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

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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 Gausss law gives: E = q enc H dA = A 2 A = 2 Notice that this diers by a factor of 2 from the equation for the electric eld at the surface of a conductor. The factor of 2 comes in because inside of a conductor, the electric eld is zero and hence the total area which contributes to the ux 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 rst consider the eld outside of the sphere. Because of symmetry, the eld must point radially outward and must be constant at equal distances from the sphere. The total 9...
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