{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1bch22 - Chapter 22 Here are sample problems that...

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

View Full Document Right Arrow Icon
Chapter 22 Here are sample problems that illustrate how to find electric fields using Gauss’ law. First, we show generally how to find electric fields due to spherically symmetric and then cylindrically symmetric charge distributions. Consider first a spherically symmetric charge distribution, where all the charge is centered around a common origin. Suppose we wish to find the electric field E ( r ) at a distance r from the center of this distribution (the origin). Then we draw a Gaussian sphere of radius r centered at this origin. Let Q encl ( r ) denote the total charge enclosed within this sphere. From symmetry, the electric field E ( r ) is the same everywhere on the surface of this sphere, and is everywhere perpendicular to the surface (directed radially outward). Therefore, the electric flux through the sphere is the product of the electric field E ( r ) and the sphere’s surface area. That is, φ E = Z S E · dA = E ( r ) · 4 πr 2 = Q encl ( r ) ² 0 E ( r ) = Q encl ( r ) 4 π² 0 r 2 (1) Written as a vector, this electric field is E ( r ) = Q encl ( r ) 4 π² 0 r 2 ˆ r (2) where ˆ r is the unit vector pointing radially outward from the common origin of the distri- bution. Consider now a cylindrically symmetric charge distribution, where each unit length of charge is centered around a common central axis, and at a given radial distance, the charge remains constant along the length of this axis. Suppose we wish to find the electric field E ( r ) at a distance r from the common central axis of this distribution. Then we draw a Gaussian cylinder of radius r and length L centered about the central axis. Let encl ( r ) denote the total charge enclosed within this cylinder. From symmetry, the electric field E ( r ) is the same everywhere on the surface of this cylinder, and is everywhere perpendicular to the surface (directed radially outward from the central axis). Therefore, the electric flux through the cylinder is the product of the electric field E ( r ) and the cylinder’s surface area. That is, φ E = Z S E · dA = E ( r ) · 2 πrL = encl ( r ) ² 0 E ( r ) = λ encl ( r ) 2 π² 0 r (3) Written as a vector, this electric field is 1
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
E ( r ) = λ encl ( r ) 2 π² 0 r ˆ r (4) where ˆ r is the unit vector pointing radially outward from the common central axis of the distribution. λ encl ( r ) is the charge per unit length enclosed within a radial distance r from the central axis. Now here are some sample problems illustrating how to apply these expressions to find the electric fields. The difficulty of each problem is on a scale from 1 to 10, 1 being least difficult, and 10 being most difficult. Each problem with a spherical distribution is immediately followed by an exactly analogous problem with a cylindrical distribution. In this way, I hope to demonstrate the parallels between the analyses for spherical and cylindrical distributions.
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