A disk of radius R with uniform positive charge density σ is placed...

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,

$cz \ge \dfrac{\sigmaq}{2{ϵ}_{0}}\left[R - \left(\sqrt{{R}^{2} + {z}^{2}} - z\right)\right] ⇒ \dfrac{c2{ϵ}_{0}}{\sigmaq} \ge \dfrac{\left[R - \left(\sqrt{{R}^{2} + {z}^{2}} - z\right)\right]}{z} ⇒ \beta \ge \dfrac{\left[R - \left(\sqrt{{R}^{2} + {z}^{2}} - z\right)\right]}{z}$

For option A

$\dfrac{\left[R - \left(\sqrt{{R}^{2} + {z}^{2}} - z\right)\right]}{z} = \dfrac{R - R\left(\sqrt{1 + \dfrac{625}{49}} - \dfrac{25}{7}\right)}{R\dfrac{25}{7}} ≃ 0 . 242$

Since, $0 . 25 > 0 . 242$

Option A is correct.

For option B

$\dfrac{\left[R - \left(\sqrt{{R}^{2} + {z}^{2}} - z\right)\right]}{z} = \dfrac{R - R\left(\sqrt{1 + \dfrac{9}{49}} - \dfrac{3}{7}\right)}{R\dfrac{3}{7}} ≃ 0 . 79$

Since, $0 . 25 < 0 . 79$

Option B is incorrect

For option C

$\dfrac{\left[R - \left(\sqrt{{R}^{2} + {z}^{2}} - z\right)\right]}{z} = \dfrac{R - R\left(\sqrt{1 + \dfrac{1}{3}} - \dfrac{1}{\sqrt{3}}\right)}{R\dfrac{1}{\sqrt{3}}} ≃ 0 . 73$

Since, $0 . 25 < 0 . 73$

Therefore, the particle returns to ${z}_{0}$

Hence, option C is correct.

For option D,

For any value of $z > 0$

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

Hence, option D is correct.