Surface charge density of conducting disc

[Titan97]

Homework Statement

This is problem 3.4 from Prucell and Morin if you have the book.

Homework Equations

None

The Attempt at a Solution

Electric field inside a conducting sphere is zero. Let P be a point on one of its equatorial plane. The field along the plane is zero. So I know the charge distribution that can produce zero electric field along a disc.

Let the charge density of the shell be $\sigma$. It is a constant function.

Consider a patch like the red line in the below diagram. The yellow surface is the disc.

Let the thickness of the patch be $dr$. Let its area be $A$.
$A$ is a projection of the circle on the shell having the red line as diameter.

If the area of that circle is $A_1$, then area of the patch is $A_1\cos\theta=A$
Hence $A=\frac{A_1}{\cos\theta}$

Here, $$A_1=2\pi R\cos\theta\cdot\frac{dr}{\cos\theta}$$

Now, $$Q=\int_0^R{\sigma A}=\sigma\int\frac{2\pi R^2}{\sqrt{R^2-r^2}}dr$$.

Is my understanding correct? I am not getting the correct answer.

 

[haruspex]

The area of your band A does not depend on theta. (Ref Archimedes.)
But anyway, the band A is specific to the location of P. Clearly, if there is a field from a rotationally symmetric field it will be in the radial direction, and the band A will not have a component in that direction at P.
You need to consider the field at P due to the entire shell.