Writing surface charge density in 3D

Question

Suppose I have a disk of radius $R$ centered in the $xy$-plane. Suppose that I plaster a uniform surface charge density $\sigma_0$ on this disk. How do I write this surface charge density as a volume charge density $\rho$?

Certainly, the total charge on the disk should be $\pi R^2 \sigma_0$, i.e. $$\int_\mathbb{R^3} dV \rho = \pi R^2 \sigma_0.$$

Naively, I write $$\rho = \sigma_0 \delta(\theta - \frac{\pi}{2}) H(R- r)$$ where $H$ is the Heaviside step function. However, $$\int_0^\infty r^2 dr \int_0^{2\pi}d\phi \int_0^\pi \sin\theta d\theta \rho=\frac{2}{3}\pi R^3\sigma_0,$$ which is different from $\pi R^2 \sigma_0$ by a factor of $\frac{3}{2R}$. What is going on?

Answer 1

A simple check of the dimension of your "naive" formula $\sigma_0 \delta(\theta-\pi/2) H(R-r)$ (having dimension charge/area) reveals that this expression fails to be a space charge density, which should have dimension charge/volume.

An easy way to find the correct result is by starting in cartesian coordinates, where the space charge density is given by $$\rho(x,y,z)= \sigma_0 \delta(z) H\left(R-\sqrt{x^2+y^2}\right)=\sigma_0 \delta(z) H\left(R-\sqrt{x^2+y^2+z^2}\right).$$ Transforming to spherical coordinates, one obtains $$\rho(r,\theta,\phi)=\sigma_0 \delta(r \cos \theta) H(R-r)=\frac{\sigma_0}{r}\delta(\cos \theta)H(R-r)=\frac{\sigma_0}{r \sin \theta} \delta(\theta-\pi/2)H(R-r),$$ where $r\gt 0$ and $0\lt \theta \lt \pi$ were taken into account. Now, the integral $$\int\limits_0^R\! dr \,r^2 \int\limits_{-1}^1 \!d \cos \theta \int\limits_0^{2\pi} \! d \phi \, \frac{\sigma_0}{r} \delta(\cos \theta) =\sigma_0 R^2 \pi $$ gives indeed the desired result for the total charge sitting on the disk.

Answer 2

Volume density. You can easily do it using cylindrical coordinates, $(r,\theta, z)$, with the $z$ coordinate along the axis of the disk, and $z=0$ being the $x-y$ plane. Volume density reads

$$\rho(r,\theta,z) = \sigma \delta(z) H(R-r) \ .$$

Physical dimensions. The physical dimensions are ok, since $[\rho] = \frac{\text{mass}}{\text{length}^3}$, $[\sigma] = \frac{\text{mass}}{\text{length}^2}$, and $[\delta(z)] = \frac{1}{\text{length}}$, and $[H] = 1$. You can check the physical dimension of the Dirac's delta using its "definition", $1 = \int_{z=-a}^a \delta(z) \, dz$: the result of the integral is non-dimenisonal, integration over $z$ provides a length dimension, so that $\delta(z)$ must be $\frac{1}{\text{length}}$.

Mass, and inertia. Integrating over a volume $V$ containing the disk, here the whole space:

• mass reads

  $$\begin{aligned} m = \int_{V} \rho & = \int_{z=-\infty}^{+\infty} \int_{\theta=0}^{2\pi} \int_{r=0}^{+\infty} \sigma \delta(z) H(R-r) \, r \, dr \,d\theta \, dz = \\ & = \int_{\theta=0}^{2\pi}\int_{r=0}^R \sigma\, r \, dr \,d\theta = \\ & = \sigma 2\pi \frac{R^2}{2} = \sigma \pi R^2 = \sigma A \end{aligned}$$

• polar inertia w.r.t. around the axis of the disk,

  $$\begin{aligned} I_G = \int_{V} \rho \, r^2 & = \int_{z=-\infty}^{+\infty} \int_{\theta=0}^{2\pi} \int_{r=0}^{+\infty} \sigma \, r^2 \, \delta(z) H(R-r) \, r \, dr \,d\theta \, dz = \\ & = \int_{\theta=0}^{2\pi}\int_{r=0}^R \sigma\, r^3 \, dr \,d\theta = \\ & = \sigma 2\pi \frac{R^4}{4} = \sigma \pi \frac{R^4}{2} = \frac{1}{2} \, \sigma \, A \,R^2 \ . \end{aligned}$$

Answer 3

The idea we want to capture is that we should only integrate over $r\in[0,R]$. This requires a multiplication by an indicator function over this region or equivalently, the difference of two Heaviside functions: $\mathrm{H}(r) - \mathrm{H}(r-R)$. It is unclear to me how you arrived at the Dirac delta function that you did but that factor is unnecessary. Personally, I would keep $\sigma_0$ a surface charge density to get $\rho = \sigma_0[\mathrm{H}(r) - \mathrm{H}(r-R)]$. Then we have that the total charge is \begin{align} Q &= \int_0^{2\pi} \int_0^{\infty} \sigma_0[\mathrm{H}(r) - \mathrm{H}(r-R)]~ r \mathrm{d}r\mathrm{d}\theta \\ &= \int_0^{2\pi} \int_{0}^R \sigma_0~ r \mathrm{d}r\mathrm{d}\theta \\ &= \pi R^2 \sigma_0,\end{align} as desired.

If you really want to describe this distribution in three dimensions instead, then you would have $\rho = \sigma_0 \delta(z) \mathrm{H}(R - \sqrt{x^2 + y^2})$ which you can integrate over $\mathbb{R^3}$ by switching to cylindrical coordinates; you would arrive at the same result.