Finding the Capacitance of a Conducting disc

[RJLiberator]

Homework Statement

Consider a disc (negligible thickness) of radius R with its center at the origin. Assume that the disc has a radially symmetric surface charge density given by ##σ(s) = \frac{A}{\sqrt{R^2-s^2}}## at a distance s from the origin, where A is some constant. You can take it for granted that this charge density leads to a potential that is constant everywhere on the surface of the disc, and hence, is suitable for describing the electrostatics of a conducting disc, and hence, find the capacitance of a conducting disc of radius R.

[Hint: In principle, the potential at any position on the disc can be calculated directly from the given charge density. Write down that expression for a general point which is at a distance r from the center. Note that since the charge distribution is cylindrically symmetric, so will V(r), which means you can place r along the x-axis in the (s, φ) polar coordinate system. You will realize that the integrations that you need to do are not trivial (do not attempt to perform the integrations for a general r, as their solutions involve elliptic functions). However, since you are given that the potential is constant anywhere on the disc, i.e. it is independent of r, it suffices to find the potential due to the given charge density at a special point on the disc, where the integrations are very easy to perform. This will allow you to relate the constant A to the potential V. Then you can find the total charge Q and relate Q and V to each other.

Homework Equations

C = Q/V
##V = \frac{1}{4 \pi \epsilon_0} \int \frac{ \rho}{r}d \tau##

The Attempt at a Solution

The Polar coordinate idea is killing me, I think.

But, we know what the charge distribution is, that's good.
The hint tells me to place r on the x axis... this means that in the integrations, ##r = \sqrt{s-s'}## for the distance between some point s and r?

Now, I think the idea from the hint is in the following set up:

[tex] V = \frac{1}{4 \pi \epsilon_0} \int_∞^s \int_0^{2 \pi} \frac{A}{\sqrt{R^2-s^2}} \frac{1}{\sqrt{(s-s')^2}} s dφds[/tex]

To let s' be some number that makes the integration simple compared to impossible to integrate regularly.

As far as the dφ integration goes, we simply get a 2pi out of it.

Am I setting this up correctly?

 

[mfb]

To get the potential via an integral you need the electric field strength. Where do you have this expression?

 

[RJLiberator]

Hi, thank you for the reply to my question.

I'm not sure I follow. Why would I need the electric field to get potential? The equation in the "relevant equations" part of this topic does not involve E. I have the charge density (given) from the problem. I would just need script r. I believe my problem here is in the script r, and I think it now needs to be understood with the law of cosines.

something like: ##\sqrt{s^2+s'^2-2ss'}##.

 

[mfb]

Ah, I misunderstood what you wanted to integrate. If you integrate over the charge density, why do you integrate from ##\infty## to s?
In addition, it is confusing to use s both as integration variable and integration bound.

You'll need something like the square root in your last post, yes.