Introduction: Gauss's law quantifies the idea that the amount of electric field emanating from a region tells you something about how much charge is in that region. If one can add up all of the electric flux around a closed surface then one knows how much charge is inside that surface. Mathematically, this reads E . dA = Venc Eo An alternative approach is to turn Gauss's law around. One can use symmetry arguments to determine where the electric field is constant, thus allowing Gauss's law to tell you what the electric field is on that surface if you know how much charge is inside the surface. 1. Describe the three characteristics that matter in determining electric flux through a surface. 2. Write down the two important and special features of a Gaussian surface which allow you to simplify and solve the Gauss's law integral for the electric field. 3. Discuss the following question from the text. Justify your answer. STOP TO THINK 24.1 A uniformly charged rod has a finite length (e) L. The rod is symmetric under rotations about the axis and under reflection in any plane containing the axis. It is not symmetric under translations or under reflections in a plane perpendicular to the axis unless that plane bisects the rod. Which field shape or shapes match ## x the symmetry of the rod? (a) - 9-4. Consider an infinite line of charge with linear charge density 1. a. Draw the direction of the electric field vectors and justify your answer with a symmetry argument. b. Draw a Gaussian surface which has the same symmetry as the field whose surface goes through a point which is a distance r away from the line charge. C. Two students are discussing how to apply Gauss's law to the line charge: Student 1: The Gaussian surface has to go all the way around the charges because Gauss's law requires a closed surface. So that means that the Gaussian surface won't have ends on it for an infinite line. Student 2: In order to have a closed surface, the surface has to have ends on it, but it shouldn't matter where they are because the field is perpendicular to those end surfaces. On what points are the students correct and on what points are they not quite right? 5. Use Gauss's law to determine the strength of the electric field a distance r away from the line charge. a. Show that Gauss's law reduces to Qenc EgA for an appropriately chosen Gaussian surface. b. What is the surface area (that matters) of your Gaussian surface? C. Using the linear charge density 1, how much charge is enclosed within your Gaussian surface? d. Solve for the electric field.6. Consider a uniformly charges plane of surface charge density n, which is charge per unit area. a. Use a symmetry argument to draw the direction of the electric field above and below the plane of charge. b. Draw in the appropriate Gaussian surface which goes through points a distance of way from the plane. (Hint: Make sure that both ends of your surface are symmetric so that they are guaranteed to have the same electric field strength) c. Determine how much charge is enclosed within your Gaussian surface. d. Determine the electric field on the top and bottom of the plane