22 - MasteringPhysics: Assignment Print View

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Unformatted text preview: MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assig... 1 of 13 17/4/07 14:59 [ Assignment View ] E lisfra i 2, vor 2007 22. GAuss' LAw ssignment is due At 2:00Am on WednesdAy, JAnuAry 31, 2007 Credit for problems submitted lAte will decreAse to 0% After the deAdline hAs pAssed. The wrong Answer penAlty is 2% per pArt. Multiple choice questions Are penAlized As described in the online help. The unopened hint bonus is 2% per pArt. You Are Allowed 4 Attempts per Answer. Gauss' Law Gauss's Law Learning Goal: To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written where is the permittivity of vacuum. Part A How should the integral in Gauss's law be evaluated? ANSWER: Answer not displayed Part B Part not displayed Gauss's Law in 3, 2, and 1 Dimension Gauss's law relates the electric flux through a closed surface to the total charge enclosed by the surface: . You can use Gauss's law to determine the charge enclosed inside a closed surface on which the electric field is known. However, Gauss's law is most frequently used to determine the electric field from a symmetric charge distribution. The simplest case in which Gauss's law can be used to determine the electric field is that in which the charge is localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in three-dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates. Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely along some direction, say along the z axis. In this case, the point has been extended to a line, namely, the z axis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only with x and y , or with and in cylindrical coordinates. A one-dimensional problem may be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane, and consequently, it has planar symmetry. Three dimensions Consider a point charge in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all directions. To find , the magnitude of the field at distance from the charge, the logical [ MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assig... 2 of 13 17/4/07 14:59 Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of the electric field and an infinitesimal surface element involves . The flux integral is therefore reduced to , where is the magnitude of the electric field on the Gaussian surface, and is the area of the surface....
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