5.3.1: Illustrations

Illustration 1: Flux and Gaussian Surfaces

The bar graph shows the flux, \(\phi\), through four Gaussian surfaces: green, red, orange and blue (position is given in meters, electric field strength is given in newtons/coulomb, and flux is given in \(\text{N}\cdot\text{m}^{2}/\text{C}\)). Restart. Note that this animation shows only two dimensions of a three-dimensional world. You will need to imagine that the circles you see are spheres and that the squares you see are actually boxes. Flux, \(\phi\), is a measure of the amount of electric field through a surface. Gauss's law relates the flux to the charge enclosed (\(q_{\text{enclosed}}\)) in a Gaussian surface:

\[\phi=q_{\text{enclosed}}/\varepsilon_{0}\qquad\text{and}\qquad\phi=\int\mathbf{E}\cdot d\mathbf{A}=\int E\cos\theta dA\nonumber\]

where \(\varepsilon_{0}\) is the permittivity of free space (\(8.85\times 10^{-12}\text{ C}^{2}\text{/N}\cdot\text{m}^{2}\)), \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is the unit area normal to the surface, and \(\theta\) is the angle between the electric field vector and the surface normal.

Begin by moving the green Gaussian surface around. What is the flux when the surface encloses the point charge? What is the flux when the point charge is not inside the surface? What about the red surface? Since the flux is the electric field times the surface area, why doesn't the size of the surface matter? As long as the point charge is enclosed, the flux is the same and is equal to \(q_{\text{enclosed}}/\varepsilon_{0}\)_. When the charge is not enclosed, the flux is always zero. Notice that both the green and red Gaussian surfaces can be moved to either enclose or not enclose the charge. Therefore, the two fluxes should, and do, agree. However, only when these surfaces are centered on the charge can you use them to determine the electric field.

The orange surface has a different symmetry from the point charge (and its electric field). With the orange surface, why doesn't the shape matter in finding the flux? Again, what matters is whether the charge is enclosed or not. Move the surface to a point where the flux is zero. Is the electric field zero at the surface of the box? If the electric field is not zero, why is the flux zero? If you think about flux as a flow of electric field through an area (a bit like fluid flow), which was the early analogy for electric field and flux, then when there is no charge inside, the electric field that comes into the box must also leave. There is no source of electric field inside the box. However, the cubical box no longer has the same symmetry (a spherical symmetry) of the point charge. While the flux is zero for these scenarios, the value of the flux cannot be used to determine the electric field. The integral \(\int E \cos\theta dA\) is not equal to the integral \(E\int \cos\theta dA\) because \(E\) is not uniform across the Gaussian surface.

Finally, try two charges using the blue surface. What happens when the blue surface encloses just one charge? What happens when it encloses both charges?

Illustration authored by Anne J. Cox.
Script authored by Mario Belloni.

Illustration 2: Near and Far View of a Filament

The different configurations of the animation show different views of the same charged filament: an intermediate view, a close view, and a view of the filament from far away (position is given in meters, electric field strength is given in newtons/coulomb, and flux is given in \(\text{N}\cdot\text{m}^{2}\text{/C}\)). A dragable Gaussian "surface" (you must imagine each of these to be three dimensional; boxes are cubes, circles are spheres) mimics the symmetry of the charge distribution as closely as possible. Restart.

Compare the fluxes through the Gaussian surfaces in both the intermediate and far views when the Gaussian surface encloses the entire filament. Why are they the same? In the far view, why is the flux the same even if the detector is not centered on the charge? The same amount of charge is enclosed by the Gaussian surface in both cases; therefore, the flux is the same.

For each view, what do you think the electric field lines will look like? Check your answers using the "show \(E\)-field" links. Why have we chosen the Gaussian surfaces to have different shapes for the near and far views? In the intermediate view the electric field does not have a symmetry, while in the near view the electric field has an approximate rectangular symmetry (if you are close enough to the charge distribution), and in the far view the electric field has an approximate spherical symmetry (if you are far enough away from the charge distribution). This means that you can only use Gauss's law for the near view and the far view. Given that there are two different symmetries, the electric field you find using Gauss's law will be different for the two cases. That is okay, because Gauss's law allows you to calculate the electric field on the Gaussian surface (not anywhere else).

Move the surfaces for the "show \(E\)-field" links so that the electric field vectors are either perpendicular to or parallel to the surfaces for the near and far views. Can you do the same for the intermediate view? No. This lets you know that the symmetry for the intermediate case does not allow you to use Gauss's law to calculate the electric field at the surface. Gauss's law still holds for all three configurations. You can still calculate the flux through the surface (it is proportional to the charge enclosed). However, for the intermediate view, because the electric field varies at different spots on the surface (points in different directions relative to the surface), you cannot use Gauss's law to calculate the electric field at the surface (you must use Coulomb's law). Thus, although true, Gauss's law is only useful for calculating the electric field for certain symmetrical charge distributions (spherical, cylindrical, and planar).

Illustration authored by Anne J. Cox and Mario Belloni.
Script authored by Mario Belloni.

Illustration 3: A Cylinder of Charge

In this animation each charge is a line or rod of charge into and out of the screen. You can create charge distributions and see the electric field that results (position is given in meters and electric field strength is given in newtons/coulomb). Restart.

Begin by adding one line charge. Notice the field lines. Imagine that this charge extends both into and out of the computer screen. Clearly, there is cylindrical symmetry. You could imagine putting a tube centered on this charge. How could you convince another student that the magnitude of the electric field at any point on the tube would be the same as any other point on the tube? Well, there is nothing special about the placement of your tube in the direction out of the screen. There is nothing special about how long that tube is.

Now, add another line charge. What happens to the symmetry? Far away from the two charges, there would still be cylindrical symmetry. Why? If you are far enough away it looks (approximately) like a single rod. Look at what happens to the field as you add \(10\) more line charges and then half a cylinder. When you create a full cylinder, what is the symmetry? The field everywhere inside the cylinder is zero. What Gaussian surface could you use to find the electric field inside the cylinder or outside the cylinder? The full cylinder has cylindrical symmetry about the middle of the cylindrical shell of line charge. Since there is a symmetry, we can use Gauss's law to calculate the electric field. Since there is no charge enclosed by a Gaussian surface of radius \(1.65\text{ m}\), the flux is zero, and because of the symmetry we can say that \(E\) is zero inside the cylindrical shell. We could also use a cylindrical Gaussian surface to calculate the electric field outside of the cylindrical shell of line charge.

As you work problems using Gauss's law, you will need to be able to identify symmetry (look at the symmetry of the charge configuration), as well as recognize the direction of the field and where it cancels out. Be sure that you can do that for this configuration.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.