# Can Gauss' Law in differential form apply to surface charges?

I'm calculating the electric field outside a coaxial cable using only Gauss' Law in differential form.

The charge density on the interior solid conducting cylinder is exactly cancelled by the surface charge of the exterior shell conducting cylinder.

The problem is, Gauss' Law in differential form is:

$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$

How do I turn this into an equation regarding $$\sigma$$ the surface charge, instead? Every time I try to do it, I wind up with the wrong units. I though perhaps it might involve the Delta function, but couldn't get that to work.

(I've been working in cylindrical units)

Should this have a homework tag?

If the radius of the cable is $r_0$, the charge density $\rho = \alpha\delta(r -r_0)$. You need a constant of proportionality that makes the total charge per unit length be $\lambda$.

The total charge in length $l$ would be $q = \lambda l = \iiint \rho dx$, where the integral is over length $l$ along the z axis.

If you work out the integral, you should get an expression for $\alpha$ in the right units.

Then you need a relationship between $\lambda$ and $\sigma$. That would be $\lambda l = \sigma 2\pi r_0l$

• Should I add that? This is an aspect of a homework question, I've done all the rest but this one part is holding me back. Commented Apr 6, 2014 at 5:02
• I'm not sure I fully understand. I've found what I think is $\alpha$, except that's in units of $C/m^2$, as in surface charge. But can I just chuck that in, in place of $\rho$ in Gauss' Law? I get the right answer, but I don't understand how I can just replace volume charge which what seems to be a surface charge with the delta function attached, and not screw up the units. Commented Apr 6, 2014 at 5:13
• As you travel radially outward, you encounter a sheet of charge. The charge per length is a $\delta$ function. In Cartesian coordinates, a point charge would be $\rho = q\delta(x-x_0)\delta(y-y_0)\delta(z-z_0)$. For a sheet, the charge density would be $\lambda = a\delta(x-x_)$. Commented Apr 6, 2014 at 5:23
• Ah, so the delta function isn't dimensionless then, it acts as if it has units of $1/m$? Commented Apr 6, 2014 at 5:36
• Yes. Sorry. $\rho = \alpha \delta(x - x_0 )$. $\rho$ has units of $C/m^3$. $\alpha$ would have units of $C/m^2$. Commented Apr 6, 2014 at 5:40