# Electric field and potential of a point charge in a (strictly) 2D 'world'

I was trying to figure out how the electric potential and electric field are different in a 3D system versus in a 2D system (I take such a 2D 'world' to be the $$xy$$-plane, i.e. $$z=0$$, in a Cartesian coordinate system). The context/motivation has to do with the Coulomb interactions that electrons/holes experience in a 2D or a 3D metal, how this is different and how this results in the Coulomb screening being different between the two. (I want to emphasise that such a 2D material is a truly 2D system, i.e. the charges living in such a material do not know about a third dimension.)

I have been looking on the internet a lot, and I know that the answer has to be something like $$V_\text{2D}(r) \sim \log(r) \qquad E_\text{2D}(r) \sim \frac{1}{r},$$ and of course $$V_\text{3D}(r) \sim \frac{1}{r} \qquad E_\text{3D}(r) \sim \frac{1}{r^2},$$ but when trying to reproduce the 2D result I am struggling. (PS. I am using SI units)

Starting with Gauss' law,

$$\nabla \cdot \mathbf{E}\left(\mathbf{r}\right) = \frac{\rho\left(\mathbf{r}\right)}{\varepsilon_0}$$

I transformed it to its integral form in 2D and 3D using the divergence theorem in "$$n$$" dimensions, $$\int_U \text{d}^{n}{r} \, \nabla\cdot\mathbf{A} = \oint_{\partial U} \text{d}^{n-1}r \, \mathbf{A} \cdot \mathbf{\hat{n}}.$$

which gives

$$\oint_{\partial S} \text{d}r \, \mathbf{E}\left(x,y\right) \cdot \mathbf{\hat{n}} = \frac{1}{\varepsilon_0} \int_S \text{d}^2 r \, \rho\left(x,y\right), \quad \text{(2D)}$$ $$\oint_{\partial V} \text{d}^2 r \, \mathbf{E}\left(x,y,z\right) \cdot \mathbf{\hat{n}} = \frac{1}{\varepsilon_0} \int_V \text{d}^3 r \, \rho\left(x,y,z\right). \quad \text{(3D)}$$

Here I explicitly show that $$\mathbf{E}$$ and $$\rho$$ are in 2D and 3D. PS. in 2D, $$S$$ is the surface you integrate over and $$\partial S$$ is its boundary 'line'; in 3D, $$V$$ is the volume you integrate over and $$\partial V$$ is its boundary surface

Now suppose we have a point charge $$Q$$ at the origin, i.e. $$\rho\left(\mathbf{r}\right) = Q \, \delta^{(3)}\left(\mathbf{r}\right)$$ (where $$\delta^{(3)}$$ is the Dirac delta function in 3 dimensions, i.e. with units $$\text{m}^{-3}$$). I am using this in 2D and in 3D, since $$\rho\left(\mathbf{r}\right)$$ must have dimensions of $$\text{C} \, \text{m}^{-3}$$ by virtue of Gauss' law, regardless the dimensionality.

In 3D, we use as the integration volume a sphere with radius $$r$$ centred at the origin (just like any elementary electromagnetism textbook does it).

In 2D, we use as the integration surface a circle with radius $$r$$ centred at the origin $$(x,y)=(0,0)$$ (remember, the 2D 'world' is the $$xy$$-plane, i.e. $$z=0$$ plane).

By symmetry considerations, it's clear that in 2D and in 3D the field only has a radial component. I will mark this as $$E_r$$ in both cases, although I want to emphasise that "$$r$$" in 2D means $$\sqrt{x^2+y^2}$$ while in 3D it means $$\sqrt{x^2+y^2+z^2}$$. This gives

$$2\pi r E_r = \frac{Q}{\varepsilon_0} \int_S \text{d}^2 r \, \delta^{(3)}\left(\mathbf{r}\right) = \frac{Q}{\varepsilon_0} \delta\left(z\right), \quad \text{(2D)}$$ $$4 \pi r^2 E_r = \frac{Q}{\varepsilon_0} \int_V \text{d}^3 r \, \delta^{(3)}\left(\mathbf{r}\right) = \frac{Q}{\varepsilon_0}, \quad \text{(3D)}$$

where I made use of $$\delta^{(3)}\left(\mathbf{r}\right) = \delta(x)\delta(y)\delta(z)$$. Rewriting gives my final result for the electric field:

$$\mathbf{E}\left(x,y\right) = \frac{Q}{2\pi\varepsilon_0} \frac{1}{r} \delta\left(z\right) \mathbf{\hat{r}}, \quad \text{(2D)}$$ $$\mathbf{E}\left(x,y,z\right) = \frac{Q}{4 \pi \varepsilon_0} \frac{1}{r^2} \mathbf{\hat{r}}, \quad \text{(3D)}$$

Disturbingly, I'm left with a Dirac delta function in my electric field, although I do get the $$E_\text{2D} \sim 1/r$$ dependence I expected. I guess it makes some sense that the $$\delta$$-function is there, since I'm only concerned with $$z=0$$ anyways, but it of course shouldn't be the case that a physical quantity like the electric field is merely a mathematical 'distribution' i.e. delta function. However, I cannot just leave out the $$\delta$$-function, since it carries a unit of $$1/\text{meter}$$ which is crucial to make the units consistent.

