PHYS208 Fundamentals of Physics II
Application of Gauss's Law to confirm "Coulomb's Law" for a Point Charge
We already have examined the electric field from a point charge as a consequence of Coulomb's law.
However here I would like to consider the single point charge as the simplest conceptual
situation where Gauss's law may be used for evaluation of the electric field.
The following series of steps are generally associated with application of Gauss's law
for the evaluation of the electric field arising from a charge distribution
1. Understand the geometry
You cannot get much simpler that a point, so there is not much to understand in this case.
Generally, however, you must make sure that you understand exactly how the charge is
distributed: is it a uniform distribution?
is it distributed over a volume, a surface, or along a line?
For a point charge there is not really a distribution of charge; it is concentrated in
a region so small that no structure is evident -- that is, a "point!"
2. Understand the symmetry
The point charge has spherical symmetry. The charge appears the same from any relative
angle of observation, thus the electric field at a point will depend only on how far it
is away from the point charge. There is no angular dependence of the electric field.
From our previous experience we expect that the electric field
will point radially away from the point; an any point of observation, the electric field will
point along the line connecting the point charge and the evaluation point.
By symmetry arguments alone, we expect that the electric field about a point charge
will have the following functional dependence:
3. Construct the gaussian surface
Recall the recommendations for selection of a gaussian surface to make application
of Gauss's law tractable:
Some reflection on the symmetry of the charge distribution and its field suggests
that a spherical surface, centered on the point charge, will satisfy these requirements.
- All sections of the gaussian surface should be chosen so that they are either
parallel or perpendicular to E.
- |E| should be constant on each surface having non-zero flux.
4. Examine the gaussian surface
At all points on the spherical surface, the vector associated with the infinitesimal surface
element dA will point in the same direction as the electric field at that point,
i.e. radially outward. Thus the integrand of the surface integral for evaluating
electric flux will reduce from E . dA to E dA.
Furthermore, the magnitude of the electric field will be identical at all points on the spherical
surface. The sphere is centered about the charge; each point on the surface is an
identical distance from the center (by definition of a spherical surface). Thus
the electric field may be moved outside the integral; the surface integral reduces to the
product of the electric field and the surface area of a sphere.
5. Evaluate the electric flux through the gaussian surface
Do not forget the formula for the surface area of a sphere!
6. Evaluate the charge enclosed by the gaussian surface
In this case the charge enclosed is simply the charge of the point charge, Q.
Of course not all evaluations of the charge enclosed by the gaussian surface will be so
simple. Perhaps only a fraction of the charge distribution will be enclosed; furthermore,
the charge distribution may not be uniform and an integral may need to be evaluated
involving a charge density.
7. Apply Gauss's law for the result!
This is identical to our earlier result, once the connection between k of Coulomb's law
and the permittivity of free space is made. Remember that the direction of the electric field at a point
was established earlier!
Last updated March 4, 1998.
Copyright George Watson, Univ. of Delaware, 1997.