Question

#8 Gauss's Law and The Shell Theorem Consider a hollow sphere with charge uni- formly distributed on its surface. Suppose the total charge is Q, where Q may be positive or negative Recall that Gauss's law as we have seen it is: Qenclosed ΣΕ A = EO where A = 47tr2 is the total area of the Gaussian surface Suppose the sphere radius is Ro and r > Ro. In terms of Gauss's Law, the reason why the electric field strength outside the sphere is given by 1 Q (5) 4760 72 (i.e. that of a point charge with r the distance from the sphere center) is: O A. There is no net electric flux from the charged spherical shell O B. The electric flux through a sphere of radius r is 47rE O C. Electric Flux flows out of the sphere for Q < 0 and into the sphere for Q < 0 O D. Gauss's Law should not be applied here due to the high degree of symmetry

Answer 1

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