## 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:
1. All sections of the gaussian surface should be chosen so that they are either parallel or perpendicular to E.
2. |E| should be constant on each surface having non-zero flux.
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.

#### 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!

"http://www.physics.udel.edu/~watson/phys208/gauss-point.html"
Last updated March 4, 1998.
Copyright George Watson, Univ. of Delaware, 1997.