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.
Best Answer
In the passage you quoted, McDonald is describing the very edge of the thin conducting disk. Recall that although the disk is thin, it does have a finite thickness. Therefore, its edge looks like this, with the conductor drawn in grey:
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 $3\pi/2$.