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Unformatted text preview: Chapt. 24 Gausss Law 189 just outside of the shell is and everywhere inside is . Now say you remove a differential bit of surface area dA from the shell to create a hole. Now the electric field just at the hole is . HINT: Use Gausss law for the first part and then use Gausss law for the differential area dA exploiting the planar symmetry at the dA location for a differentially small box enclosing dA . a) r ; r ; 2 r b) r ; 0 ; 2 r c) 2 r ; 0 ; 2 r d) 2 r ; 0 ; r e) 2 r ; r ; 2 r Full-Answer Problems 024 qfull 00100 2 5 0 moderate thinking: Gauss law and a cube 1. Consider a charged conducting cube. The cube has net charge q negationslash = 0. The cube is isolated from any electric fields not arising from itself. The system is electrostatic. a) Can Gausss law be used to derive a simple exact expression for the electric field around the cube? Why or why not?...
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- Fall '06