(e) In the region b< r< c, since the shell is conducting, the electric field is zero. Thus, for r = 2.30a, we have E= 0. (f) For r > c, the charge enclosed by the Gaussian surface is zero. Gauss’ law yields 4002πrEE=⇒ =. Thus, E= 0 at r= 3.50a. (g) Consider a Gaussian surface that lies completely within the conducting shell. Since the electric field is everywhere zero on the surface, GGEdA⋅=z0 and, according to Gauss’ law, the net charge enclosed by the surface is zero. If Qiis the charge on the inner surface of the shell, then q1+ Qi= 0 and Qi= –q1= –5.00 fC. (h) Let Qobe the charge on the outer surface of the shell. Since the net charge on the shell is –q, Qi+ Qo= –q1. This means Qo= –q1– Qi= –q1–(–q1) = 0. 51. At all points where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is to be found. The field is
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