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Unformatted text preview: Reminder: Homework Assignments: PHYS 0175 LONCAPA HW #02 Due on Sunday(@10pm), January 25 th . PHYS 0175 LONCAPA HW #03 Now Open: Due on Sunday(@10pm), February 1st Lecture 8 (Jan. 23, 2009): Illustrative examples involving Gauss Law Applying Gauss Law: cylindrical symmetry Applying Gauss Law: planar symmetry Applying Gauss Law: spherical symmetry Chapter 23: Gauss Law (contd) Gausss Law: encl E Q E dA = = r r encl E Q E dA = = r r The total electric flux through a closed surface is equal to the total (net) charge inside the surface divided by : enc q E dA = = r r Illustrative example 8.1: What is the electric flux through this disk of radius r = 0.10 m if the uniform electric field has a magnitude E = 2.0x10 3 N/C? 3 2 3 2 2 cos (2.0 10 ) ( ) cos30 (2.0 10 ) (3.14 10 ) 0.866 54 / E EA r N m C  = = = = Illustrative example 8.2: (a) Calculate the electric flux through each face of a cube that is placed inside a uniform electric field so that two of the cubes faces are perpendicular to the electric field. (b) Now rotate the cube through an angle about a vertical axis. What will now be the electric flux through each face of the cube? Illustrative example 8.3: a point charge qIllustrative example 8....
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 Spring '08
 Koehler
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