23_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

# 23_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

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Electric Forces and Electric Fields 23 15.43 We choose a spherical gaussian surface, concentric with the charged spherical shell and of radius r . Then, ( ) 22 cos 4 cos0 4 EA E r r E θπ π Σ= ° = . (a) For r (that is, outside the shell), the total charge enclosed by the gaussian surface is Qq . Thus, Gauss’s law gives a > 0 q =+ − = 2 4 0, or rE E = = 0. (b) Inside the shell, r , and the enclosed charge is a < = + . Therefore, from Gauss’s law, 2 2 4 , or E r 00 4 qq = == ∈∈ 2 e kq r The field for is r < a 2 directed radially ou e r = E G tward . 15.44 Construct a gaussian surface just barely inside the surface of the conductor, where E 0 = . Since E inside, Gauss’ law says 0 = 0 = 0 Q inside. Thus, any excess charge residing on the conductor must be outside our gaussian surface (that is, on the surface of the conductor). 15.45 0 at all points inside the conductor, and c E = os cos90 0 θ = °= on the cylindrical surface. Thus, the only flux through the gaussian surface is on the outside end cap and Gauss’s law reduces to cos cap o Q EA EA = . The charge enclosed by the gaussian surface is
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