Gauss's Law
Example: Point Charge
Consider the E-field from a point charge Q and the flux through a sphere centered on that charge. By symmetry (there it is, again!) we know the E-field is radially outward (assuming the charge Q is positive) so it is perpendicular to the surface. That means
E
dA = E dA
So the integral over the surface is
And Gauss's Law is that this flux through the surface is equal to the charge enclosed by the surface divided by epsilon,
Therefore,
Of course, this is just Coulomb's Law. We would expect this since we started with Coulomb's Law when we "derived" Gauss's Law. The two are intimately connected!
Return to Ch24 ToC (c) Doug Davis, 2002; all rights reserved
