**Java Security Update:** Oracle has updated the security settings needed to run Physlets.

Click here for help on updating Java and setting Java security.

# Exploration 24.1: Flux and Gauss's Law

green surface | red surface | blue surface

Please wait for the animation to completely load.

In this Exploration we will calculate the flux, Φ, through three Gaussian surfaces: green, red, and blue **(position is given in meters and electric field strength is given in newtons/coulonb)**. 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. Restart.

Flux is a measure of the electric field through a surface. It is given by the following equation:

Φ = ∫** _{surface} E · **d

**A**= ∫

**E cosθ dA,**

_{surface}where **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.

Move the test charge along one of the Gaussian surfaces (you must imagine that it is a sphere even though you can only see a cross section of it).

- What is the magnitude of the electric field along the surface?
- In what direction does it point?
- What direction is normal to the Gaussian surface?

If the electric field, **E**, and the normal to the Gaussian surface, **A**, always point in the same direction relative to each other, and the electric field is constant, then the equation for flux becomes: Φ = Ecosθ ∫dA = EAcosθ.

- In the case of the point charge in (a)-(c), what is the angle between the electric field and the normal to the surface?

This means that cosθ = 1. Therefore, for this case, Φ = EA.

- Calculate the flux for the surface you've chosen (remember that the surface area of a sphere is 4πR
^{2}). - Calculate the flux for the other two surfaces.

Because the electric field decreases as 1/r^{2}, but the area increases as r^{2}, the flux is the same for all three cases. This is the basis of Gauss's law: The flux through a Gaussian surface is proportional to the charge within the surface. With twice as much charge, there is twice as much flux. Gauss's law says that Φ = q_{enclosed}/ε_{0}.

- What is the magnitude and sign of the point charge?

Exploration authored by Anne J. Cox.

Script authored by Mario Belloni.

« previous

next »