What am I doing wrong here? As you can see, I tried to do this derivation rigorously by starting from Gauss' law in coordinate-free form, and then converting using the divergence theorem; in this step, the difference between 2D and 3D should arise. I have seen other questions on this website where people would instead simply say that the right-hand side becomes the "enclosed charge" $$Q_\text{encl} = \int_V \text{d}^3 r \,\rho\left(\mathbf{r}\right)$$ even in 2D (e.g. this answer), but they end up (as far as I can see) with inconsistent units (i.e. they miss a $$1/\text{meter}$$ which I have in my $$\delta\left(z\right)$$). Another common method of getting rid of this I see is that people use a 'line charge density' $$\lambda$$ (units $$\text{C} \, \text{m}^{-1}$$) (e.g. this answer), but that is only if you have an infinitely long line with a charge per unit length $$\lambda$$. This shouldn't be valid for a point charge I think.

I hope anyone can help me here. If anything is unclear, please let me know and I will try to clarify it.

• Why don't you use surface charge density. Divergence theorem is connecting integral of region to boundary. Flow of river or water needs divergence in two dimension only. Commented Dec 14, 2022 at 12:53
• @NeilLibertine what form of Gauss' law (differential or integral form) would I use that contains a surface charge density? Because I believe you would run into the same problem, since $\rho\left(x,y,z\right) = \sigma\left(x,y\right) \delta\left(z\right)$ for an arbitrary surface charge density $\sigma\left(x,y\right)$ in the $z=0$ plane. Commented Dec 14, 2022 at 13:05
• Integration of delta function gives unity. What is the problem. Commented Dec 14, 2022 at 13:10
• @NeilLibertine if I integrate $\nabla \cdot \mathbf{E}\left(\mathbf{r}\right) = \frac{1}{\varepsilon_0} \sigma\left(x,y\right) \delta\left(z\right)$ over the entire $xy$-plane, I am not integrating over $z$. So, the $\delta\left(z\right)$ would not disappear, that's the problem. Commented Dec 14, 2022 at 13:13
• But that assume that z=0., and if you want to include delta in z direction then integrate it separate with z direction of volume variable. If you are integrating a region in plane then why are you defining z. Commented Dec 14, 2022 at 13:22

To model a 2D system in a 3D world, your quantities should be independent of $$z$$. In other words, every slice of constant $$z$$ should be identical. You use the density $$\rho(x,y,z)=Q\delta(x)\delta(y)\delta(z)$$, which does depend on $$z$$.

How do we fix this? The answer is simple. Just make the system translation invariant in the $$z$$ direction. For example $$\rho(x,y,z)=\lambda \delta(x)\delta(y)$$, which represents a wire with constant line density. The invariant variable is now $$x^2+y^2$$ (polar coordinates), instead of $$x^2+y^2+z^2$$. What do you get when you use this? What test surface would you use for a problem with cylindrical symmetry?

• Thanks for your answer! I now understand the issue that you're addressing: there shouldn't be any $z$-dependence, so you just use an infinitely long wire that lies on the $z$-axis. Then you would use a circular integration surface (radius $r$) that is centred at the origin, and indeed I then get $\mathbf{E}(r) = \frac{\lambda}{2\pi \varepsilon_0} \frac{1}{r} \mathbf{\hat{r}}$ (in polar coordinates). But how can I then interpret this $\lambda$ as being the charge of a point charge? Say, the point charge has a charge of 1 Coulomb, then what would $\lambda$ be? Commented Dec 14, 2022 at 13:28
• @Thijs With the test surface I was hinting at a cylinder in 3D. A cylinder has zero flux through the top and bottom because of mirror symmetry and so the only contribution is from the walls. $\lambda$ is a line density so has units of coulomb/length. For all intents and purposes, the field generated by this wire acts like a point charge for a 2D slice. I'm sorry but I don't know the proper way to equate this line density with a 2D point density. Maybe this could be a separate question? Commented Dec 14, 2022 at 16:13
• Ah yes, of course, if you do the integral in 3D you use a cylinder, if you do it in 2D you use a circle. I tried it and found that you get the same result in both cases, i.e. the length of the cylinder that you use, cancels out, as it does in textbook examples. I am indeed quite tempted to make a separate question about the meaning of $\lambda$ in terms of a point charge for such a point charge when you only consider a slice of this line charge, I think it's very interesting. Commented Dec 15, 2022 at 11:15

