*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

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.

© 2004 by Prentice-Hall, Inc. A Pearson Company