Motion of Charged Particle in Disc Electric Field and Vertical Force

To determine the correct statement(s), we analyze the potential energy and net force acting on the positively charged particle of mass $m$ and charge $q$ along the $z$-axis ($z \ge 0$).

1. Potential Energy Formulation

The electrostatic potential along the $z$-axis due to the uniformly charged disc is given by: $V(z) = \frac{\sigma}{2\epsilon_0}\left(\sqrt{R^2 + z^2} - z\right)$

The electrostatic potential energy of the particle is: $U_{\text{elec}}(z) = q V(z) = \frac{q\sigma}{2\epsilon_0}\left(\sqrt{R^2 + z^2} - z\right)$

The additional constant vertical force $\vec{F} = -c\hat{k}$ corresponds to a potential energy: $U_{\text{ext}}(z) = cz$

Given $\beta = \frac{2c\epsilon_0}{q\sigma}$, we express $c$ as $c = \beta \frac{q\sigma}{2\epsilon_0}$. The total potential energy $U(z)$ of the particle is: $U(z) = U_{\text{elec}}(z) + U_{\text{ext}}(z) = \frac{q\sigma}{2\epsilon_0}\left[\sqrt{R^2 + z^2} - (1 - \beta)z\right]$

---

2. Force Analysis

The net force acting on the particle along the $z$-axis is: $F_z(z) = -\frac{dU}{dz} = \frac{q\sigma}{2\epsilon_0}\left[(1 - \beta) - \frac{z}{\sqrt{R^2 + z^2}}\right]$

---

3. Analysis of Options

For $\beta = \frac{1}{4}$:

The potential energy and net force become: $U(z) = \frac{q\sigma}{2\epsilon_0}\left[\sqrt{R^2 + z^2} - \frac{3}{4}z\right]$ $F_z(z) = \frac{q\sigma}{2\epsilon_0}\left[\frac{3}{4} - \frac{z}{\sqrt{R^2 + z^2}}\right]$

Setting $F_z(z_{\text{eq}}) = 0$ gives the equilibrium position: $\frac{z_{\text{eq}}}{\sqrt{R^2 + z_{\text{eq}}^2}} = \frac{3}{4} \implies z_{\text{eq}} = \frac{3}{\sqrt{7}}R$

• For $z > z_{\text{eq}}$, $F_z(z) < 0$ (force is directed towards the origin, $-\hat{k}$).
• For $z < z_{\text{eq}}$, $F_z(z) > 0$ (force is directed away from the origin, $+\hat{k}$).

Thus, $U(z)$ has a local minimum at $z = z_{\text{eq}}$.

---

Option A: $\beta = \frac{1}{4}$ and $z_0 = \frac{25}{7}R$

Since $z_0 = \frac{25}{7}R > z_{\text{eq}} \approx 1.134 R$, the initial force pushes the particle towards the origin. To see if it reaches the origin $z = 0$, we compare $U(z_0)$ with $U(0)$:

$U(0) = \frac{q\sigma}{2\epsilon_0} R$

$U(z_0) = \frac{q\sigma}{2\epsilon_0}\left[\sqrt{R^2 + \left(\frac{25}{7}R\right)^2} - \frac{3}{4}\left(\frac{25}{7}R\right)\right] = \frac{q\sigma}{2\epsilon_0} R \left(\frac{\sqrt{674}}{7} - \frac{75}{28}\right)$

Evaluating the bracketed term: $\frac{\sqrt{674}}{7} - \frac{75}{28} = \frac{4\sqrt{674} - 75}{28}$

Notice that $(4\sqrt{674})^2 = 16 \times 674 = 10784$ and $(103)^2 = 10609$. Since $10784 > 10609$, we have $4\sqrt{674} > 103$. Therefore: $4\sqrt{674} - 75 > 103 - 75 = 28 \implies \frac{4\sqrt{674} - 75}{28} > 1$

Thus, $U(z_0) > U(0)$. As the system is conservative, the initial total mechanical energy $E = U(z_0)$ is greater than $U(0)$, providing sufficient kinetic energy for the particle to cross the potential hill between $z_{\text{eq}}$ and $0$ and reach the origin.

Option A is correct.

---

Option B: $\beta = \frac{1}{4}$ and $z_0 = \frac{3}{7}R$

Since $z_0 = \frac{3}{7}R < z_{\text{eq}} = \frac{3}{\sqrt{7}}R$, the net force at $z_0$ is positive ($F_z(z_0) > 0$). When released from rest, the particle accelerates in the $+\hat{k}$ direction (away from the origin) and never moves towards $z = 0$.

Option B is incorrect.

---

Option C: $\beta = \frac{1}{4}$ and $z_0 = \frac{R}{\sqrt{3}}$

Since $z_0 = \frac{R}{\sqrt{3}} \approx 0.577 R < z_{\text{eq}}$, the force at $z_0$ is directed in the $+\hat{k}$ direction. The particle accelerates up to $z_{\text{eq}}$, overshoots it due to kinetic energy, reaches a upper turning point $z_1 > z_{\text{eq}}$ where $U(z_1) = U(z_0)$, and then reverses direction, eventually returning back to $z = z_0$. Under conservative forces, it undergoes periodic oscillatory motion between $z_0$ and $z_1$, continually returning to $z = z_0$.

Option C is correct.

---

Option D: $\beta > 1$ and $z_0 > 0$

For $\beta > 1$, we have $(1 - \beta) < 0$. The net force is: $F_z(z) = \frac{q\sigma}{2\epsilon_0}\left[(1 - \beta) - \frac{z}{\sqrt{R^2 + z^2}}\right]$

Since $\frac{z}{\sqrt{R^2 + z^2}} > 0$ for all $z > 0$, it follows that $F_z(z) < 0$ for all $z > 0$. The force is strictly directed towards the origin at all points along the positive $z$-axis. Hence, when released from rest at any point $z_0 > 0$, the particle continuously accelerates towards the origin and always reaches it.

Option D is correct.

---

Conclusion

The correct options are A, C, and D.