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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 (cont’d) Gauss’s Law: encl E Q E dA ε Φ = • = ∫ r r Ñ encl E Q E dA ε Φ = • = ∫ r r Ñ The total electric flux through a closed surface is equal to the total (net) charge inside the surface divided by ε : enc q E dA ε Φ = • = ∫ r r Ñ 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 φ π Φ = = × × × = × × × × = ⋅ 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 cube’s faces are perpendicular to the electric field. (b) Now rotate the cube through an angle θ 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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This note was uploaded on 04/06/2009 for the course PHYS 0175 taught by Professor Koehler during the Spring '08 term at Pittsburgh.
 Spring '08
 Koehler
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