ch24_p101 - 101. (a) For r > r2 the field is like that...

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101. (a) For r > r 2 the field is like that of a point charge and V Q r = 1 4 0 π ε , where the zero of potential was taken to be at infinity. (b) To find the potential in the region r 1 < r < r 2 , first use Gauss’s law to find an expression for the electric field, then integrate along a radial path from r 2 to r . The Gaussian surface is a sphere of radius r , concentric with the shell. The field is radial and therefore normal to the surface. Its magnitude is uniform over the surface, so the flux through the surface is Φ = 4 π r 2 E . The volume of the shell is 43 2 3 1 3 π b gc h rr , so the charge density is ρ = 3 4 2 3 1 3 Q π c h , and the charge enclosed by the Gaussian surface is qr r Q = F H G I K J −= F H G I K J 4 3 3 1 3 3 1 3 2 3 1 3 π c h . Gauss’ law yields 4 4 0 2 3 1 3 2 3 1 3 0 3 1 3 2 2 3 1 3 π π rE Q E Qr r rr r = F H G I K J ¡ = c h . If V s is the electric potential at the outer surface of the shell ( r = r 2 ) then the potential a distance r from the center is given by VV E d rV Q r r r dr V Q rrr r r r ss r r r r s =− F H G I K J −+− F H G I K J z z 4 1 4 1 22 02 3 1 3 1 3 2 3 1 3 2 2 2 1 3 1 3 2 π π . The potential at the outer surface is found by placing
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ch24_p101 - 101. (a) For r &gt; r2 the field is like that...

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