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# Illustration 24.1: Flux and Gaussian Surfaces

green surface | red surface | orange surface | blue surface

Please wait for the animation to completely load.

The bar graph shows the flux, Φ, 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 N·m ^{2}/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, Φ, is a measure of the amount of electric field through a surface. Gauss's law relates the flux to the charge enclosed (q

_{enclosed}) in a Gaussian surface:

Φ = q_{enclosed}/ε_{0} and Φ = ∫** E · **d**A **= ∫ E cosθ dA

where ε_{0} is the permittivity of free space (8.85 x 10^{-12} C^{2}/N·m^{2}), **E** is the electric field, d**A** is the unit area normal to the surface, and θ 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_{enclosed}/ε_{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 ∫ E cosθ dA is not equal to the integral E ∫ cosθ 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.

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