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in 3D, electric field of a piont charge is inversely proportional to the square of distance while the potential is inversely proportional to distance. We can derive it from Coulomb's law. however, I don't known how to derive the formula in 2D and 1D. I read in a book that electric potential of a point charge in 2D is proportional to the logarithm of distance. How to prove it?

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The trick is to use Gauss' law.

Suppose space is a 2d plane (Flatland!), and that there's a charge $q$ sitting at the origin. Gauss' law says that if we enclose the charge in a 1-sphere $S$ (aka, a circle), then we must have $\int_S \langle \vec{E} , \vec{n}\rangle = 2 \pi q$ (in convenient units), where $\vec{n}$ is the normal vector to the circle. If you assume $\vec{E}$ is rotationally symmetric, i.e., $\vec{E} = E(r) \hat{r}$, this turns into $E(r) 2\pi r = 2\pi q$, implying that $E(r) = q/r$. Integrating a field that goes like $1/r$ gives you a logarithmic potential.

You can also uses Gauss' law in 1d, enclosing the charge in a $0$-sphere (two points, equidistant from the origin). I'll leave it to you to try that one.

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Thank you! But I am still confused: How is Gauss's law derived? In 3D, the most basic experimental law is Coulomb's law from which we can derive Gauss's law. Is there a basic law in 2D and 1D plays the role of Coulomb's law in 3D? – hlew Nov 18 '12 at 15:54
Gauss' law is one of Maxwell's equations: $\vec{\nabla}\dot{}\vec{E} = \mbox{constant} \times \rho$. It's fundamental, not derived. If you integrate both sides of this equation over a volume, and then apply the divergence theorem, you'll get the integral I used. – user1504 Nov 18 '12 at 16:05
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NO, Maxwell's equations are not fundamental. The fundamental experimental law is Coulomb's law, Faraday's law of induction... Base on Coulomb's law in 3D, we define electric field of a piont charge and obtain the formula of the electric field. By integrating both sides of this formula over a closed surface, and then applying the divergence theorem, we can get Gauss's law(ie one of Maxwell's equations) – hlew Nov 18 '12 at 16:48
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@trhyt: You seem awfully sure of yourself here. But Maxwell's equations apply in situations where the Newtonian force law implicit in Coulomb's law has long since broken down. – user1504 Nov 18 '12 at 21:06
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@Chris: "In convenient units" – user1504 Jan 30 at 1:22
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Irrespective of the dimensions, Poisson equation is always true. That is, if $\phi$ is the electric potential and $\rho$ is the charge density then, $\nabla^2\phi = \rho/\epsilon_0$. The Green's function of this equation satisfies $\nabla^2 G(\vec{x},\vec{x}^\prime) = \delta(\vec{x} - \vec{x}^\prime)$.

A Fourier transform of this equation is $k^2G(\vec{k}) = 1$ or $G(\vec{k}) = 1/k^2$. A inverse 3d transform will give $1/r$ and an inverse 2d transform will give $\log r$ dependence. One can do the mathematics for 1d case as well to get a $r$ dependence.

The key point is that the Fourier transform of the Green's function of the Laplacian in any dimensions is $1/k^2$. The potential due to a point charge is just the inverse Fourier transform of $1/k^2$ in appropriately dimensioned space.

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Given a little tensor machinery, you can give $\nabla$ its own vector Green's function also. Much less circuitous, in my mind, than having to backtrack to potentials. – Muphrid May 14 at 20:23
@Amey Joshi Why is Poisson equation always true in any dimension? – hlew May 15 at 4:28
@hlew We believe Poisson equation to be valid in any dimensions because it follows from a) Gauss law and b) the fact that electrostatic field can be expressed as a gradient of a scalar potential. The latter two facts are assumed to be true in any dimensions. – Amey Joshi May 15 at 6:22
@Muphrid I don't know the technique you mentioned. Would you mind giving some more details? – Amey Joshi May 15 at 6:27
@Amey Joshi If Gauss law is valid in any dimension, then we can just use Gauss' law to answer the question. The approach by Poisson equation seems tortuous. – hlew May 15 at 7:50
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