# What would the $\nabla^2 V$ be at the border of a conductor?

We are studying laplace's and Poisson's equation for electrostatics in my class. It was cursorily said that $\nabla^2 V = \sigma / \epsilon_o$ at the surface of the conductor. This makes no sense as it is dimensionally incorrect.

Moreover it raises to me what is an interesting question: What is the volumetric density when you cross a surface distribution of charge? My mathematical intuition says its zero or "hard to calculate". How do you clarify such a concept? After all the volumetric density of a point charge is defined via the Dirac Delta Function...

Thanks.

• You're very right in thinking this is dimensionally incorrect. You are missing the Dirac-delta functions, as you suggested for a point-charge. $\nabla^2 V = (\sigma / \epsilon_0)\,\, \delta^2(\mathbf{r}-\mathbf{r}')$, where $\mathbf{r}'$ is any point on the surface and $\delta^2(\mathbf{r})$ is the two-dimensional Dirac delta function. – Arturo don Juan Sep 15 '15 at 21:56

The way to write the surface charge density $\sigma(\vec r)$ on a surface $S$ as (volume) charge density is: $$\rho(\vec r) = \int_S d\Sigma(\vec r')\, \sigma(\vec r') \delta(\vec r - \vec r').$$ Where $d\Sigma$ is the surface measure. The integral runs over the surface $S$.
Intuitively, one represents the surface charge as a sum of infinitesimal-strength point charges distributed over the surface (each point of the surface carries a point charge $d\Sigma(\vec r) \sigma(\vec r)$ and $d\Sigma$ is infinitesimally small).
As a simple plausibility check one can calculate the total charge: \begin{align*} Q &= \int d^3r\,\rho(\vec r) = \int d^3r \int_S d\Sigma(\vec r')\,\sigma(\vec r') \delta(\vec r - \vec r') \\ &= \int_S d\Sigma(\vec r')\, \sigma(\vec r') \underbrace{\int d^3r\, \delta(\vec r - \vec r')}_{=\,1}. \end{align*} This is obviously the expected result. ($\delta(\vec r - \vec r')$ integrated over the entire space is obviously $1$ for any $\vec r'$.)
I want to add there is more to know about this topic if one is interested in rigour: In more mathematical terms $\rho$ is not a function but a measure (just like probability distributions are not functions, but measures). To handle this one has to define what a measure on the right-hand side of a differential equation is supposed to mean (and that gets you deeply into the theory of generalized functions, distributions and weak derivatives).
• Sorry for accepting your answer so late. I have to say it makes mathematical sense but I don't see any physical interpretation of $\rho(\vec{r})$. Is there any? Thanks. – DLV Sep 28 '15 at 3:33