Electric field and electric potential of a point charge in 2D and 1D

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.