solutions 3 - Patel, Kinal – Homework 3 – Due: Jan 29...

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Unformatted text preview: Patel, Kinal – Homework 3 – Due: Jan 29 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points A uniformly charged ring of radius 9 . 1 cm has a total charge of 88 μ C. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Find the magnitude of the electric field on the axis of the ring at 0 . 72 cm from the center of the ring. Correct answer: 7 . 4863 × 10 6 N / C. Explanation: Let : a = 9 . 1 cm = 0 . 091 m , Q = 88 μ C = 8 . 8 × 10- 5 C , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , and x = 0 . 72 cm = 0 . 0072 m . The electric field is E = k e xQ ( x 2 + a 2 ) 3 / 2 = (8 . 98755 × 10 9 N · m 2 / C 2 )(0 . 0072 m) (0 . 0072 m 2 + 0 . 091 m 2 ) 3 / 2 × (8 . 8 × 10- 5 C) = 7 . 4863 × 10 6 N / C . 002 (part 2 of 4) 0 points Find the magnitude of the electric field on the axis of the ring at 6 . 89 cm from the center of the ring. Correct answer: 3 . 66451 × 10 7 N / C. Explanation: Let : x = 6 . 89 cm = 0 . 0689 m . E = (8 . 98755 × 10 9 N · m 2 / C 2 )(0 . 0689 m) (0 . 0689 m 2 + 0 . 091 m 2 ) 3 / 2 × (8 . 8 × 10- 5 C) = 3 . 66451 × 10 7 N / C . 003 (part 3 of 4) 0 points Find the magnitude of the electric field on the axis of the ring at 16 cm from the center of the ring. Correct answer: 2 . 02913 × 10 7 N / C. Explanation: Given : x = 16 cm = 0 . 16 m . E = (8 . 98755 × 10 9 N · m 2 / C 2 )(0 . 16 m) (0 . 16 m 2 + 0 . 091 m 2 ) 3 / 2 × (8 . 8 × 10- 5 C) = 2 . 02913 × 10 7 N / C . 004 (part 4 of 4) 10 points Find the magnitude of the electric field on the axis of the ring at 136 . 985 cm from the center of the ring. Correct answer: 418706 N / C. Explanation: Let : x = 136 . 985 cm . E = (8 . 98755 × 10 9 N · m 2 / C 2 )(1 . 36985 m) (1 . 36985 m 2 + 0 . 091 m 2 ) 3 / 2 × (8 . 8 × 10- 5 C) = 418706 N / C . Comment: The interesting thing to note here is that the field first increases from (a) to (b), then decreases from (b) to (c) and also from (c) to (d). Thus, we should have a maximum somewhere between x = 0 . 72 cm and x = 16 cm. In fact, the maximum is at= 16 cm....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.

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solutions 3 - Patel, Kinal – Homework 3 – Due: Jan 29...

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