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Gauss's Law for Spherical Symmetry
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[QUOTE="Vectorspace17, post: 4524773, member: 419071"] So from the center to the edge of the sphere (R = 1), I know the enclosed charge should be: from 0 to 1, ∫ρ(r)4(pi)r^2 dr = 4pi[(1/4)r^4] from 0 to 1 = pi When put into E(r), this does give k(pi)/r^2, but I'm confused about this result because this is supposed to be the final answer to the region r > 1, not 0 to R = 1. From 1 to r > 1, its surrounded by air, which means the charge density should be constant, so from 1 to r, the integral to find the enclosed charge is: ∫4(pi)r^2 dr = 4pi[(1/3)r^3] from 1 to pi = (4/3)(pi)[r^3 - 1] This is the integral that I though would give the charge enclosed for r > 1, but it yields a really ugly solution. The only thing I can think of (and I'm not sure if it makes any sense) is that my first result would be the charge on the surface of the sphere, and since its surrounded be air, that charge would remain constant as you move away from the surface of the sphere. [/QUOTE]
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Gauss's Law for Spherical Symmetry
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