Gauss's Law



" Equations are just the boring part of mathematics. I attempt to see things in terms of geometry."
Stephen Hawking



defn of E flux
so that the total electric flux passing through a surface, S, is given by,

elec gauss eqn2
  • Gauss's Law then states that,
red tick    elec gauss 3    red tick
where the circle on the integral means that the surface is closed and qinside is the net charge inside this closed surface.
  • exclamation A closed surface has a definite inside and outside differentiated by the surface, e.g. the surface of a sphere.
  • exclamation The dA vector of a closed surface is always directed from the inside to the outside of the surface.
  • exclamation The exact location of the charges inside the closed surface is not important, all that matters is the net charge.
  • exclamation ε0 is the "Permittivity of the Vacuum" a constant whose value is 8.85 x 10-12 C2/(N.m2) where the Coulomb constant, k = 1/(4πε0).  Note that if the charges are not located in vacuum ε0 must be replaced by the permittivity of the medium in question. 
  • exclamation The proof of Gauss's Law is beyond the scope of this course.  Suffice to say the inverse square dependence on distance of Coulomb's Law is critical.
  • Before using Gauss's Law to evaluate electric fields a brief qualitative discussion is worthwhile.  Consider the situation of two point charges below.  Application of Gauss's Law over each of the closed surfaces:
  • elec gauss figure 2 S1:  At every point on this surface both E and dA are directed "outwards", such that the scalar product E·dA = EdAcosθ is always positive.  Thus the integral over the surface S1 will be positive, as it must be if Gauss's Law is to be satisfied, since the net charge enclosed is positive.
  • S2 :  E is directed "inwards", dA "outwards", leading to a negative value for the flux through S2, consistent with the fact that the net charge enclosed is negative.
  • S3:  Some of this surface has E directed "inwards" the remainder has E directed "outwards".  dA is "outwards" everywhere on the surface.  Therefore the flux integral has both positive and negative contributions.  Since there is no net charge enclosed by S3 by Gauss's Law the net flux will be zero.
  • S4: Once again there are negative and positive contributions to the flux integral, so that we can write Gauss's Law,
elec_gauss_eqn4


Did you hear about the French post-doc who went to work at the Fermi Lab, but never went in because the sign over the door always said it was closed.


 

Dr. C. L. Davis
Physics Department
University of Louisville
email: c.l.davis@louisville.edu