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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,
$$\begin{equation}\tag{2}F=\frac{1}{2\pi\epsilon}\cdot\frac{q_1q_2}{r}\end{equation}$$
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?

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3 Answers 3

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:

http://en.wikipedia.org/wiki/Green%27s_function#Table_of_Green.27s_functions

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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.

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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.

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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

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