# Does the electric field in 1D exist in nature?

Physics Asked by Ali Esquembre Kucukalic on May 8, 2021

This is a question related to:

Electric field and electric potential of a point charge in 2D and 1D

Let’s study in more detail the 1D case. In the 3D case we integrate a sphere of radius $$r$$ and get $$E(r) = frac{1}{4piepsilon_0}frac{Q}{r^2}$$. Integrating a circumference of radius r in 2D gives $$E(r) = frac{1}{2piepsilon_0}frac{Q}{r}$$. This leads to my first question:

• What are the dimensions for the electric field in 1D, i.e., should we modify the units of the permittivity so that we get again the field in units of $$force/charge$$?

If we go to 1D, then it seems there is no possible shape over which to do the integral, i.e., we are integrating over a $$0$$-sphere. Thus, the Gauss law should give something like:

$$int Ecdot dS^0 = E = Q/epsilon_0$$

This is, in fact, what many books claim: the field in 1D goes as $$1/r^0$$. So my second question is

• Does this mean that in a 1D space particles would interact with the same strength at every possible distance? Do we have such an example in nature?

Electrostatics in less than 3D is a fascinating subject. However, one should be aware that the extension to spaces with dimensionality different than three is not unique.

On the one hand, there is the most obvious choice of introducing a smaller dimensionality as the effect of dealing with special 3D charge distributions that require less than three coordinates to be specified. In this first case, the lower dimensionality corresponds to a reduction of the degrees of freedom required to describe a spatial configuration. This is the case of uniformly charged infinite parallel lines and uniformly charged parallel planes. In the former case, two coordinates are enough to specify the position of one of these lines; in the latter, one is enough to specify the position of a plane. Therefore, the dimensionality of the space configuration space is $$2$$ and $$1$$, respectively. All the relations and physics dimensions remain the same as in $$3D$$ dimensional case. In particular, it remains meaningful to speak about point-like charges interacting through the usual Coulomb potential. The interaction laws between infinite parallel charged lines and planes could be derived from the usual Coulomb law, in the limit of continuous line and surface charge densities. Of course, the Gauss theorem still requires a 3D volume and a 3D surface. Therefore your formula $$E(r)=frac{1}{2pi epsilon_0}frac{Q}{r}$$ is not correct if $$Q$$ represents a 3D point-like charge. It should be written as $$E(r)=frac{1}{2pi epsilon_0}frac{lambda}{r},$$ where $$lambda$$ is a line charge density, such that its 1D integral gives a result with the same physical dimension as a 3D charge.

On the other hand, one can imagine a different case, not based on the physical 3D charges, but much closer to the formal structure of the 3D electrostatics: a model based on the validity of Gauss's law in D dimensions. In such a case, it is not Coulomb's potential in real space that remains valid, but its Fourier transform proportional to $$frac{1}{k^2}$$. Therefore, the equivalent of Coulomb's law becomes begin{align} phi(r) &=-q_1 q_2ln r ~~~~~~~~~ {mathrm{in~~2D} } phi(r) &=~~q_1 q_2 left| x right| ~~~~~~~~~ {mathrm{in~~1D} } end{align} In such a case, there is no reason, and it impossible to maintain the same physical dimensions for charges as in 3D. However, this is a more coherent, fully 1D or 2D theory.

Notice that, although it is impossible to have real 1D or 2D electric charges, other physical systems can be well described by such a lower-dimensional electrostatics. The simplest example I can think of is the interaction between line vortexes with a logarithmic behavior.

Correct answer by GiorgioP on May 8, 2021

One can make a 1D situation by extending all the charges into infinite parallel planes or sheets, just as one can make a 2D situation by replacing the charges with parallel lines. Since everything is the same if one translates along the extended directions, the field does not depend on the added dimension(s).

Yes, the result implies that in an 1D universe every single charged particle feels a constant force due to every other charged particle.

Answered by Anders Sandberg on May 8, 2021

Think of it in terms of electric field lines emanating from a charged particle.
In one dimension where can the electric field lines (packed on top of one another??) go other than along the line defining the one dimension?
So the electric field strength will be constant $$proptofrac{1}{r^0} Rightarrow$$ constant.

My guess is that this is not something one comes across in practice other than in a limiting case?

For a much more detailed analysis read Electrodynamics in 1 and 2 Spatial Dimensions.

Answered by Farcher on May 8, 2021