Sci-fi physics — 2D electrostatics

Problem

Sci-fi physics: 2D electrostatics. In a universe with 2 spatial dimensions, how would the electric field of a point charge vary with distance? And the potential? Can $V=0$ still be fixed at infinity? Hint: in the 2D case the Gaussian “surface” is a circle of perimeter $2\pi r$.

Solution

Electric field. In three dimensions Gauss’s theorem is applied on a sphere of area $4\pi r^2$, and from $E\cdot 4\pi r^2 = q/\varepsilon_0$ follows the familiar $E\propto 1/r^2$. In two dimensions the Gaussian surface degenerates into a circle of “perimeter” $2\pi r$, which plays the role of the area. The two-dimensional flavour of Gauss’s theorem then gives: $E\cdot 2\pi r = \frac{q}{\varepsilon_0} \quad\Longrightarrow\quad E = \frac{q}{2\pi\varepsilon_0}\,\frac{1}{r}.$ The field decays as $\ev{E \propto 1/r}$, no longer as $1/r^2$. This is exactly the same behaviour as the field of an infinite straight wire in our 3D world: a point in 2D “is” a wire seen along the suppressed third dimension.

Potential. The potential is obtained by integrating the field: $V(r) = -\int E\,\dd r = -\frac{q}{2\pi\varepsilon_0}\int \frac{\dd r}{r} = -\frac{q}{2\pi\varepsilon_0}\,\ln r + \text{const}.$ So $\ev{V \propto -\ln r}$: a logarithmic potential, no longer $\propto 1/r$.

Reference at infinity. The logarithm diverges as $r\to\infty$: it does not tend to a finite value. It is therefore not possible to fix the constant by imposing $V=0$ at infinity, as is done in 3D. One must choose an arbitrary reference radius $r_0$ and set $V(r_0)=0$, giving $\ev{V(r) = -\dfrac{q}{2\pi\varepsilon_0}\ln\dfrac{r}{r_0}}$. The disappearance of the natural reference at infinity is the most notable physical consequence of moving to two dimensions.

Connections

Topics: Electric field and potential Concepts: Electric field · Electric potential Skills: Applying Gauss’s theorem Methods: Gauss’s theorem Objects: Infinite straight wire