# Capacitance of a conducting disk

I'm reading this(PDF) derivation of the capacitance of a thin conducting disk. The surface charge density of such a disk can be shown to be:

$\sigma(r) = \frac{Q}{4\pi a\sqrt{a^2 - r^2}}$ (in Gaussian units)

where r is a point on the disk, and a is the radius of the disk.

The text at the end of the first page states:

"If we let s measure the distance radially inward from the circumference of the disk, then the charge density (3) varies as $\frac{1}{\sqrt{s}}$ for small s. However, the charge density near the edge of a conductor whose surfaces intersect at an exterior angle of $\frac{3π}{2}$, as in the present problem, is known to vary as $\frac{1}{\sqrt[3]{s}}$, for s measured normal to the edge along either surface."

I am having trouble visualizing what the author means by this. Could someone help me understand the geometry better? Thanks.

-

One way to describe this geometry is to say the internal angle is $\pi/2$. McDonald chooses to instead say that the exterior angle is $3\pi/2$. He does this to maintain consistency with the reference he cites (Jackson): if you look up figure 2.13 on page 78, you'll notice this angle being described as $3\pi/2$.