Question: Find the flux through the uncharged disc of radius R due to point charge '+q' at a distance x from the center of disc and on its axis.

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[![diagram](https://i.stack.imgur.com/Wmq9C.png)](https://i.stack.imgur.com/Wmq9C.png)

So I tried to solve this by gauss law. First I started by connecting the points on the periphery of the disc to the point charge so as to obtain a cone (with α as semi vertical angle) . Now to apply gauss law I considered a gaussian sphere with +q charge at its center. Now charge q acts for whole sphere but I only need the part of the sphere where disk is present.

> 
> But here is the problem as you can see in the diagram that the disc
> and curved surface of area are not coinciding as there is some part left.
> (Pls correct me If I am visualizing wrongly)
> 
> 
> 

So my question is can I consider that part to be negligible and move on taking solid angle as 2π(1-cosα)? or else I wont be able to solve this by gauss law.

Answer: 
Consider the following gaussian surface: the curved part of the sphere$(S\_1)$ attached to the flat disc at end of cone$(S\_2)$.

[![enter image description here](https://i.stack.imgur.com/RQgGt.png)](https://i.stack.imgur.com/RQgGt.png)

What is the flux through this surface? Zero (Why?). Now, the surface integral of flux through this surface can be written in the following way:

$$ \int\_{S\_1} \vec{E} \cdot\vec{ dS }+\int\_{S\_2} \vec{E} \cdot \vec{dS}=0$$

In the above considered surface, the flat disc has a normal pointing towards the charge, to flip this add a negative sign into the equation(look at the picture). Finally the result is that the flat disc and curved surface has equal flux passing through them.
If the volume between the flat surface and the curved surface is charge free, then any flux which enters through the flat surface must leave through the curved surface.