At first glance, the equation may appear quite daunting. However, it is surprisingly easy to understand if you first understand electric flux and surface area.

For the purpose of this lesson, I will use images to demonstrate each example. In these images, there will be lines to represent flux. Please note that, within a figure, each line represents an equal amount of flux.

Electric flux is basically a measure of the electric field over an area. Since electric field is measured in N/C, electric flux is measured in webers or Nm²/C (or you could say Vm²/m, but that is confusing). Imagine, if you will, a square perpendicular to an electric field. If the square was 1 meter on a side, and the electric field was 1 N/C, then the amount of electric flux through the square would be 1 Nm²/C. Lets try something harder: A circle with a radius of 2 meters is perpendicular to a 4 N/C field, what then would be the flux through the circle? With that setup, you would have about 12.25 (4p) m² of area, times 4 N/C would bring 49 Nm²/C.

Another interesting thing about flux is that it is conserved. If you have a sphere that is a certain distance away from an electrically charged point, there will be a uniform electric field through the entire sphere. If you double the radius of the sphere, then the electric field through the sphere is quartered; however, the surface area of the sphere is quadrupled, making the total electric flux unchanged.

Electric flux lines always start on a positive charge, and end on a negative charge. So if there is a volume that does not contain any charge, then electric flux should neither start nor stop inside that volume. Therefore, any flux that goes into that volume must also go out. Therefore, its net flux is zero. For example, lets take a cube (1 m³) that has two faces incident with an electric field (1 N/C). On one face, there will be an electric flux of 1 Nm²/C going in. On the opposite face, there will be an electric flux of 1 Nm²/C going out. The sum of these is, of course, zero.

In all of the previous examples, the areas have been perpendicular to the electric field. What if they aren’t? In that case, you must find the component of the electric field that is incident to the area. For example, you have a square (1 m²) that forms a 30° angle with an electric field (1 N/C). Since the flux is the product of area and incident electric field, you need to find the component of the electric field that is perpendicular to the square. If you draw it out, you will see that this component is equal to the electric field (1 N/C) multiplied by the sine of the angle, sin (30°) = 0.5. Therefore the incident electric field is 0.5 N/C, and the electric flux is 0.5 Nm²/C. This, by no coincidence, is the same flux going through an area half that of the square, but perpendicular to the field. If you were to take these two areas just described, and turned them into two sides of a triangular prism, you would see that the two fluxes would have to be equal. Since all of the other sides would be parallel to the electric field (with, therefore, 0 flux), then the flux going into one face would have to equal the flux going out of the other face.

When applying Gauss’ Law, it is usually possible to ignore the ‘incident’ part altogether. As long as all surfaces you are working with are perpendicular to the electric field (Gaussian surfaces), then the flux is equal to the area times the strength of the electric field. If you wish to make these into volumes, then make the rest of your surfaces parallel to the electric field, and have no flux.

Enough with the preliminaries, let's get on to Gauss’ Law. Simply put, Gauss’ Law states that the net flux out of a closed surface is equal to the enclosed charge divided by the permittivity of free space. To take a simple case, imagine a sphere with a point charge within it. This sphere has a radius r, and a surface area of 4pr². The point has a charge of q. The net electric flux out of the sphere is q/ε_o. If the point is in the very center of the sphere, the surface of the sphere is always incident with the point charge, and with any electric field emanating from the point, therefore you can divide both sides of the equation by the area of the sphere to find the electric field strength. Doing this gives the resulting field strength . This is also known as Coulomb's Law.

Gauss’ Law can be used for more than just proving what you already know. Gauss’ Law can find the field strength emanating from other geometries.

• For instance, take a wire with a linear charge density (C/m) of λ. Assuming that the wire is quite long, the electric field will always be perpendicular to the line of the wire. This means that a cylinder that uses the wire as its central axis would have its main side as a Gaussian surface, and its top and bottom would be parallel to electric field. If the cylinder has a radius r and a length of L, its side would have an area of 2prL. The enclosed charge would be equal to λL, so the electric field would be .
• Now, assume that we have a surface with a surface charge density (C/m²) of r. If we were to take two squares, each with an area of A, put them on either side of the charged surface so they were both Gaussian surfaces, and connect them with squares that were parallel to the electric field, we would have a box. This box would have an enclosed charge of rA, and flux would be going through an area of 2A. Using Gauss’ Law, you can find that the electric field through these two sides is always

The most common use for Gauss’s Law is finding the capacitance of geometrical capacitors. The only way to calculate the capacitance of capacitors is to find an equation that relates charge and voltage, since capacitance is charge per voltage. Gauss’ Law can be used to find the electric field strength, and voltage is electric field integrated with respect to displacement. We will look at three simple kinds of capacitor arrangements.

• The first, and most well known, is the parallel plate capacitor. Using Gauss’ Law of a charged surface (remembering that this time electric field only goes through one of the sides of the cube, yields . Integrating electric field with respect to displacement, we find since displacement is unrelated to either permittivity or charge. Replace ρ with q/a, and you get , thus relating voltage and charge. Using the equation for capacitance, , which you probably already knew, but this proves that the Gaussian method works.
   
• Slightly more complicated is the coaxial capacitor. This consists of a small cylinder encased in a larger cylinder (for the purpose of demonstration, these cylinders will have respective radii of a and b). The electric field between these two cylinders can be found using Gauss’ Law for a charged wire (the smaller cylinder). This gives . Since voltage is electric field integrated by displacement, we get

 

Whew. Replace λ with q/L, we get

.

Looking at the capacitance equation, we get

.
 

• Finally, the most complicated capacitor type that we are going to look at is the spherical capacitor. This consists of two spheres, one inside of the other (and, creatively enough, having radii of a and b). We already know that the electric field of a spherical Gaussian surface is . Voltage is electric field integrated with respect to displacement, so

 .

Using the equation for capacitance,

Gauss’ Law has many other applications that I am not even going to begin thinking about, at least not in this guide. For more practice, check out the Hecht Physics Calculus, Chapter 17, pp 680-684, MC questions 18&19, and problems 42-69.

Contributed by Lawrence Camarota, August 2004