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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)).

Figure a shows an irregular 3 dimensional shape labeled S. A small circle with a plus sign, labeled q is outside it. Three arrows labeled vector E originate from q and pass through S. The patches where the arrows pierce the surface of S are highlighted. The patch where one arrow enters the shape is labeled dA1 and the patch where the arrow emerges from the shape is labeled dA2. Figure b shows an oval with two small circles inside it. These are labeled plus and minus. Three arrow from outside the oval point to the circle labeled minus. Three arrows point from plus to minus. Three arrows point from plus to outside the oval. Figure c has an irregular shape labeled S2. Within it is a circle named S1. At its center is a small circle labeled plus. Six arrows radiate outward from here in different directions.
Understanding the flux in terms of field lines. (a) The electric flux through a closed surface due to a charge outside that surface is zero. (b) Charges are enclosed, but because the net charge included is zero, the net flux through the closed surface is also zero. (c) The shape and size of the surfaces that enclose a charge does not matter because all surfaces enclosing the same charge have the same flux.

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 E through any closed surface, also called a Gaussian surface    , is equal to the net charge enclosed ( q enc ) divided by the permittivity of free space ( ε 0 ) :

Φ Closed Surface = q enc ε 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 or S ' is negative.

Figure a has an irregular shape labeled S. Within it is a circle labeled S prime. At its center is a small circle labeled plus. Six arrows radiate outward from here in different directions. Figure b has the same irregular shape S and circle S prime. At its center is a small circle labeled minus. Six arrows from different directions radiate inward to minus.
The electric flux through any closed surface surrounding a point charge q is given by Gauss’s law. (a) Enclosed charge is positive. (b) Enclosed charge 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 Φ = ( q 1 + q 2 + q 5 ) / ε 0 . Note that q 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.

Figure shows an irregular shape labeled S. Within it are charges labeled positive q1 and negative q2 and q5. Outside S are charges labeled positive q3, q4, q6 and q N minus 1 and negative q7 and q N.
The flux through the Gaussian surface shown, due to the charge distribution, is Φ = ( q 1 + q 2 + q 5 ) / ε 0 .

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.

Practice Key Terms 1

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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