Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We can notice that in the Coulomb's law equation,
$$\begin{equation}\tag{1}F=\frac{1}{4\pi\epsilon}\cdot\frac{q_1q_2}{r^2}\end{equation} $$

$4\pi r^2$ factor in the denominator expresses directly the surface of a virtual sphere with radius $r$. Actually we can look at this equation as it was for $3$ dimensional objects. If we suppose want to consider for $2$ dimensional objects, can we modify the equation as,
Here we can think of $2\pi r$ as area of virtual circle. I don't really know whether it works or not. So, can we have equation (2) as the modified equation for electrostatic force between two $2$ dimensional uniformly charged objects?

share|cite|improve this question

Physically speaking, the laws of electrodynamics are 3-dimensional and so you have to take these as starting point and see what they imply for any charge configuration of interest. A force $F$ of form $\propto\frac{1}{4\pi}\frac{1}{r^2}$ falls faster than one which goes as $\propto\frac{1}{2\pi}\frac{1}{r}$ and so without futher information, the physics which apply is the known behaviour $\propto\frac{1}{4\pi}\frac{1}{r^2}$, which you can also write as $\propto\frac{\partial}{\partial r}\left(\frac{1}{4\pi}\frac{-1}{r}\right)$

Mathematically speaking, what what you do is to compute $F\propto\text{grad}(G)$, where the force $F$ is the gradient of a potential $G$ which is given from the Poisson equation in $n$ dimensions, and where there is only one charge in the center of the coordinate system. Your two dimensional force is $F\propto \frac{1}{2\pi}\frac{1}{r}= \frac{1}{2\pi}\frac{\partial }{\partial r}\mathrm{ln}(r)$, i.e. $G= \frac{1}{2\pi}\mathrm{ln}(r)$. A list of similar potentials is given here, only the fifth of which corresponds to electrostatics in 3 dimensions:

share|cite|improve this answer

Gauss law is the most general form of equation to describe the electric field. Columb law for an arbitrary electric field states F=q*E. Gauss law in its integral form reads

enter image description here

D is the electric flux density, dS is the surface normal element, rho is the charge density and dV is the volume element. What that equation physically says is, the charge confined in a volume is equal to the surface integral of flux normal to the surface of that volume. As you see it is 3D by definition as it includes volume and surface. If you tested equation 2 you wrote against Gauss law, you will see it is inconsistent. That is why equation 2 doesn't describe a point charge under any circumstance, simply because the flux across the "circle" as you described it is part of the total flux through the sphere.

As a general rule, Gauss law applies to 3D, when you want to use in 2D or 1D you should start from 3D and make necessary simplifications. For 2D usage think of it as taking a slice to convert the 3D to 2D. The law will remain the same.

For the record, equation 2 has a r-dependence that describes an infinitely long charged line. That is one of the common exercises student do in elementary electromagnetic class, which is finding the electric field of an infinitely long charged line using Gauss law.

Have a look here for general description of Gauss law. In page 6 you see the example I am speaking about.

share|cite|improve this answer

Well surely you can consider it for 2 dimensional chardes, but to check it out experimentally would simply not be possible. As no charge known to us is 2 dimensional in its existence and its electric influence is also spread in the 3 dimensions we know of, experiencing and experimenting with 2d is not possible to date and hence your hypothesis can not be tested for validation.

Seeing the analogy your extrapolation seems correct and i believe similarly we can get results even for a single dimensional world or even multiple dimensional worlds. But again all of these can neither be proven nor disproven.

share|cite|improve this answer
There are analogous systems in condensed matter, like Abrikosov vortices in superconductors, which have effective 2D interactions. – Michael Brown Oct 23 '13 at 9:52
I do not really know about abrikosov vortices but is that(condensed matter) only where the analogue would be applicable, nowhere else ? – Rijul Gupta Oct 23 '13 at 9:55

While, the answer is yes, you can get the same result if you start from the quantum field theory. The result gotten from the quantum field theory is the force is inverse to the distance of dimension minus one power. In two dimension, 2-1=1, so the force inverse to r. In N dimension, the force is inverse to $r^{N-1}$.

share|cite|improve this answer

What is 2D charge? It is not a good definition, and you should be more carefully. The force will be proportional to the charge or the square of charge? 2D charge in the vortex is the topological charge? The thing in 2D makes different.

share|cite|improve this answer
What's wrong with two-dimensional charges? We can define gravity in two dimensions, so why not electric charges? – Kyle Kanos Oct 12 '15 at 13:28
What is the connection between gravity and charges? – pring Oct 12 '15 at 14:36
The key one is the mathematical connection: both forces are built on inverse-square laws, $F_{grav}=GMm/r^2$ & $F_{elec}=kQq/r^2$. Obviously fundamental differences in what they describe, but the mathematics are pretty similar. – Kyle Kanos Oct 12 '15 at 14:39
The concept mass can be interpreted by Higgs mechanism, will the charge obeys the same reason? – pring Oct 13 '15 at 1:17
What is being discussed is classical mechanics (specifically the use of Green's function, as mentioned by NikolajK) without any need of going into particle physics. Mathematically, the Green's function can allow you to solve the gravitational and electric forces in 1, 2 & 3 dimensions. – Kyle Kanos Oct 13 '15 at 10:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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