 032 m) 2 = 0 . 0010258 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 . 0027 m radicalbig (0 . 0027 m) 2 + (0 . 032 m) 2 = 0 . 915924 , E = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 0010258 C / m 2 ) × (0 . 915924) = 5 . 30572 × 10 7 N / C so bardbl vector E bardbl = 5 . 30572 × 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: 5 . 79276 × 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 . 0010258 C / m 2 ) = 5 . 79276 × 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 24 cm from the center of the disk. Correct answer: 5 . 08146 × 10 5 N / C. Explanation: Let : x = 24 cm . E = 2 π k e σ = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 0010258 C / m 2 ) × parenleftBigg 1 0 . 24 m radicalbig (0 . 24 m) 2 + (0 . 032 m) 2 parenrightBigg = 5 . 08146 × 10 5 N / C .
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