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Unformatted text preview: here), so the total electric ux through the Gaussian surface is zero and the net charge within it is zero (by Gauss law). Since the central rod is known to have charge q , then the inner surface of the shell must have charge q . And since the shell is known to have total charge 2 q , it must therefore have charge q on its outer surface. (c) inally, we consider a cylindrical Gaussian surface whose radius places it between the outside of conducting rod and inside of the shell. Similarly to part (a), the ux through the Gaussian surface is = 2 rLE , where E is the Feld at this Gaussian surface, in the region between the rod and the shell. The charge enclosed by the Gaussian surface is only the charge q on the rod. Therefore, Gauss law yields 2 rLE = q = E = q 2 Lr . The positive sign indicates that the Feld points outward....
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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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