# Net flux

1. Jan 3, 2007

### physgirl

"The net flux through ANY closed surface surrounding a point charge q is given by q/"permittivity of free space" and is independent of the shape of that surface."

I'm having a little trouble understanding this... when my book derived this formula, it did so by using the surface area of a sphere "4*pi*r^2" which made things cancel out and simplify out to q/"permittivity of free space"... so how does this equation still apply to ANY closed surface even if it's not a sphere? because I thought flux was proportional to the number of electric field lines going through a certain surface area, so the magnitude of the surface area still should matter...

a clarification would be great!

2. Jan 4, 2007

### Mindscrape

Let's see here:

$$\iiint_V \nabla \cdot \mathbf{E} dV = \iiint_V \frac{\rho}{\epsilon_0} dV$$

Pretty easy to see for an electric field you end up with the volumes canceling, which in general is not true. But I take it you don't know these mathematics. In words it just means that if you take a point charge, and it must be a point charge, that the flux is independent of the arbitrary volume. Sort of hard to explain without the mathematics. The statement is in fact true.

Draw out the field, and some easy volumes. Then note how no matter what volume you draw, as long as it is simple and closed, that the stuff coming out is the same. See if that helps. There are probably some cool java things online somewhere that will demonstrate this because it is so fundamental.

3. Jan 4, 2007

### Staff: Mentor

I'm not sure that I'm in-sync with Mindscrape's response. To the original post (OP) -- I'd try out the integrations over a few of the simplest 3-D boundaries that are available/useful. As an exercise, compare and contrast the solutions you get by integrating over the one uniform surface of the sphere versus the 6 surfaces of the cube...

4. Jan 4, 2007

### Hootenanny

Staff Emeritus
Don't forget that the density of field lines (no. of field lines per unit volume) is proportional to the field strength which obeys the inverse square law...

5. Jan 4, 2007

### tim_lou

Any close surface can be broken down into tiny pieces of spheres at varying distance from the center. So that Gauss's law is true for the whole surface.

or think about it in another way, pick a sphere that is enclosed by that random surface, then project little pieces of that random surface onto the sphere, you'll see that

$$\vec{E}\cdot\vec{dA}=E'dA'$$

where E' and dA' is evaluated on the sphere.

I know this justification is kind of sketchy... but hey, the proof of Gauss's law requires Stokes' theorem which I don't even know the proof after I finished Calc III...

Last edited: Jan 4, 2007
6. Jan 5, 2007

### Galileo

Nah, you need the divergence theorem for Gauss' Law. Intuitively, the difference in flux from using a (small) sphere and another arbitrary surface is the flux through the outer surface minus the flux through the small sphere (make the sphere fit inside the arbitrary one). But those surfaces enclose a volume where the divergence is zero (which you can check from Coulomb's law, the divergence is zero, except at the location of the point charge), so by the divergence theorem the difference in flux through the arbitrary surface and the sphere is zero.