1. Jul 4, 2007

### lugita15

According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed by the surface divided by epsilon_0. This obviously excludes charges outside the surface. But what about *on* the surface? That is, it lies neither inside nor outside. Does this charge count as being "enclosed" in the sense of Gauss's Law? In other words, does the electric field of a charge "on" a closed surface create a nonzero flux through the surface?

Any help would be greatly appreciated.

2. Jul 4, 2007

### Ronnin

I believe by definition all charge must be enclosed, therefore there is no surface charge.

3. Jul 4, 2007

### Meir Achuz

A Gaussian surface is a mathematical surface you decide on.
To apply Gauss's theorem, you must choose a surface that has charge either within or without, but not on your Gaussian surfce.

4. Jul 4, 2007

### Nancarrow

If a charge was on the surface, the flux would be undefined at that point (since the field itself is undefined at the charge itself)