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Physics 1 Problem Solutions 193

Physics 1 Problem Solutions 193 - Chapt 24 Gausss Law 189...

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Chapt. 24 Gauss’s 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 Gauss’s law for the first part and then use Gauss’s law for the differential area dA exploiting the planar symmetry at the dA location for a differentially small box enclosing dA . a) σ ε 0 ˆ r ; σ ε 0 ˆ r ; σ 2 ε 0 ˆ r b) σ ε 0 ˆ r ; 0 ; σ 2 ε 0 ˆ r c) σ 2 ε 0 ˆ r ; 0 ; σ 2 ε 0 ˆ r d) σ 2 ε 0 ˆ r ; 0 ; σ ε 0 ˆ r e) σ 2 ε 0 ˆ r ; σ ε 0 ˆ r ; σ 2 ε 0 ˆ 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 Gauss’s law be used to derive a simple exact expression for the electric field around the cube? Why or why not?
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