Gauss’ Law can be used to prove that any excess charge on a conductor must be on its surface. Here’s how we do this: Define a surface just inside the surface of a conductor. Gauss’ Law relates the charge inside that surface to the electric field on the surface. Because the electric field inside a conductor is zero, the charge enclosed by any surface inside the conductor is zero. Therefore any charge on the conductor must reside on its surface. If a cavity were carved out of the inside of a conductor, there would be no electric field in the surrounding conductor and no charge in the cavity. Therefore there would be no electric field in the cavity. A person inside a metal box would not be shocked even if the box were charged to a high voltage. Let a hollow conductor have no net charge, but place a charge Q in the hollow region. Construct a Gaussian surface outside of the cavity but inside the conductor. Gauss’ Law relates the electric field on this surface to the charge enclosed by the surface. The electric field is zero at this Gaussian surface, which is inside the conductor. Therefore the net charge enclosed by the conductor must also be zero. There must be a charge on the inner surface of the conductor, Q sub s, which is opposite that of the charge Q inside the cavity. The net charge on the conductor is zero. Therefore an equal and opposite charge must exist on the outer surface of the conductor The charge on the inner surface of the cavity confines the electric field to the cavity. The location of the cavity therefore does not affect the distribution of charge on the surface. The charges on the outer surface arrange themselves in such a way that the electric field inside the conductor is zero. The external electric field is the same as if the charge on the outer surface of the conductor were the only charge. A metal sphere is located inside a cavity of a second, larger sphere. The inner sphere has 0.10 coulombs of charge and the outer sphere has a net charge of 0.08 coulombs. Next we demonstrate the simple relation between the charge density and the electric field at the surface of a conductor. Define a closed cylindrical Gaussian surface enclosing a small volume near the surface of the conductor. The flat caps of this cylinder have area A. The charge per unit area on the surface is sigma. In general, sigma varies from point to point over the surface. The charge, q, enclosed by the cylinder is sigma A. However we know from Gauss' Law that the charge enclosed by the cylinder is equal to the flux of the electric field through the surface of the cylinder." The electric field inside the conductor is zero, and because the electric field must be perpendicular to the surface, no flux pierces the sides of the cylinder. Only the outer cap of the cylinder contributes to the integral. The electric field is perpendicular to this cap. Therefore, the flux through the Gaussian surface is equal to the magnitude of the electric field, E, times the area of the cap, A. Substitute the charge in terms of sigma into this equation. Solve for the magnitude of the electric field near the surface. It is proportional to the surface charge density. Air breaks down electrically and a conductor will emit sparks, when the electric field reaches about 3 times 10 to the sixth volts per meter. What local surface charge density would be required for the electric field to reach this level on a conductor?

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