Solution 6 - homework 06 KANUNGO ANIL Due 4:00 am Question...

Info icon This preview shows pages 1–3. Sign up to view the full content.

homework 06 – KANUNGO, ANIL – Due: Sep 16 2007, 4:00 am 1 Question 1, chap 24, sect 1. part 1 of 1 10 points A spherical shell of radius 8 . 2 m is placed in a uniform electric field with magnitude 4130 N / C. Determine the total electric flux through the shell. Correct answer: 0 N · m 2 / C (tolerance ± 1 %). Explanation: Let : r = 8 . 2 m and E = 4130 N / C . The uniform field enters the shell on one side and exits on the other, so the total flux is Φ = contintegraldisplay vector E · d vector A = 0 . Question 2, chap 24, sect 3. part 1 of 2 10 points A thin spherical shell of radius 5 . 16 m has a total charge of 6 . 45 C distributed uniformly over its surface. Let : k e = 8 . 988 × 10 9 N · m 2 / C 2 . 5 . 16 m vector E + + + + + + + + + + + + + + + + + + + + Find the electric field vector E 9 . 9 m from the center of the shell (outside the shell). Correct answer: 5 . 91497 × 10 8 N / C (tolerance ± 1 %). Explanation: Let : a = 5 . 16 m , Q = 6 . 45 C , and r = 9 . 9 m . If we construct a spherical Gaussian surface of radius r > a , concentric with the shell, then the charge inside this surface is Q . Therefore the field at a point outside the shell is equiva- lent to that of a point charge Q at the center. For r > a , E 1 = k Q r 2 = (8 . 988 × 10 9 N · m 2 / C 2 )(6 . 45 C) (9 . 9 m) 2 = 5 . 91497 × 10 8 N / C . Question 3, chap 24, sect 3. part 2 of 2 10 points Find the electric field E 2 2 . 53 m from the center of the shell (inside the shell). Correct answer: 0 N / C (tolerance ± 1 %). Explanation: The electric field inside the spherical shell is zero. This also follows from Gauss’ law ap- plied to a spherical surface of radius r < a . Since the net charge inside the surface is zero, and because of the spherical symmetry of the charge distribution, application of Gauss’ law shows that E = 0 in the region r < a . The same result can be obtained using Coulomb’s law and integrating over the charge distri- bution. This calculation is, however, rather complicated. Question 4, chap 24, sect 2. part 1 of 3 10 points Consider a charge q located at the origin and at the corner of a cube as shown in the figure. The sides of the cube have length a . y x z a
Image of page 1

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

homework 06 – KANUNGO, ANIL – Due: Sep 16 2007, 4:00 am 2 Find the total flux due to the charge q .
Image of page 2
Image of page 3
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