Gauss's Law

Example: Spherical Shell of Charge What about a shperical

shellof charge? This can be a thin shperical shell surrounding an empty sphere. Or, more likely, it may be the charge on a conductor. Since the charge is free to move in a conductor, it will repel itself as far away as possible -- and that means it will move to the outside surface of the conductor.Again, a direct application of Coulomb's Law and a calculation of the electric field and its vector nature sounds almost painful to comtemplate. However, with Gauss's Law we can solve this

by inspectionfor we have already done all of the work necessary.Outside the sphere, for r > a, this is just like the sphere with

uniformcharge distribution. That is, the electric field is just like that for a point charge,E = k Q / r ^{2}for r > a

Inside the sphere, for r < a, the charge enclosed by a sphere of radius r is zero. So the flux must also be zero. And the flux is proportional to the electric field. So the electric field must be zero

E = 0 for r < a

That also means the electric field

inside any conductoris zero!

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(c) Doug Davis, 2002; all rights reserve