Gauss' theorem and inverse square law

In summary, Gauss's Law states that the flux of the electric field through a closed surface is Q/ε, and this holds true for electrostatics in simply connected regions of space. However, for time-dependent fields, the leading order of the electric field goes like 1/r, but since it is perpendicular to the direction of propagation, the total flux is zero. This is why it is correct to calculate the electric field caused by an infinite long charged rod or an infinite plane charged surface, even though they do not obey the inverse square law. This is because individually, the charges on the rod or surface do obey the inverse square law, and it is only the sum of these electric fields that are not inverse square.
  • #1
physics user1
So, I know that the gauss law states that the Flux of the electric field through a closed surface is Q/ε , but does the gauss theorem works also for non inverse square law Fields?

I think not because in order to not have a Flux depending on distance but a constant one we need that r^2 of the surface has to cancel with something that is r^2 too, am I right?

And why then is correct to calculate the electric field caused by a infinite long charged rod or an infinite plane charged surface, the electric field there doesn't obey to the inveresults square law but we assume that the gauss theorem works...

I thought about this and I came to the conclusion that individually charges on the rod or the surface obeys to the inverse square law, it's just the sum of all these electric fields of all the charged enclosed in the gauss surface that are not inverse square so it's OK because that, is my "theory" Right?
 
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  • #2
Gauss's Law reads (in Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{E}=\rho.$$
For electrostatics you also have
$$\vec{\nabla} \times \vec{E}=0$$
and this implies that there exists (in any simply connected region of space) an electric potential such that
$$\vec{E}=-\vec{\nabla} \phi.$$
This implies
$$\Delta \phi=-\rho,$$
and the Treen's function for the (negative) Laplace operator in 3D Euclidean space is
$$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|}.$$
From a long distance you can characterize any charge distribution just by its total charge, and the corresponding monopole contribution to the potential reads
$$\phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}|} \; \Rightarrow \; \vec{E}=-\vec{\nabla} \phi=\frac{Q}{4 \pi |\vec{x}|^2} \frac{\vec{x}}{|\vec{x}|}.$$
So the leading order of the multipole expansion indeed always decays with ##1/r^2## with distance (in three dimensions).

Of course, this does not hold for the full Maxwell equations, i.e., for time-dependnet fields. There the leading order of the em. waves goes like ##1/r##.
 
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  • #3
The radiated electric field depends on 1/r but it's perpendicular to the direction of propagation so the total flux is zero. If you take an sphere centered on an accelerating particle the electric field will be perpendicular to the surface everywhere so the flux will be zero.
 

Related to Gauss' theorem and inverse square law

1. What is Gauss' theorem?

Gauss' theorem, also known as Gauss's law, is a fundamental principle in electromagnetism that relates the flow of an electric field through a closed surface to the charge enclosed within that surface. It states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space.

2. How does Gauss' theorem relate to the inverse square law?

Gauss' theorem is mathematically equivalent to the inverse square law, which states that the strength of an electric field is inversely proportional to the square of the distance from the source of the field. This means that as the distance from the source increases, the strength of the field decreases by the square of that distance.

3. What are some real-world applications of Gauss' theorem and the inverse square law?

Gauss' theorem and the inverse square law are used in a variety of fields, including electrical engineering, physics, and astronomy. They are used to calculate the force between charged particles, the strength of electric fields generated by electronics, and the intensity of light from distant stars and galaxies.

4. How does Gauss' theorem apply to non-uniform electric fields?

Gauss' theorem can be applied to non-uniform electric fields by dividing the surface into small, infinitesimal elements and calculating the flux through each element. The total flux through the entire surface can then be found by summing the individual fluxes. This method is known as the differential form of Gauss' theorem.

5. What is the relationship between Gauss' theorem and Coulomb's law?

Coulomb's law, which describes the force between two charged particles, is a special case of Gauss' theorem in which the electric field is constant and the surface enclosing the charge is a sphere. By using Gauss' theorem, the force between two charged particles can be calculated for more complex situations, such as non-spherical charge distributions.

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