A disk of radius $$R$$ with uniform positive charge density $$\sigma$$ is placed on the $$xy$$ plane with its center at the origin. The Coulomb potential along the z-axis is

$$V(z) = \dfrac{\sigma}{2\epsilon_0}\left(\sqrt{R^2 + z^2} - z\right)$$

A particle of positive charge $$q$$ is placed initially at rest at a point on the z axis with $$z = z_0$$ and $$z_0 > 0$$. In addition to the Coulomb force, the particle experiences a vertical force $$\vec{F} = -c\hat{k}$$ with $$c > 0$$. Let $$\beta = \dfrac{2c\epsilon_0}{q\sigma}$$. Which of the following statement(s) is(are) correct?

The electrostatic potential produced by the charged disk is given as
$$V(z)=\dfrac{\sigma}{2\epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right).$$

For a particle of charge $$q$$ the corresponding electrostatic potential energy is
$$U_{\text{C}}(z)=qV(z)=\dfrac{q\sigma}{2\epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right).$$

The particle is also acted upon by a constant downward force $$\vec F=-c\hat k$$. Since $$F_z=-\dfrac{dU_g}{dz}=-c$$, the mechanical potential energy associated with this force is $$U_g(z)=c\,z$$ (we take $$U_g(0)=0$$).

Hence the total potential energy of the particle is
$$U(z)=U_{\text{C}}(z)+U_g(z) =\dfrac{q\sigma}{2\epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right)+c\,z.$$

Put $$\alpha=\dfrac{q\sigma}{2\epsilon_{0}},\qquad \beta=\dfrac{2c\epsilon_{0}}{q\sigma}=\dfrac{c}{\alpha} \;(\beta\gt 0).$$ Then

$$U(z)=\alpha\Bigl[\sqrt{R^{2}+z^{2}}-z+\beta z\Bigr] =\alpha\Bigl[\sqrt{R^{2}+z^{2}}+(\beta-1)z\Bigr].$$

The motion is one-dimensional along the $$z$$-axis. Because only conservative forces act, the mechanical energy $$E=U(z)+K$$ is conserved. At the instant $$t=0$$ the particle is released from rest at $$z=z_0$$, so $$E=U(z_0).$$

If the particle is to reach the origin, its kinetic energy at $$z=0$$ must be non-negative:
$$U(z_0)\ge U(0).$$ Since $$U(0)=\alpha R,$$ this gives the single inequality

$$\sqrt{R^{2}+z_0^{2}}+(\beta-1)z_0\;\ge\;R.\qquad -(1)$$

Let us examine each option one by one.

$$\beta=\dfrac14,\; z_0=\dfrac{25}{7}R \;(=3.571R).$$ Compute $$\sqrt{R^{2}+z_0^{2}}=R\sqrt{1+\Bigl(\dfrac{25}{7}\Bigr)^{2}} =R\sqrt{\dfrac{674}{49}}=\dfrac{\sqrt{674}}{7}R\approx3.709R,$$ $$(\beta-1)z_0=-\dfrac34\cdot\dfrac{25}{7}R=-2.678R.$$ Thus $$\sqrt{R^{2}+z_0^{2}}+(\beta-1)z_0\approx3.709R-2.678R=1.031R\gt R.$$ Inequality $$(1)$$ is satisfied, so the particle reaches $$z=0$$. Option A is correct.

$$\beta=\dfrac14,\; z_0=\dfrac{3}{7}R\;(=0.4286R).$$ $$\sqrt{R^{2}+z_0^{2}}=R\sqrt{1+\Bigl(\dfrac{3}{7}\Bigr)^{2}} =1.088R,$$ $$(\beta-1)z_0=-\dfrac34\cdot\dfrac{3}{7}R=-0.321R.$$ Sum $$=1.088R-0.321R=0.767R\lt R,$$ violating $$(1)$$. Hence the particle cannot reach the origin. Option B is wrong.

$$\beta=\dfrac14,\; z_0=\dfrac{R}{\sqrt3}\;(=0.577R).$$ $$\sqrt{R^{2}+z_0^{2}}=R\sqrt{1+\dfrac13}=1.155R,$$ $$(\beta-1)z_0=-\dfrac34\cdot0.577R=-0.433R.$$ Sum $$=1.155R-0.433R=0.722R\lt R,$$ so the particle cannot reach $$z=0$$.

To see what happens, look at the force $$F(z)=-\dfrac{dU}{dz} =-\alpha\left[\dfrac{z}{\sqrt{R^{2}+z^{2}}}+(\beta-1)\right].$$ For the given $$z_0$$, evaluate the bracket: $$\dfrac{z_0}{\sqrt{R^{2}+z_0^{2}}}+(\beta-1)=\dfrac{1}{2}-\dfrac34=-\dfrac14.$$ Hence $$F(z_0)=+\dfrac{\alpha}{4}$$, i.e. a force in the $$+z$$ direction. The particle first moves upward, its kinetic energy increases, and because the potential $$U(z)$$ grows as $$z\to\infty$$ (for $$\beta=\tfrac14$$, $$U\sim\alpha\beta z$$), it finally comes to rest at some $$z=z_1$$ where $$U(z_1)=U(z_0)$$, and then returns to $$z_0$$. Thus the particle executes oscillatory motion between $$z_0$$ and $$z_1$$ and indeed returns to its starting point. Option C is correct.

Take any $$\beta\gt1$$ and any $$z_0\gt0$$. Because $$0\lt\dfrac{z_0}{\sqrt{R^{2}+z_0^{2}}}\lt1,$$ we get $$\sqrt{R^{2}+z_0^{2}}+(\beta-1)z_0 \;>\;R+0=R.$$ Inequality $$(1)$$ is satisfied for every positive $$z_0$$; therefore the particle always acquires enough mechanical energy to reach the origin. Option D is correct.

Hence the correct choices are:
Option A, Option C and Option D.