CourseNotes.9

# CourseNotes.9 - the endcaps of the cylinder Plugging all of...

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Figure 4: Line of charge with cylindrical Gaussian surface. 4.4.2 A Charged Plane: Planer Symmetry Figure 5: Charged plane with cylindrical Gaussian surface. In this case, we again want to choose a Gaussian surface which mimics the symmetry of the problem. We will choose a cylinder as we did in the last problem, but we could just as easily have chosen a box or a cube. By symmetry, the electric ﬁeld must point away from the plane (normal) and must be constant over the entire surface. Furthermore, symmetry implies that the electric ﬁeld must be the same on opposite sides of the surface at equal distances from the plane. Since the ﬁeld points normally away from the plane, the only ﬂux through our Gaussian surface will be through the endcaps. The charge enclosed by the surface will be q enc = σ A where A is the are of
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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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