The Dirac Delta Function and Densities

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Section 9.4 The Dirac Delta Function and Densities

The total charge/mass in space should be the same whether we consider it to be distributed as a volume density or idealize it as a surface or line density. See Section 9.1.

The Dirac delta function relates line and surface charge densities (which are really idealizations) to volume densities. For example, if the surface charge density on a rectangular surface is \(\sigma(x,y)\text{,}\) with dimensions \(C/L^2\text{,}\) then the total charge on the slab is obtained by chopping up the surface into infinitesimal areas \(dA = dx\,dy\) and summing up (integrating) the charge \(\sigma(x,y) dA\) on each piece, \(\int\int \sigma(x,y) \, dx \, dy\text{.}\) Equivalently, one can recognize that this surface charge density is actually a volume charge density, idealized to be concentrated at, say, \(z=0\text{.}\) Thus,

\begin{equation} \rho(x,y,z) = \sigma(x,y) \,\delta(z)\tag{9.4.1} \end{equation}

and integrating this over a solid region yields

\begin{equation} \int\!\int\!\int \rho(x,y,z) \,dz\,dx\,dy = \int\!\int\!\int \sigma(x,y) \,\delta(z) \,dz\,dx\,dy = \int\!\int \sigma(x,y) \,dx\,dy\tag{9.4.2} \end{equation}

which yields the same answer as before. Recall that \(\delta(z)\) has dimensions of inverse length, so that \(\rho\) has the correct dimensions, namely \(CL^{-3}\text{.}\)