ch23-p051 - 51. At all points where there is an electric...

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(e) In the region b < r < c , since the shell is conducting, the electric field is zero. Thus, for r = 2.30 a , we have E = 0. (f) For r > c , the charge enclosed by the Gaussian surface is zero. Gauss’ law yields 400 2 π rE E =⇒ = . Thus, E = 0 at r = 3.50 a . (g) Consider a Gaussian surface that lies completely within the conducting shell. Since the electric field is everywhere zero on the surface, G G EdA ⋅= z 0 and, according to Gauss’ law, the net charge enclosed by the surface is zero. If Q i is the charge on the inner surface of the shell, then q 1 + Q i = 0 and Q i = –q 1 = 5.00 fC. (h) Let Q o be the charge on the outer surface of the shell. Since the net charge on the shell is – q, Q i + Q o = –q 1 . This means Q o = –q 1 – Q i = –q 1 –(– q 1 ) = 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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