Now suppose we have a point charge $$Q$$ at the origin, i.e. $$\rho\left(\mathbf{r}\right) = Q \, \delta^{(3)}\left(\mathbf{r}\right)$$ (where $$\delta^{(3)}$$ is the Dirac delta function in 3 dimensions, i.e. with units $$\text{m}^{-3}$$). I am using this in 2D and in 3D, since $$\rho\left(\mathbf{r}\right)$$ must have dimensions of $$\text{C} \, \text{m}^{-3}$$ by virtue of Gauss' law, regardless the dimensionality.

This is your problem. By the definition of density, you must have $$\int \rho(\mathbf{r}) \, d^n \mathbf{r} = Q_\text{enc}$$ where $$Q_\text{enc}$$ has the units of charge and $$n$$ is the number of dimensions. This implies that dimensionally you must have $$[\rho] [L^n] = [Q] \quad \Rightarrow \quad [\rho] = \frac{[Q]}{[L^n]}.$$ In other words, the charge density in 2-D must have units of $$\text{C/m}^2$$, not $$\text{C/m}^3$$.

Note that what does change between dimensions is the dimensions of $$\epsilon_0$$; we must also have from Gauss's Law $$\oint \mathbf{E} \cdot d^{n-1}\mathbf{a} = \frac{Q_\text{enc}}{\epsilon_0} \quad \Rightarrow \quad [\epsilon_0] = \frac{[Q]}{[E][L^{n-1}]}.$$ If we want $$\mathbf[E]$$ to have the same units in all dimensions (which seems reasonable) then the easiest route is to say that the units of $$\epsilon_0$$ differ between the various dimensions. So in 3 dimensions it has units of $$\text{C}^2/\text{N}\cdot\text{m}^2$$; in 2 dimensions it would instead have units of $$\text{C}^2/\text{N}\cdot\text{m}$$. If we modify $$\epsilon_0$$ in this way between the dimensions, then everything is nice & dimensionally consistent.

• Thanks for your answer. I get that the density should indeed have units of $\text{C} \, \text{m}^{-2}$, thanks for clarifying that. I also see that redefining $\varepsilon_0$ to have an extra unit of $\text{meter}$, that this issue would be solved, although I find it hard to get my head around what it means to redefine $\varepsilon_0$ in that way. In fact, in this answer, this 'dimensionality' issue is resolved by using $\lambda$ $[\text{C} \, \text{m}^{-1}]$ instead of $Q$, which also solves the issue. (1/3) Commented Dec 14, 2022 at 13:52
• I think those 2 answers approach the problem from a different angle, by either 'redefining' $\varepsilon_0$ to have an extra unit of meter, or by redefining $Q$ to have an extra unit of $1/\text{meter}$. In the final answer, $\mathbf{E}(\mathbf{r}) = \frac{\lambda}{2\pi \varepsilon_0} \frac{1}{r}$, you could also use the charge $Q$ instead of $\lambda$, and then use the newly defined $\varepsilon'_0$ $[\text{C}^2 \, \text{N}^{-1} \, \text{m}^{-1}]$, meaning that $\mathbf{E}(\mathbf{r}) = \frac{Q}{2\pi \varepsilon'_0} \frac{1}{r}$ would be another correct answer. (2/3) Commented Dec 14, 2022 at 13:52
• But then the question is: how would you "define" either $\lambda$ or $\varepsilon'_0$ such that the electric potential physically describes a point charge with a certain charge $Q$ of, say, 1 Coulomb? I am having a hard time trying to understand where I could get an extra unit of 'meter' into $\lambda$ or $\varepsilon'_0$, since there is no length scale anywhere. (PS I'm new to this website, so I hope splitting up a comment like this is acceptable...) (3/3) Commented Dec 14, 2022 at 13:53
• @Thijs: From the perspective of my answer, $\epsilon_0$ is just a proportionality constant that tells you (for instance) the magnitude of the force between two unit charges separated by one unit of distance. It has whatever units it needs to have so that the force law$$|\mathbf{F}| = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^{n-1}}$$ has the correct dimensions. In the "dimensional reduction" perspective, one way of doing it is to define your space such that the $z$ direction is compact (i.e., it wraps around, such that $z = 0$ and $z = L$ are actually the same point in space.) ... Commented Dec 14, 2022 at 14:52
• ... From this latter perspective, the pertinent length scale is the size of the extra dimension, and $\epsilon'_0 \propto L \epsilon$. However, it's not clear how to easily apply this perspective to the $L \to \infty$ limit. Commented Dec 14, 2022 at 14:55