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 and look at the directions involved. By symmetry (there it is, again!) we know the E-field is radially outward (assuming the charge Q is positive) so it is

perpendicularto the surface. That means

EdA= E dA

So the integral over the surface is

And

Gauss's Lawis that this flux through the surface is equal to the charge enclosed by the surface divided by epsilon,Therefore,

That is, the electric field

outsidethe sphere is exactly the same as if there were only apoint chargeQ.

Return to Ch24 ToC(c) Doug Davis, 2002; all rights reserved