# Is there a way to see linear and surface charge density as a “special case” of volume charge density?

When deriving Gauss’s law in differential form (GLDF), $$\nabla \cdot \mathbf E = \frac{\rho}{\epsilon_0},$$ from Gauss’s law in integral form (GLIF) we get a tidier formula, which is however less general (to my understanding). In fact, while we can freely apply GLIF to point charges, linear charge distributions, surface charge distributions and volume charge distributions, we can only apply GLDF to the latters.

However, if we consider a point charge $$q$$ (placed at the origin of our chosen frame of reference for simplicity), we can define $$\rho(\mathbf x) = q \delta(\mathbf x)$$ and this makes GLDF true for point charges as well. In fact we are now able to recover GLIF in the case of point charges by integrating GLDF.

Is there something similar that we can do in the case of linear charge distributions and surface charge distributions?

You can always think of line- and surface-densities as a collection of point charges. If you already have no problem in using Dirac deltas, then we can easily just extend them to more complex geometries. Weird deltas might be harder to handle, but in principle it is possible.

For example you could write, for a series of $$N$$ charges of value $$q$$ along a line at positions $$x_i$$ along the $$x$$-axis, something like

$$\rho(\vec{x})=\Sigma^N_i q \delta(x-x_i)$$

clearly this is not problematic. It's just a sum of point charges.

If you extend this reasoning to continuum space [see below for a quick derivation], you can use Dirac's delta functions that are $$0$$ everywhere except along a line. For example, if we choose the $$x-axis$$ you could write $$\rho(\vec{x})=\lambda(x) \delta(y)\delta(z)$$

where instead of the vector $$\vec{x}$$ I used the individual components and $$\lambda(x)$$ is the charge density along the $$x$$ axis. Those deltas are zero except where both $$y=0$$ and $$z=0$$ i.e. along the $$x-axis$$.

If you integrate that over the full space (integrals all go from $$0$$ to $$\infty$$), you get

$$\int\rho(\vec{x}) d\vec{x}= \int \lambda(x) \delta(y)\delta(z) dx dy dz =\int \lambda(x) dx \int \delta(y)dy\int\delta(z)dz$$ and because $$\int \delta(y)dy=1$$ and the same for z

$$\int\rho(\vec{x}) d\vec{x}=\int \lambda(x) dx =Q$$ which gives you the total charge as expected (which can be infinite, is $$\lambda$$ is taken constant, although non-physical).

In the same way, we can define $$\rho(\vec{x}) = \sigma(x, y) \delta(z)$$ which is only non-zero if $$z=0$$ i.e. on the full $$xy$$-plane.

You can then create any charge density you want, using combined delta functions. You could also define it "by hand", just saying "here is constant and here is zero" and then integrating each domain separatly if necessary, but that will give you a bit more of a problem in dealing with discontinuities, so you need to careful.

For example, of course you can also limit your charge distribution in space by defining

$$\rho(\vec{x})=\lambda(x)\delta(y)\delta(z)$$ if $$|x| and 0 otherwise, which would give you a thin wire along the the $$x$$-axis of length $$L$$ - or you could define $$\lambda(x)$$ using step functions. That would give you a total charge of $$Q$$ along such wire which, if assumed at constant density, would give you $$\lambda(x)=Q/L$$.

So yes, you can use GLDF in a very general way. You just get the annoying thing of having to deal with $$\delta$$s.

Derivation

As an exercise, we can derive this expression. There is a bit of hand-waving but I think it sends you in the right direction.

Imagine you have charges along the $$x-axis$$ of value $$\lambda(x)dx$$ - i.e. the local density times an infinitesimal length.

We can write the density $$\rho(\vec{x})$$ as a sum of point charges at positions $$x'$$ (with $$x'$$ is a continous variable spanning the $$x$$ axis).

$$\rho(\vec{x}) = \Sigma \lambda(x')dx'\delta(\vec{x}-\vec{x'})$$

now of course, because we are dealing with infinitesimal charges, that sum actually is an integral along the line

$$\rho(\vec{x}) = \int \lambda(x')dx'\delta(\vec{x}-\vec{x'})$$

and if we write $$\delta(\vec{x}-\vec{x'})=\delta(x-x')\delta(y-y')\delta(z-z')$$ we have to perform only the integral on the variable $$x'$$

$$\rho(\vec{x}) = \delta(z-z')\delta(y-y') \int \lambda(x')\delta(x-x')dx'$$ and using the fact that $$\int f(x')\delta(x-x')dx' = f(x)$$ $$\rho(\vec{x}) = \delta(z-z')\delta(y-y')\lambda(x)$$

and because $$y'$$ and $$z'$$ are $$0$$ as we are on the $$x$$-axis, then $$y-y'\to y$$ and $$z-z'\to z$$ and we get $$\rho(\vec{x}) = \delta(z)\delta(y)\lambda(x)$$

• (+1) Nice answer, but I think you made a few typos. Your second-to-last and third-to-last integrals should be taken $\mathrm d x'$ not $\mathrm d x$. You may also want to avoid using the phrase "complex delta function," as that may be mixed up with the complex delta function representation, which is $\delta (x - x') = \int_{-\infty}^\infty e^{ik(x-x')} \ \mathrm d k$. Feb 4, 2021 at 23:59
• Thank you. I think I fixed it. Feb 5, 2021 at 0:31

You can have $$\rho$$ represent any charge distribution you like, e.g. for a charged spherical shell of radius $$r_0$$ $$\rho(r) = q \delta(r-r_0)$$

• The integral of this delta is $1$ when that point ${\bf r}$ coincides with the point ${\bf r}_0$. There is no sphere or spherical shell. A delta centered at the surface of a sphere of radius $r_0$ should be written as $\delta(| {\bf r}|-r_0)$. Feb 4, 2021 at 22:40
• @GiorgioP Thanks for pointing that out. It was not meant as a vector though. I basically just copied the math notation from the OP and he had the x bolded. I have unbolded everything now to avoid any misunderstanding. It should have been clear though from the definition 'radius $r_0$' anyway. Feb 4, 2021 at 22:53