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If you know the amount of charge contained within a Gaussian surface, then the total flux of the Electric Field generated by the enclosed charge is calculated from Gauss' Law.
As a demonstration, imagine a pair of cats that have charges placed on them by their loyal masters. Although the contours of the cats' elegant frames represent a complicated geometry, calculating the flux is a simple task if the charge on the cats is known. The flux through the Gaussian surface in Figure 1 is given by Gauss' law
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(1) |
Figure 1: Gaussian Surface Encompassing Two Cats
Note that we add the charges in equation (1) because it is the net enclosed charge. For example if the charge on cat 1 is
![$10.5 \,\, [\mu C]$](http://images.physicslibrary.org/cache/objects/240/l2h/img3.png "$10.5 \,\, [\mu C]$") and the charge on the cat 2 is
![$12.2 [\mu C]$](http://images.physicslibrary.org/cache/objects/240/l2h/img4.png "$12.2 [\mu C]$") , then the total flux through  is
The reverse of this problem is another important result. If we measure the flux through a given Gaussian surface, then we can calculate the amount of enclosed charge.
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- Figure 1, The Cat Clip art is public domain and was downloaded from WP Clipart
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![[parent]](https://images.physicslibrary.org/images/uparrow.png)
![\includegraphics[scale=.6]{Cats.eps}](http://images.physicslibrary.org/cache/objects/240/l2h/img2.png "\includegraphics[scale=.6]{Cats.eps}")
![$\displaystyle \Phi = \frac{10.5\times 10^{-8} \, [C] + 12.2\times 10^{-8} \, [C] }{8.85 \times 10^{-12} \, [C^2/N m^2]}$](http://images.physicslibrary.org/cache/objects/240/l2h/img6.png "$\displaystyle \Phi = \frac{10.5\times 10^{-8} \, [C] + 12.2\times 10^{-8} \, [C] }{8.85 \times 10^{-12} \, [C^2/N m^2]}$")
![$\displaystyle \Phi = 3073.4 \,\, [N m^2/C]$](http://images.physicslibrary.org/cache/objects/240/l2h/img7.png "$\displaystyle \Phi = 3073.4 \,\, [N m^2/C]$")