dE

A

A

E

2

dA

5

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Problem: Gauss’ Law (1998)41. Gauss's law provides a convenient way to calculate the electric field outside and near each of the following isolated charged conductors EXCEPT aExpain your reasoning

Problem: Gauss’ Law (1993)38. The net electric flux through a closed surface is

D) zero if only positive charges are enclosed by the surfaceE) zero if the net charge enclosed by the surface is zeroExplain your reasoningProblem: Gauss’ Law (1993)Questions 51-52Two concentric, spherical conducting shells have radii r1and r2and charges Q1and Q2, as shown above. Let r be the distance from the center of the spheres and consider the region r1< r < r2.51. In this region the electric field is proportional to

2

/r

Show your workProblem: Gauss’ Law (1984)40.A closed surface, in the shape of a cube of side a, is oriented as shown above in a region where there is a constant electric field of magnitude E parallel to the x-axis. The total electric flux through the cubical surface is

2Show your work2/26/20126