• Electromagnetics I
• Ch 5
• Loc 5.5

# Gauss’ Law - Integral Form

Gauss’ Law is one of the four fundamental laws of classical electromagnetics, collectively known as Maxwell’s Equations. Before diving in, the reader is strongly encouraged to review Section 2.4. In that section, Gauss’ Law emerges from the interpretation of the electric field as a flux density. Section 2.4 does not actually identify Gauss’ Law, but here it is:

Gauss’ Law (Equation 5.5.1) states that the flux of the electric field through a closed surface is equal to the enclosed charge.

Gauss’ Law is expressed mathematically as follows:

where

is the electric flux density

,

is a closed surface with differential surface normal

, and

is the enclosed charge. We can see the law is dimensionally correct;

has units of C/m

, thus integrating

over a surface gives a quantity with units of C/m

m

= C, which are the units of charge.

Gauss’ Law has a number of applications in electromagnetic theory. One of them, as explored below, is as a method to compute the electric field in response to a distribution of electric charge. Note that a method to do this, based on Coulomb’s Law, is described in Sections 5.1, 5.2, and 5.4. Gauss’ Law provides an alternative method that is easier or more useful in certain applications.

### Exercise

EXAMPLE 5.5.1: ELECTRIC FIELD ASSOCIATED WITH A CHARGED PARTICLE, USING GAUSS’ LAW.

In this example, we demonstrate the ability of Gauss’ Law to predict the field associated with a charge distribution. Let us do this for the simplest possible charge distribution. A particle of charge

located at the origin, for which we already have the answer (Section 5.1).

Gauss’ Law applies to any surface that encloses the charge, so for simplicity we chose a sphere of radius

centered at the origin. Note that

on the right hand side is just

for any surface having

. Gauss’ Law in this case becomes

If we can solve for

, we can get

using

. The simplest way to solve for

is to use a symmetry argument, which proceeds as follows. In this problem, the magnitude of

can depend only on

, and not

or

. This is because the charge has no particular orientation, and the sphere is centered on the charge. Similarly, it is clear that

must point either directly toward or directly away from the charge. In other words,

. Substituting this in the above equation, we encounter the dot product

, which is simply 1. Since

and

are constants with respect to the integration, we obtain:

The remaining integral is simply

, thus we obtain:

Bringing the known vector orientation of

back into the equation, we obtain

and finally using

we obtain the expected result

Here’s the point you should take away from the above example:

Gauss’ Law combined with a symmetry argument may be sufficient to determine the electric field due to a charge distribution. Thus, Gauss’ Law may be an easier alternative to Coulomb’s Law in some applications.