2
$\begingroup$

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 $E(r) = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$. Integrating a circumference of radius r in 2D gives $E(r) = \frac{1}{2\pi\epsilon_0}\frac{Q}{r}$. 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:

$\int E\cdot dS^0 = E = Q/\epsilon_0$

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?
$\endgroup$
1
$\begingroup$

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 $$ E(r)=\frac{1}{2\pi \epsilon_0}\frac{Q}{r} $$ is not correct if $Q$ represents a 3D point-like charge. It should be written as $$ E(r)=\frac{1}{2\pi \epsilon_0}\frac{\lambda}{r}, $$ 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 $$ \begin{align} \phi(r) &=-q_1 q_2\ln r ~~~~~~~~~ {\mathrm{in~~2D} } \\ \phi(r) &=~~q_1 q_2 \left| x \right| ~~~~~~~~~ {\mathrm{in~~1D} } \end{align} $$ 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.

$\endgroup$
2
  • $\begingroup$ Thank you very much for the answer. However, how can you use Gauss Law, which is fundamental, to obtain the equation you wrote $E(r) \propto \lambda$? $\endgroup$ Apr 15 at 10:59
  • $\begingroup$ @AliEsquembreKucukalic with the usual 3D charges, the field of a uniformly charged line can be obtained by using a finite cylindric volume with the axis of the cylinder coinciding with the charged line. By symmetry, there is no flux through the two bases of the cylinder and the formula for the surface of the lateral area of the cylinder allows to get the result. $\endgroup$
    – GiorgioP
    Apr 15 at 11:21
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Very clear answer, thank you very much. The case of the infinite planes was known to me, but it implies a 2D universe, right? I am wondering if we can get the 1D Field in an actual 1D system. $\endgroup$ Apr 14 at 9:51
  • $\begingroup$ The thing is that a 3D world where everything is a plane is equivalent to a 1D world where everything is a point. One can of course consider particles on quantum wires surrounded by insulators and conductors that might approximate the 1D case, but there are many extraneous issues there (the E field extends in 3D locally). $\endgroup$ Apr 14 at 14:41
0
$\begingroup$

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 $\propto\frac{1}{r^0} \Rightarrow$ 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.

$\endgroup$
2
  • $\begingroup$ Yes, but thinking of field lines in this case only shows that there is only a single possible direction, right? They do not show that the space dependence does not exist, or if they do, I am not seeing it right now. And thank you very much for the reference :) $\endgroup$ Apr 14 at 9:53
  • $\begingroup$ A field line cannot just disappear so once it leave a charge it must continue on confined to the one dimensional space it is enclosed in. $\endgroup$
    – Farcher
    Apr 14 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.