This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ð lisfra ð 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 threedimensional 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 onedimensional 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 threedimensional 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....
View
Full Document
 Spring '09
 All
 Electrostatics, Magnetic Field, Electric charge, Surface charge

Click to edit the document details