Homework 3-solutions - soto(rrs946 Homework 3 Homann(56605...

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

View Full Document Right Arrow Icon
soto (rrs946) – Homework 3 – Hoffmann – (56605) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points In a particular region of the earth’s atmo- sphere, the electric field above the earth’s surface has been measured to be 152 . 5 N / C downward at an altitude of 270 m and 173 N / C downward at an altitude of 413 m. The permittivity of free space is 8 . 85 × 10 12 C 2 / N · m 2 . Neglecting the curvature of the earth, cal- culate the volume charge density of the at- mosphere assuming it to be uniform between 270 m and 413 m. Correct answer: - 1 . 26871 × 10 12 C / m 3 . Explanation: Let : 1 = 270 m , 2 = 413 m , E l 1 = 152 . 5 N / C , E l 2 = 173 N / C , and ǫ 0 = 8 . 85 × 10 12 C 2 / N · m 2 . Choose a closed cylinderical surface for this portion of the atmosphere. The total flux is integraldisplay S E n dA = Q net ǫ 0 ( E l 1 - E l 2 ) A = Q net ǫ 0 Q net = ( E l 1 - E l 2 ) A ǫ 0 . Thus the volume charge density is ρ = Q net V = ( E l 1 - E l 2 ) A ǫ 0 A l 1 = ( E l 1 - E l 2 ) ǫ 0 2 - 1 = 152 . 5 N / C - 173 N / C 413 m - 270 m × (8 . 85 × 10 12 C 2 / N · m 2 ) = - 1 . 26871 × 10 12 C / m 3 . The curvature of the earth can be neglected because the maximum helight is < 0 . 01% of the radius of the earth. 002 10.0points A charge of 5 . 6 μ C is 20 cm above the center of a square of side length 40 cm. The permittivity of free space is 8 . 85 × 10 12 C 2 / N · m 2 . Find the flux through the square. Correct answer: 1 . 05461 × 10 5 N · m 2 / C. Explanation: Let : q = 5 . 6 μ C = 5 . 6 × 10 6 C , = 20 cm = 0 . 2 m , s = 40 cm = 0 . 4 m , and ǫ 0 = 8 . 85 × 10 12 C 2 / N · m 2 . The point charge is at the center of the cube. Applying Gauss’ law, the flux through one face of the square is Φ square = 1 6 Φ total = 1 6 q inside ǫ 0 = 1 6 parenleftbigg 5 . 6 × 10 6 C 8 . 85 × 10 12 C 2 / N · m 2 parenrightbigg = 1 . 05461 × 10 5 N · m 2 / C . keywords: 003 10.0points A point charge 6 . 1 μ C is located at the center of a uniform ring having linear charge density 4 . 9 μ C / m and radius 1 . 21 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