Lemuel physics 2 5 - Wong Lemuel Homework 5 Due 7:00 pm...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 0 = 5 . 9 × 10 - 6 C 8 . 85419 × 10 - 12 C 2 / N · m 2 = 666351 N · m 2 / C . keywords: 003 (part 1 of 3) 10 points The charge per unit length on a long, straight filament is 55 μ C / m.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern