Unformatted text preview: 40. We imagine a spherical Gaussian surface of radius r centered at the point charge + q . From symmetry consideration E is the same throughout the surface, so I ~ E · d ~ A = 4 πr 2 E = q encl ε , which gives E ( r ) = q encl 4 πε r 2 , where q encl is the net charge enclosed by the Gaussian surface. (a) Now a < r < b , where E = 0. Thus q encl = 0, so the charge on the inner surface of the shell is q i = − q . (b) The shell as a whole is electrically neutral, so the outer shell must carry a charge of q o = + q . (c) For r < a q encl = + q , so E ¯ ¯ ¯ ¯ r<a = q 4 πε r 2 . (d) For b > r > a E = 0, since this region is inside the metallic part of the shell. (e) For r > b q encl = + q , so E ¯ ¯ ¯ ¯ r<a = q 4 πε r 2 . The field lines are sketched to the right. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

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