Unformatted text preview: the endcaps of the cylinder. Plugging all of this into Gauss’s law gives: E = q enc ± H dA = σ A ± 2 A = σ 2 ± Notice that this diﬀers by a factor of 2 from the equation for the electric ﬁeld at the surface of a conductor. The factor of 2 comes in because inside of a conductor, the electric ﬁeld is zero and hence the total area which contributes to the ﬂux is only A instead of 2 A . Otherwise, the derivation is the same. 4.4.3 A Uniformly Charged Sphere: Spherical Symmetry In this case we choose a spherical Gaussian surface centered at the center of the sphere to mimic the symmetry. Lets ﬁrst consider the ﬁeld outside of the sphere. Because of symmetry, the ﬁeld must point radially outward and must be constant at equal distances from the sphere. The total 9...
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 Spring '08
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 Physics, Charge, gaussian surface, cylindrical Gaussian surface

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