A disk of radius R with uniform positive charge density σ is placed on the x y plane with its center at…

For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore,

$$\begin{gathered} cz\ge \frac{\sigma q}{2{ϵ}_0}\left[R-\left(\sqrt{R^2+z^2}-z\right)\right] \\ \Rightarrow \frac{c2{ϵ}_0}{\sigma q}\ge \frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z} \\ \Rightarrow \beta \ge \frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z} \end{gathered}$$

For option A

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}=\frac{R-R\left(\sqrt{1+{\displaystyle \frac{625}{49}}}-{\displaystyle \frac{25}{7}}\right)}{R{\displaystyle \frac{25}{7}}}\simeq 0.242$

Since, $0.25>0.242$

Option A is correct.

For option B

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}=\frac{R-R\left(\sqrt{1+{\displaystyle \frac{9}{49}}}-{\displaystyle \frac{3}{7}}\right)}{R{\displaystyle \frac{3}{7}}}\simeq 0.79$

Since, $0.25<0.79$

Option B is incorrect

For option C

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}=\frac{R-R\left(\sqrt{1+{\displaystyle \frac{1}{3}}}-{\displaystyle \frac{1}{\sqrt{3}}}\right)}{R{\displaystyle \frac{1}{\sqrt{3}}}}\simeq 0.73$

Since, $0.25<0.73$

Therefore, the particle returns to $z_0$

Hence, option C is correct.

For option D,

For any value of $z>0$

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}<1$

Hence, option D is correct.