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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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  • $\begingroup$ 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? $\endgroup$ – hlew Nov 18 '12 at 15:54
  • $\begingroup$ 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. $\endgroup$ – user1504 Nov 18 '12 at 16:05
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    $\begingroup$ 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) $\endgroup$ – hlew Nov 18 '12 at 16:48
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    $\begingroup$ @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. $\endgroup$ – user1504 Nov 18 '12 at 21:06
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    $\begingroup$ @Chris: "In convenient units" $\endgroup$ – user1504 Jan 30 '13 at 1:22
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The Coulomb potential has the following forms for a positive charge in each dimensionality: \begin{align} \Phi_{\operatorname{1-d}}(r) &= -\frac{\sigma}{2\epsilon_0} r, \\ \Phi_{\operatorname{2-d}}(r) &= -\frac{\lambda}{2\pi \epsilon_0} \ln(r),\ \mathrm{and} \\ \Phi_{\operatorname{3-d}}(r) &= \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r}\right). \end{align} The reason for this is that the electric field, defined as $-\nabla\Phi$ in general and $-\dfrac{\partial\Phi}{\partial r} \hat{r}$ in this case, times the measure of the boundary of a "ball" has to be a constant. In 1-d, a ball is a line and the measure of its boundary is just the number of points on its ends (i.e. 2). In 2-d the ball is a circle and the measure of its boundary is the circumference (i.e. $2\pi r$). In 3-d the ball is a sphere and the measure of its boundary is the sphere's surface area ($4\pi r^2$). Notice that those are exactly the quantities in the denominator when we calculate the electric field from the potentials: \begin{align} \vec{E}_{\operatorname{1-d}}(r) &= \frac{\sigma}{\epsilon_0} \left(\frac{\hat{r}}{2}\right), \\ \vec{E}_{\operatorname{2-d}}(r) &= \frac{\lambda}{\epsilon_0} \left(\frac{\hat{r}}{2\pi r}\right),\ \mathrm{and} \\ \vec{E}_{\operatorname{3-d}}(r) &= \frac{q}{\epsilon_0} \left(\frac{\hat{r}}{4\pi r^2}\right). \end{align}

The name of the law that implies this is known as Gauss's law.

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  • $\begingroup$ I was wondering about the divergence of -ln(r) after r = 1 in 2D. Can we still use electrostatic after r = 1 ? What does it mean? $\endgroup$ – projesh Jun 20 '18 at 12:58
  • $\begingroup$ @projesh the divergence happens at $r=0$, just like in 3D. $r=1$ is where the natural logarithm switches sign, which is meaningless because it can be moved by adding a constant to the potential. $\endgroup$ – Sean E. Lake Jun 20 '18 at 14:01
  • $\begingroup$ Thanks, but I am still confused. Even if you add a large number, at $x \to \infty, -ln(x)$ is undefined and does not go to zero like 3D. Does it mean it can only be used up to a certain length scale depending on the constant? $\endgroup$ – projesh Jun 20 '18 at 21:40
  • $\begingroup$ @projesh That problem happens in 1-d, too. The potential goes to zero at infinity only for 3-d or higher. It is just a fact that 1-d is zero at $r=0$ and goes to $\infty$ as $r\rightarrow\infty$, 3-d and higher goes to $0$ as $r\rightarrow\infty$ and goes to $\infty$ as $r\rightarrow 0$, and 2-d goes to infinity at both ends. $\endgroup$ – Sean E. Lake Jun 21 '18 at 17:24
  • $\begingroup$ @SeanE.Lake, thanks a lot for the detailed explanation. Can you please comment on the numerical values and on the dimensional analysis of the "charges" $\sigma$, $\lambda$ and $q$ and of the vacuum permittivity $\epsilon_0$, i.e. how these changes with the dimensionality of the system. From what you've written, it seems that $\epsilon_0$ is always the same (both its numerical value and its dimension, i.e. 8.854 F/m, no matter of the dimensionality of the system). $\endgroup$ – AndreaPaco Apr 13 '19 at 8:43
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Deriving the electric field for a 2D world can be done in several ways. It would depend on what behavior of the electrostatic interaction you want to preserve in that world.

I you asked Coulomb, when he published his expression for the interaction, he would probably have said that the expression should be the same ($1/r^2$) just that the distance $r$ would only involve the $x$ and $y$ dimensions $r^2=x^2+y^2$.

However, when Gauss found that the flux integral is proportional to the charge enclosed by it, then is not so easy to answer. Because if you assume a 2D electron to have a $1/r^2$ field, then such world would not obey Gauss law. And if you impose that world to follow Gauss law, then Coulomb living in this 2D world would have found a $1/r$ law instead.

So which of the two properties is more fundamental? In my opinion, Gauss law is, but I have no way to prove that since there is no 2D world to experiment with.

My answer to your question the book you read based its statement about the point charge's electric potential in a 2D by tacitly assuming that Gauss law holds for any world regardless of the dimensions. But there is no proof of its veracity.

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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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  • $\begingroup$ 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. $\endgroup$ – Muphrid May 14 '13 at 20:23
  • $\begingroup$ @Amey Joshi Why is Poisson equation always true in any dimension? $\endgroup$ – hlew May 15 '13 at 4:28
  • $\begingroup$ @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. $\endgroup$ – Amey Joshi May 15 '13 at 6:22
  • $\begingroup$ @Muphrid I don't know the technique you mentioned. Would you mind giving some more details? $\endgroup$ – Amey Joshi May 15 '13 at 6:27
  • $\begingroup$ @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. $\endgroup$ – hlew May 15 '13 at 7:50

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