# Gauss's Law - Irregular Surfaces

1. Dec 30, 2011

### Rarugged

Gauss's Law -- Irregular Surfaces

I don't fully understand why Gauss's Law holds for any Gaussian surface. My textbook clearly derives Gauss's Law from Coulomb's Law using a spherical surface, but it then extends the result to any Gaussian surface without sufficient explanation.

Why does Gauss's Law hold for irregular surfaces as well?

Explanations I have tried to understand include terms like: field lines, projections.

2. Dec 30, 2011

### clem

Re: Gauss's Law -- Irregular Surfaces

For a point charge, $$\oint{\bf dS\cdot E}=q\oint{\bf dS\cdot{\hat r}}/r^2 =q\oint d\Omega=4\pi q,$$, because $$d\Omega={\bf dS\cdot{\hat r}}/r^2$$.

Last edited: Dec 30, 2011
3. Dec 30, 2011

### Snicker

Re: Gauss's Law -- Irregular Surfaces

This is a very good question! Although your teacher didn't go over it, however your textbook may have a "proof". Search there.

The fact that Gauss' law works for irregular closed surfaces in addition to spheres lies in the fact that they both have the same measure of "solid angle" when measured from their insides.

It is best to think of this with a two-dimensional analogue. Consider a rubber band in a coordinate system with the origin at the band's interior. Suppose we took a ray from the origin and swung it across the plane keeping it fixed at the origin. How much "angle" does rubber band take up? That is, how much angle is swept by the ray until it no longer intersects the rubber band? An entire revolution. That is, $$2\pi$$. This is because the rubberband surrounds the origin and is closed. It doesn't matter what shape it is. It could be a circle or a hexagon or any shape as long as it surrounds the origin. If the origin lied outside the rubber band. Then the angle swept out by the rubber band would be less than that of a revolution. Can you visualize why?

Mathematically, we would say that

$$\oint_\Gamma {d\phi } = 2\pi$$.​

Here, gamma is any closed curve with the origin at the interior and phi is the angle parameter of the ray from the origin doing the sweeping.

However, this can be generalized in three dimensions. Here, we consider "solid angles." Recall that an angle corresponds to the length of an arc on the unit circle. In the same manner, solid angles correspond to the surface area of some patch of the unit sphere (total surface area = $$4\pi$$). We then have
$$\oint_\Delta {d\Omega = 4\pi }$$.​

Here, delta is some closed surface with the origin in the interior and omega is a parameter describing the solid angle. Can you visualize this?

A very special thing happens when you take the flux integral of an inverse-square field. Remember that an angle is the length of an arc divided by the radius. Similarly, a solid angle is the area of a patch divided by the SQUARE of the radius (the solid angle must be unitless.). Consider a charge q and a closed surface surrounding it. Taking the flux of the charge's electric field through that surface, we then have (in Gaussian units)
$$\oint {{\bf{E}} \cdot {\bf{dS}}} = q\oint {\frac{{d{\bf{S}}}}{{{r^2}}}} = q\oint {d\Omega = 4} \pi q$$

I hope this helps.

Last edited: Dec 30, 2011
4. Dec 31, 2011

### Born2bwire

Re: Gauss's Law -- Irregular Surfaces

In addition to the above, Gauss' Law is basically just an application of the divergence theorem from vector calculus. Any mathematical proof of the divergence theorem would be sufficient for your question.

5. Dec 31, 2011

### fonz

Re: Gauss's Law -- Irregular Surfaces

In addition to the above in a more colloquial sense think about a charge inside a Gaussian sphere, it has lines of flux which extend outward.
Gauss's law states that the amount of flux depends only on the magnitude of the enclosed charge. So to be very crude if you were measuring flux on the number of lines penetrating this spherical Gaussian surface and it turned out to be x amount of flux lines then does it matter if you drew a sphere or a rectangle or some abstract shape? So long as the surface is closed then the number of lines penetrating it is the same.