homework 04-solutions

homework 04-solutions - howard(cah3459 homework 04...

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howard (cah3459) – homework 04 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 4) 10.0 points Consider a disk oF radius 3 . 3 cm with a uni- Formly distributed charge oF +4 . 1 μ C. Compute the magnitude oF the electric feld at a point on the axis and 2 . 4 mm From the center. The value oF the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 6 . 2766 × 10 7 N / C. Explanation: Let : R =3 . 3 cm = 0 . 033 m , k e =8 . 98755 × 10 9 N · m 2 / C 2 , Q =4 . 1 μ C = 4 . 1 × 10 - 6 C , and x =2 . 4 mm = 0 . 0024 m . The surFace charge density is σ = Q π R 2 = 4 . 1 × 10 - 6 C π (0 . 033 m) 2 =0 . 00119841 C / m 2 . The feld at the distance x along the axis oF a disk with radius R is E π k e σ ± 1 - x x 2 + R 2 ² , Since 1 - x x 2 + R 2 =1 - 0 . 0024 m ³ (0 . 0024 m) 2 + (0 . 033 m) 2 . 927464 , E π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00119841 C / m 2 ) × (0 . 927464) =6 . 2766 × 10 7 N / C so ± ± E ± = 6 . 2766 × 10 7 N / C , 002 (part 2 oF 4) 10.0 points Compute the feld From the near-feld ap- proximation x ² R. Correct answer: 6 . 76749 × 10 7 N / C. Explanation: x ² R , so the second term in the parenthe- sis can be neglected and E approx e σ π ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00119841 C / m 2 ) = 6 . 76749 × 10 7 N / C close to the disk. 003 (part 3 oF 4) 10.0 points Compute the electric feld at a point on the axis and 37 cm From the center oF the disk. Correct answer: 2 . 67572 × 10 5 N / C. Explanation: Let : x = 37 cm . E e σ π (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 00119841 C / m 2 ) × ´ 1 - 0 . 37 m ³ (0 . 37 m) 2 + (0 . 033 m) 2 µ = 2 . 67572 × 10 5 N / C .
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homework 04-solutions - howard(cah3459 homework 04...

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