Gauss's Law

Example: Uniform Spherical Charge Consider a

uniformspherical distribution of charge. This must be charge held in place in an insulator. Charge on a conductor would be free to move and would end up on the surface. This charge density isuniformthroughout the sphere.Charge Q is

uniformlydistributed throughout a sphere of radiusa. Find the electric field at a radius r.First consider r > a; that is, find the electric field at a point

outsidethe sphere.Just as before (for the point charge), we start with Gauss's Law

Just as for the point charge, we find

and we know

which means

E = k Q / r ^{2}That is, the electric field

outsidethe sphere is exactly the same as if there were only apoint chargeQ.Now, move inside the sphere of uniform charge where r < a. The volumetric charge density is

The charge contained within a sphere of radius r is

That is, the electric field

insidethe sphere of uniform charge iszeroat the center and increaseslinearlywith radius r:Of course, the two expressions for the electric field match -- have the same value -- at the surface of the sphere, for r = a.

Using Gauss's Law here made the "calculation" almost easy. A more direct application of Coulomb's Law -- with a detailed integration over the volume

anda careful consideration of thevector natureof Coulomb's Law would have beenfar more difficult.

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(c) Doug Davis, 2002; all rights reserve