6.2 Explaining gauss’s law  (Page 2/4)

You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typical field line enters the surface at $d{A}_1$ and leaves at $d{A}_2.$ Every line that enters the surface must also leave that surface. Hence the net “flow” of the field lines into or out of the surface is zero ( [link] (a)). The same thing happens if charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero (part (b)). A surface that includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge (part (c)).

Statement of gauss’s law

Gauss’s law generalizes this result to the case of any number of charges and any location of the charges in the space inside the closed surface. According to Gauss’s law, the flux of the electric field $\overrightarrow{\text{E}}$ through any closed surface, also called a Gaussian surface    , is equal to the net charge enclosed $(q_{\text{enc}})$ divided by the permittivity of free space $({\epsilon }_0)$ :

This equation holds for charges of either sign , because we define the area vector of a closed surface to point outward. If the enclosed charge is negative (see [link] (b)), then the flux through either $S\text{or}S\text{'}$ is negative.

The Gaussian surface does not need to correspond to a real, physical object; indeed, it rarely will. It is a mathematical construct that may be of any shape, provided that it is closed. However, since our goal is to integrate the flux over it, we tend to choose shapes that are highly symmetrical.

If the charges are discrete point charges, then we just add them. If the charge is described by a continuous distribution, then we need to integrate appropriately to find the total charge that resides inside the enclosed volume. For example, the flux through the Gaussian surface S of [link] is $\text{Φ}=(q_1+q_2+q_5)\text{/}{\epsilon }_0.$ Note that $q_{\text{enc}}$ is simply the sum of the point charges. If the charge distribution were continuous, we would need to integrate appropriately to compute the total charge within the Gaussian surface.

Recall that the principle of superposition holds for the electric field. Therefore, the total electric field at any point, including those on the chosen Gaussian surface, is the sum of all the electric fields present at this point. This allows us to write Gauss’s law in terms of the total electric field.