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Gauss's Law for Spherical Symmetry
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[QUOTE="gneill, post: 4524818, member: 293536"] You've calculated the total charge on the sphere to be ##\pi## Coulombs. This is correct. Gauss' law tells us we can treat a spherically symmetric charge distribution as a point charge when we're outside of it, so the external field for r > 1 obeys Coulombs law, which is what you've arrived at. Clearly there are two distinct regions to deal with, one where the charge density is nonzero (within the non-conducting sphere), and the other where the charge density is zero (within the surrounding air). So your charge density function is discontinuous at r = 1. $$ \rho (r) = \left \{ \begin{array}{l} r & : 0 \leq r \leq 1 \\ 0 & : r > 1 \end{array} \right . $$ mathematically you deal with this by breaking the sum into two separate integrals. Since the charge density function is uniformly zero for the second region, that integral must vanish. [/QUOTE]
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Gauss's Law for Spherical Symmetry
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