The relation between Gauss's law and Coulomb law and why is it important that the electric field decrease proportionally to $\frac{1}{r^{2}}$?

Question

My question relates to the third MIT's video lecture about Electricity and Magnetism, specifically from $21:18-22:00$ : http://youtu.be/XaaP1bWFjDA?t=21m18s

I have watched the development of Gauss's law, but I still don't quite understand the link between Gauss's law and Coulomb law: How does Gauss's law change if Coulomb law would of been a different one.

I also don't understand why is it so important for Gauss's law that the electric field decrease proportionally to $\frac{1}{r^{2}}$ ?

For example, what would of happened if the electric field decrease proportionally to $\frac{1}{r}$ , or $\frac{1}{r^{3}}$ ?

Answer 1

Gauss' law and Coulomb's law are equivalent - meaning that they are one and the same thing. Either one of them can be derived from the other. The rigorous derivations can be found in any of the electrodynamics textbooks, for eg., Jackson. For eg., consider a point charge q. As per Coulomb's law, the electric field produced by it is given by $$\vec{E} = \frac{kq}{r^2}\hat{r}$$, where $k = \frac{1}{4 \pi \epsilon_0}$. Now, consider a sphere of radius $r$ centred on charge q. So, for the surface $S$ of this sphere you have: $$ \int_{S}\vec{E}.\vec{ds} = \int_{S}\frac{kq}{r^2}ds = \frac{kq}{r^2}\int_{S}ds = \frac{kq}{r^2}(4\pi r^2) = 4\pi k q = \frac{q}{\epsilon_0}$$, which is Gauss' law. Note that if the $r^2$ in the expression for the surface area of the sphere in the numerator did not exactly cancel out the $r^2$ in the denominator of Coulomb's law, the surface intergral would actually depend on $r$. Hence you would not have the result that the surface integral is independent of the area of the surface, which is what is implied by Gauss' law. Though this result has been derived for a sphere, it can be derived for any arbitrary shape and size of the surface, you can refer to Jackson for eg., for the rigorous derivation. Note that by performing these steps in reverse, you can also derive Coulomb's law from Gauss' law, thus demonstrating that they are equivalent.

Answer 2

The Gauss' law/inverse square connection is explained by Frederic above. But allow me to expand on the fundamental issue.

In a nutshell, if the inverse square law doesn't hold, then the photon must have mass, and hence a finite lifetime. This is explained well in Jackson's Classical Electrodynamics (look in the index for Proca Lagrangian, in the second edition). The use of a massive photon model has been pushed through theoretically and the resulting observations allow one to put an upper bound on the photon mass. If the photon mass were measureable, we would have to rethink all sorts of fundamental issues, especially in cosmology.

This is dry, but it's readily available. http://en.wikipedia.org/wiki/Proca_lagrangian

Interestingly there is a deep connection with pure mathematics too; the residue of a function is the 1/z term of the Laurent expansion. Why? That's because this is the only term that survives to give a finite contribution to an integral at infinity. 1/r would be the force associated with a 1/r^2 potential.

Answer 3

The $1/|r|^2$ dependence is fundamentally geometrical in nature, stemming from our 3d world.

Differential equations, even PDEs, are sometimes said to have Green's functions. Consider the differential equation

$$\nabla \cdot K = \alpha$$

for some vector field $K$ and some scalar field $\alpha$. This is structurally identical to Gauss's law. This equation has a Green's function $G$ that satisfies

$$\nabla \cdot G = \delta$$

where $\delta$ is the Dirac delta function. This describes, in essence, a point source $\delta$ generating a field $G$.

The Green's function is given by

$$G(r) = \frac{\hat r}{4\pi |r|^2}$$

Thus, any differential equation of the form $\nabla \cdot K = \alpha$ has the same Green's function in 3d--a point source always generates the same basic field.

In short, it is a physical statement to say $\nabla \cdot E = \rho/\epsilon_0$, but once that has been said, Coulomb's law, which describes a point charge, must inevitably follow due to the mathematical structure of the equations and the geometrical nature of 3d space.

(Of course, you're free to work the other way around.)

Edit: crucially, in 2d and 1d, the Green's functions are different. You probably already know these solutions. The 2d Green's function is something familiar from the field of a uniform line charge, and has $1/|r|$ dependence. The 1d Green's function has constant magnitude but changes direction on opposite sides of the "point" charge--this is the basic feature of a uniformly charged sheet.

Answer 4

Gauss's law states that the ratio of charge and the dielectric constant is given by a (two-dimensional) surface integral over the electric field:

$$\int E\cdot dA=\frac{Q}{\epsilon_0},$$

where I have omitted vector notation for simplicity. It can be linked to Coulomb's law by assuming spherical symmetry of the electric field and performing the integration.

The other way around, you can start by assuming that Coulomb's law,

$$E=\frac{1}{4\pi}\frac{Q}{r^2},$$

holds, and take the divergence on both sides of the equation. This leads to the differential form of Gauss's law:

$$\nabla\cdot E=\frac{\rho}{\epsilon_0},$$

where $\rho$ is the charge density. In order to reproduce its integral form, just integrate both sides of the equation. Detailed calculations can be found on the wikipedia page on Gauss's law. If you really want to understand how it all comes together (which I assume to be true, otherwise you wouldn't ask here), I would recommend reproducing the calculation on your own. Then you can also try to see what happens if you assume other forms of Coulomb's law.