charge enclosed by the surface is simply the total charge on the sphere q . Hence, by Gauss’s law, the ﬁeld outside must be: E = q ±0 4 πr 2 which is exactly the same ﬁeld as a point charge centered at the center of the sphere. Inside of the sphere the same symmetry arguments hold, but the total charge enclosed by the sphere will be proportional to the volume of the sphere. q enc = ρV = q 4 3 πR 3 4 3 πr 3 = q r 3 R 3 Plugging this into Gauss’s law gives the electric ﬁeld inside of a uniformly charged sphere. E = ± q ±0 4 πR 3 ² r 4.5 Problems Problem 23.8 Figure 6 shows two nonconducting spherical shells ﬁxed in place. Shell 1 has uniform surface charge density +6 .0 μC m 2 on its outer surface and radius 3 .0 cm ; shell 2 has uniform surface charge density +4 .0 μC m 2 on its outer surface and radius 2 .0 cm ; the shell centers are separated by L = 10 cm . In unit-vector notation, what is the net electric ﬁeld at x = 2 .0 cm ?
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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.