This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 22 p. 1 CHAPTER 22 Gausss Law 1. Because the electric field is uniform, the flux through the circle is = ? E d A = E A = EA cos . ( a ) When the circle is perpendicular to the field lines, the flux is = EA cos = EA = (5.8 10 2 N/C)p(0.15 m) 2 = 41 N m 2 /C . ( b ) When the circle is at 45 to the field lines, the flux is = EA cos = EA = (5.8 10 2 N/C)p(0.15 m) 2 cos 45 = 29 N m 2 /C . ( c ) When the circle is parallel to the field lines, the flux is = EA cos = EA = (5.8 10 2 N/C)p(0.15 m) 2 cos 90 = . 2. Because the electric field is radial, it is perpendicular to the spherical surface just beyond the Earths surface. The field is also constant, so the flux through the sphere is = ? E d A = EA = (150 N/C)4p(6.38 10 6 m) 2 = 7.7 10 16 N m 2 /C . Note that E and d A are in opposite directions. 3. All field lines enter and leave the cube, so the net flux is net = 0 . We find the flux through a face from = ? E d A . There are two faces with the field lines perpendicular to the face, say one at x = 0 and one at x = . Thus for these two faces we have x = 0 = EA = (6.50 10 3 N/C) 2 = (6.50 10 3 N/C) 2 ; x = = + EA = + (6.50 10 3 N/C) 2 = + (6.50 10 3 N/C) 2 . For all other faces, the field is parallel to the face, so we have all others = 0 . 4. ( a ) Because the angle between the electric field and the area varies over the surface of the hemisphere, it would appear that we find the flux by integration. We see that the same flux must pass through the circular base of the hemisphere, where the field is constant and perpendicular to the surface. Thus we have = EA = E p R 2 . ( b ) If E is perpendicular to the axis, every field line must enter and leave the surface, so we have = 0 . 5. The total flux is depends only on the enclosed charge: = Q / , or Q = = (8.85 10 12 C 2 /N m 2 )(1.45 10 3 N m 2 /C) = 1.28 10 8 C = 12.8 nC . Use Word 6.0c or later to view Macintosh picture. Chapter 22 p. 2 6. The net flux through each closed surface is determined by the net charge inside. Thus we have 1 = ( Q 3 Q )/ = 2 Q / ....
View
Full
Document
This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.
 Spring '08
 ROSS

Click to edit the document details