Charge distribution on a conducting solid disc Imagine that you have a disc of radius $R$ with charge $Q$ on it. It is a conducting disc. What would be the charge distribution?
Is there a uniform distribution over whole area? $\sigma=\textrm{constant}$
Or is there a distribution depends on r? $\sigma=\sigma(r)=\frac{Q}{\pi r^2}$
Or there is no charge on the area but all of the charge is placed at the edge of the disc? $\lambda=\frac{Q}{2\pi R}$
 A: The charge density would not be uniform. It is highest at points and sharp edges, where (in theory) it tends towards infinity. For a disc the highest charge density would be at the rim. See Charge distribution on conductors and Why does charge accumulate at points?
Regarding this problem, Andrew Zangwill in Application 5.1 of his book on Modern Electrodynamics states that 

There is no truly simple way to calculate the surface charge density for a charged, conducting disk. In this Application we use a method which regards the disk as the limiting case of a squashed ellipsoid...

He proceeds to obtain the result
$$\sigma = \frac{Q}{4\pi R \sqrt{R^2-r^2}}$$
A: Start with a distribution of charges and calculate per charge with displacement is needed to get on this charge zero force (e.g. with Newton Raphson). Once done for all charges replace them all to the new position and repeat this proces till the sum of absolute force values becomes sufficient small. The end result will be an increasing charge density near the edges. This makes people aware that the exact determination of the capacitance between two conducting plates with parallel field lines is an approximation.   
