{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Act.02.02 - but what surface should we choose(Hint remember...

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

View Full Document Right Arrow Icon
Physics 2102 Section 002 September 8, 2010 Activity 1 The figure shows four solid spheres, each with charge Q uniformly distributed through its volume. Rank the spheres according to the magnitude of the electric field they produce at the point P: (i) all tie (ii) (a) = (b) > (c) = (d) (iii) (a) = (b) > (c) > (d) (iv) (a) > (b) > (c) > (d) (v) (d) > (c) > (b) > (a) Activity 2 The figure shows a cross section of a spherical metal shell of inner radius R. A point charge of –5.0 μ C is located at a distance R/2 from the center of the shell. The shell is electrically neutral. Our goal is to find the magnitudes and signs of the (induced) charges on the inner and outer surfaces of the metal shell. (a) First, let’s find the (induced) charge on the inner surface of the metal shell. We will use Gauss’ law –
Image of page 1

Info icon This 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.

Unformatted text preview: but what surface should we choose? (Hint: remember that the electric field inside a conductor is zero). Once you have chosen this surface, it is easy to see what the induced charge on the inner surface must be (use Gauss’ law with the surface from (a)). (b) Now you can calculate the charge on the outer surface (Hint: Remember, the metal shell is electrically neutral). (c) Next consider a similar problem. A conducting spherical shell has an excess charge of +10 C. A point charge of -15 C is placed at the center of the shell. Use Gauss’ law to calculate the charge on the inner and outer surfaces of the shell, choose one of the answers below: (i) Inner surface +15 C, Outer surface +10 C (ii) Inner surface 0, Outer surface +10 C (iii) Inner surface +15 C, Outer surface -5 C...
View Full 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