Flux Progression

@Karan Chatrath has provided a detailed solution. I will offer a few more remarks. From the plot below, it is evident that the flux increases for a while as the particle moves closer to the disk. But the angle of attack gets shallower as well, eventually leading to a steep decline in the flux as alpha increases.

Consider a point R on the X-Y plane lying withing the unit circle which has a position vector:

$\vec{R} = x \hat{i} + y \hat {j}$

The position vector of the point P is:

$\vec{P} = \frac{\alpha}{\sqrt{2}} \hat{i} + \left(2 - \frac{\alpha}{\sqrt{2}}\right) \hat{k}$

The electric field at point R due to charge at P is:

$d\vec{E} = \frac{\vec{R}-\vec{P}}{\lvert \vec{R} - \vec{P} \rvert^3}$

The area vector normal to the unit circle is:

$d\vec{S} = -(dx)(dy) \hat{k}$

The flux through this area due to this elementary field is:

$d\phi = d\vec{E} \cdot d\vec{S}$

Obtaining the flux would require integrating the above expression over the entire area of the unit circle. This expression can be found by substituting and simflifying all expressions. This gives:

$\phi(\alpha) = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{\left(2 - \frac{\alpha}{\sqrt{2}}\right)dy \ dx}{\left(\left(x - \frac{\alpha}{\sqrt{2}}\right)^2 + y^2 + \left(2 - \frac{\alpha}{\sqrt{2}}\right)^2\right)^{3/2}}$

we are required to compute the expression:

$I = \int_{0}^{\sqrt{8}} \phi(\alpha) d\alpha$

This results in the following monster which is tackled numerically, leading to the answer.

$\boxed{I = \int_{0}^{\sqrt{8}} \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(\frac{\left(2 - \frac{\alpha}{\sqrt{2}}\right)}{\left(\left(x - \frac{\alpha}{\sqrt{2}}\right)^2 + y^2 + \left(2 - \frac{\alpha}{\sqrt{2}}\right)^2\right)^{3/2}}\right)dy \ dx \ d\alpha \approx 2.1667}$