# hw4 - vo(ytv59 – homework 04 – Turner –(56705 1 This...

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Unformatted text preview: vo (ytv59) – 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 2 . 6 cm with a uni- formly distributed charge of +5 . 5 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: 1 . 31141 × 10 8 N / C. Explanation: Let : R = 2 . 6 cm = 0 . 026 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 5 . 5 μ C = 5 . 5 × 10 − 6 C , and x = 2 . 7 mm = 0 . 0027 m . The surface charge density is σ = Q π R 2 = 5 . 5 × 10 − 6 C π (0 . 026 m) 2 = 0 . 0025898 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 − . 0027 m radicalbig (0 . 0027 m) 2 + (0 . 026 m) 2 = 0 . 896709 , E = 2 π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 0025898 C / m 2 ) × (0 . 896709) = 1 . 31141 × 10 8 N / C so bardbl vector E bardbl = 1 . 31141 × 10 8 N / C , 002 (part 2 of 4) 10.0 points Compute the field from the near-field ap- proximation x ≪ R . Correct answer: 1 . 46247 × 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 . 0025898 C / m 2 ) = 1 . 46247 × 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 22 cm from the center of the disk....
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hw4 - vo(ytv59 – homework 04 – Turner –(56705 1 This...

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