Lemuel physics 2 5

Lemuel physics 2 5 - Wong Lemuel – Homework 5 – Due...

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Unformatted text preview: Wong, Lemuel – Homework 5 – Due: Feb 21 2007, 7:00 pm – Inst: Matt Mackie 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. NOTE THE SPECIAL DUE TIME: HW is due 7pm central time, which is 8pm our time. 001 (part 1 of 1) 10 points A 50 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 405000 N · m 2 / C. What is the electric field strength? Correct answer: 2 . 06265 × 10 6 N / C. Explanation: Let : r = 25 cm = 0 . 25 m and Φ = 405000 N · m 2 / C . By Gauss’ law, Φ = I ~ E · d ~ A The position of maximum electric flux will be that position in which the plane of the loop is perpendicular to the electric field; i.e. , when ~ E · d ~ A = E dA . Since the field is constant, Φ = E A = Eπ r 2 E = Φ π r 2 = 405000 N · m 2 / C π (0 . 25 m) 2 = 2 . 06265 × 10 6 N / C . keywords: 002 (part 1 of 1) 10 points A point charge 5 . 9 μ C is located at the center of a uniform ring having linear charge density 10 . 5 μ C / m and radius 1 . 48 m. q R a λ Determine the total electric flux through a sphere centered at the point charge and having radius R , where R < a , as shown. Correct answer: 666351 N · m 2 / C. Explanation: Let : q = 5 . 9 μ C = 5 . 9 × 10- 6 C , λ = 10 . 5 μ C / m = 1 . 05 × 10- 5 C / m , and a = 1 . 48 m . Only the charge inside the sphere with ra- dius R contributes to the total flux. Accord- ing to the Gauss’s Law, Φ = q ² = 5 . 9 × 10- 6 C 8 . 85419 × 10- 12 C 2 / N · m 2 = 666351 N · m 2 / C ....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Lemuel physics 2 5 - Wong Lemuel – Homework 5 – Due...

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