Step 1: Define the Total Potential Energy

The total potential energy U(z) of the particle of charge q at a height z on the z-axis is the sum of its electrostatic potential energy $U_e(z)$ and the potential energy $U_F(z)$ associated with the additional force $\vec{F} = -c \hat{k}$.

1. Electrostatic Potential Energy: $U_e(z) = qV(z) = q \frac{\sigma}{2 \epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right)$

2. Potential Energy from the constant force: The force is $\vec{F} = -c \hat{k}$. The potential energy change from the origin to a point z is $\Delta U = -W = -\int_0^z \vec{F} \cdot d\vec{l}$. With $d\vec{l} = dz' \hat{k}$, we have: $U_F(z) - U_F(0) = -\int_0^z (-c \hat{k}) \cdot (dz' \hat{k}) = \int_0^z c dz' = cz$ Assuming $U_F(0) = 0$, we get $U_F(z) = cz$.

3. Total Potential Energy: $U(z) = U_e(z) + U_F(z) = q \frac{\sigma}{2 \epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right) + cz$ $U(z) = \frac{q\sigma}{2\epsilon_0} \sqrt{R^2+z^2} + \left(c - \frac{q\sigma}{2\epsilon_0}\right)z$

Step 2: Express Potential Energy in terms of $\beta$

The problem defines $\beta=\frac{2 c \epsilon_{0}}{q \sigma}$. We can express c as $c = \frac{\beta q \sigma}{2 \epsilon_0}$. Substituting this into the expression for U(z): $c - \frac{q\sigma}{2\epsilon_0} = \frac{\beta q \sigma}{2 \epsilon_0} - \frac{q\sigma}{2\epsilon_0} = \frac{q\sigma}{2\epsilon_0}(\beta-1)$ Let's define a positive constant $U_0 = \frac{q\sigma}{2\epsilon_0}$. The total potential energy becomes: $U(z) = U_0 \left[ \sqrt{R^2+z^2} + (\beta-1)z \right]$ for $z \ge 0$.

Step 3: Analyze the Motion based on $\beta$

The force on the particle is $F_z = -\frac{dU}{dz}$. $\frac{dU}{dz} = U_0 \left[ \frac{z}{\sqrt{R^2+z^2}} + (\beta-1) \right]$ Equilibrium points $z_e$ occur where $F_z = 0$, which means $\frac{dU}{dz} = 0$: $\frac{z_e}{\sqrt{R^2+z_e^2}} = 1-\beta$

Case 1: $\beta \ge 1$ If $\beta \ge 1$, then $1-\beta \le 0$. Since $\frac{z_e}{\sqrt{R^2+z_e^2}}$ must be non-negative for $z_e \ge 0$, an equilibrium point can only exist at $z_e=0$ if $\beta=1$. For $\beta > 1$, there is no equilibrium point for $z \ge 0$. For z > 0, $\frac{dU}{dz} = U_0 \left[ \frac{z}{\sqrt{R^2+z^2}} + (\beta-1) \right] > 0$ since both terms in the bracket are positive. Thus, U(z) is a monotonically increasing function for z>0. The force $F_z = -\frac{dU}{dz}$ is always negative (directed towards the origin). Any particle released from rest at $z_0 > 0$ will accelerate towards the origin and will always reach it.

Case 2: $0 < \beta < 1$ In this case, $0 < 1-\beta < 1$. There exists a unique equilibrium point $z_e > 0$ which is a minimum of the potential energy (since $\frac{d^2U}{dz^2} = U_0 \frac{R^2}{(R^2+z^2)^{3/2}} > 0$, the potential is convex). This forms a potential well.

• If a particle is released at $z_0$ such that $0 < z_0 < z_e$, the force $F_z(z_0)$ will be positive (away from the origin), so it will never reach the origin.
• If a particle is released at $z_0 > z_e$, the force is initially towards the origin. The particle will oscillate. It can reach the origin z=0 only if its total energy $E = U(z_0)$ is greater than or equal to the potential at the origin, U(0). The condition is $U(z_0) \ge U(0)$. $U_0 \left[ \sqrt{R^2+z_0^2} + (\beta-1)z_0 \right] \ge U_0 R$ $\sqrt{R^2+z_0^2} \ge R - (\beta-1)z_0 = R+(1-\beta)z_0$ Squaring both sides (which are positive): $R^2+z_0^2 \ge R^2 + 2R(1-\beta)z_0 + (1-\beta)^2 z_0^2$ $z_0 \ge 2R(1-\beta) + (1-\beta)^2 z_0$ $z_0(1 - (1-\beta)^2) \ge 2R(1-\beta)$ $z_0(2\beta-\beta^2) \ge 2R(1-\beta)$ $z_0 \ge \frac{2R(1-\beta)}{\beta(2-\beta)}$ This critical value of $z_0$ can be shown to be greater than $z_e$. So, this is the sole condition for reaching the origin when $0 < \beta < 1$.

Step 4: Evaluate the Options

A: For $\beta=\frac{1}{4}$ and $z_{0}=\frac{25}{7} R$, the particle reaches the origin. This is Case 2. We check the condition $z_0 \ge \frac{2R(1-\beta)}{\beta(2-\beta)}$. $1-\beta = 1 - 1/4 = 3/4$ $2-\beta = 2 - 1/4 = 7/4$ $\beta(2-\beta) = (1/4)(7/4) = 7/16$ The critical distance is $z_c = \frac{2R(3/4)}{7/16} = \frac{3R/2}{7/16} = \frac{3R}{2} \times \frac{16}{7} = \frac{24R}{7}$. The condition is $z_0 \ge \frac{24R}{7}$. Given $z_0 = \frac{25R}{7}$, which satisfies the condition. So, statement A is correct.

B: For $\beta=\frac{1}{4}$ and $z_{0}=\frac{3}{7} R$, the particle reaches the origin. The critical distance is $z_c = \frac{24R}{7}$. Given $z_0 = \frac{3R}{7}$. This does not satisfy $z_0 \ge z_c$. So, statement B is incorrect.

C: For $\beta=\frac{1}{4}$ and $z_{0}=\frac{R}{\sqrt{3}}$, the particle returns back to $z=z_{0}$. For $\beta = 1/4$, we are in Case 2, where the potential U(z) has a minimum, forming a potential well. The particle is released from rest at $z_0$. This point $z_0$ is a turning point of its motion. Since the motion is conservative and bounded ($U(z) \to \infty$ as $z \to \infty$), the particle will oscillate within the potential well and will periodically return to its starting point $z_0$. We should check if $z_0$ is the equilibrium point itself. The equilibrium $z_e = R \frac{1-\beta}{\sqrt{\beta(2-\beta)}} = R \frac{3/4}{\sqrt{7/16}} = \frac{3R}{\sqrt{7}}$. Since $z_0 = R/\sqrt{3} \ne z_e$, the particle will not remain at rest, it will oscillate. So, statement C is correct.

D: For $\beta> 1$ and $z_{0}> 0$, the particle always reaches the origin. This corresponds to our analysis of Case 1. The force on the particle is always directed towards the origin. Therefore, if released from rest at any $z_0 > 0$, it will accelerate towards and reach the origin. So, statement D is correct.