Electric Forces and Electric Fields2315.43We choose a spherical gaussian surface, concentric with the charged spherical shell and of radius r. Then, ( )22cos4cos04EAErr EθππΣ=°=. (a) For r(that is, outside the shell), the total charge enclosed by the gaussian surface is Qq. Thus, Gauss’s law gives a>0q=+ −=240, or rEE==0.(b) Inside the shell, r, and the enclosed charge is a<= +. Therefore, from Gauss’s law, 224, or Er004qq===∈∈2ekqrThe field for is r<a2directed radially ouer=EGtward . 15.44Construct a gaussian surface just barely inside the surface of the conductor, where E0=. Since Einside, Gauss’ law says 0=0= 0Q∈inside. Thus, any excess charge residing on the conductor must be outside our gaussian surface (that is, on the surface of the conductor). 15.450at all points inside the conductor, and cE=oscos900θ=°=on the cylindrical surface. Thus, the only flux through the gaussian surface is on the outside end cap and Gauss’s law reduces to coscapoQEAEA=∈. The charge enclosed by the gaussian surface is
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