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

$V\left(z\right)=\frac{\sigma }{2{\epsilon }_0}(\sqrt{R^2+z^2}-z)$

A particle of positive charge q is placed initially at rest at a point on the z axis with z = z₀ and z₀ > 0. In addition to the Coulomb force, the particle experiences a vertical force $\overrightarrow{F}=-c\hat{k}$ with c > 0. Let $\lambda =\frac{2{\epsilon }_{0}c}{\sigma q}$. Which of the following statements (s) is (are) correct ?

The correct statements are:

• For $\lambda =\frac{1}{4}$ and z₀ = $\frac{25}{7}R$, the particle reaches the origin.

• For $\lambda =\frac{1}{4}$ and z₀ = $\frac{R}{\sqrt{3}}$, the particle returns back to z = z₀,

• For $\lambda >1$ and z₀ > 0, the particle always reaches the origin.

Step-by-step Explanation:

1. Potential Energy Formulation

The particle experiences two conservative forces along the z-axis:

1. The Coulomb force due to the charged disk with electric potential $V\left(z\right)$.

2. An external downward force $\overrightarrow{F}=-c\hat{k}$ with $c>0$.

The potential energy associated with the Coulomb force is:

$U_e\left(z\right)=qV\left(z\right)=\frac{\sigma q}{2{\epsilon }_0}(\sqrt{R^2+z^2}-z)$

The potential energy associated with the constant vertical force $\overrightarrow{F}=-c\hat{k}$ is:

$U_{ext}\left(z\right)=cz$

Thus, the total potential energy $U\left(z\right)$ of the particle is:

$U\left(z\right)=\frac{\sigma q}{2{\epsilon }_0}(\sqrt{R^2+z^2}-z)+cz$

Using the given parameter $\lambda =\frac{2{\epsilon }_{0}c}{\sigma q}$, we can rewrite $c=\frac{\lambda \sigma q}{2{\epsilon }_0}$. Substituting this gives:

$U\left(z\right)=\frac{}{2{\epsilon }_0}[\sqrt{R^2+z^2}+(\lambda -1)z]$

2. Force and Equilibrium Analysis

The net force acting on the particle along the z-axis is:

$F_z\left(z\right)=-\frac{dU}{dz}=\frac{\sigma q}{2{\epsilon }_0}[1-\lambda -\frac{z}{\sqrt{R^2+z^2}}]$

3. Evaluating the Cases

Case A: When $\lambda >1$ and $z_0>0$

Since $\lambda >1$, the quantity $1-\lambda <0$. Furthermore, $\frac{z}{\sqrt{R^2+z^2}}>0$ for all $z>0$. Therefore, the net force $F_z\left(z\right)$ is strictly negative (downward) everywhere along the positive z-axis.

As a result, $U\left(z\right)$ increases strictly with $z$. When the particle is released from rest at any initial point $z_0>0$, it continuously accelerates downwards and always reaches the origin $z=0$.

Thus, the fourth option is correct.

Case B: When $\lambda =\frac{1}{4}$

For $\lambda =\frac{1}{4}$, we have $1-\lambda =\frac{3}{4}$. The equilibrium position $z_{eq}$ where the net force is zero is found by setting $F_z\left(z\right)=0$:

$\frac{z_{eq}}{\sqrt{R^2+z_{eq}^2}}=\frac{3}{4}\Rightarrow z_{eq}=\frac{3}{\sqrt{7}}R\approx 1.13R$

For $z<z_{eq}$, $F_z\left(z\right)>0$ (upward force), and for $z>z_{eq}$, $F_z\left(z\right)<0$ (downward force). Thus, $z_{eq}$ represents a stable potential energy minimum.

1. For $z_0=\frac{R}{\sqrt{3}}$:

Since $z_0=\frac{R}{\sqrt{3}}\approx 0.58R<z_{eq}$, the net force at $z_0$ is upward ($F_z>0$). Upon release from rest, the particle accelerates upwards towards $z_{eq}$, passes it, reaches an upper turning point, and then turns around to return back to $z=z_0$ in an oscillatory motion.

Thus, the third option is correct.

2. For $z_0=\frac{25}{7}R$:

Since $z_0>z_{eq}$, releasing the particle from rest moves it towards the origin. To check if it reaches the origin, we compare the potential energy at $z_0$ with that at the origin $z=0$:

$U\left(0\right)=\frac{\sigma qR}{2{\epsilon }_0}$

$U\left(z_0\right)=\frac{\sigma q}{2{\epsilon }_0}[\sqrt{R^2+{\left(\frac{25}{7}R\right)}^2}-\frac{3}{4}\left(\frac{25}{7}R\right)]=\frac{\sigma qR}{2{\epsilon }_0}[\frac{\sqrt{674}}{7}-\frac{75}{28}]$

Evaluating the term inside the bracket:

$\frac{4\sqrt{674}-75}{28}\approx \frac{4\left(25.96\right)-75}{28}=\frac{28.84}{28}>1$

Since $U\left(z_0\right)>U\left(0\right)$, the kinetic energy of the particle as it reaches $z=0$ is strictly greater than zero ($K\left(0\right)=U\left(z_0\right)-U\left(0\right)>0$). Therefore, the particle successfully reaches the origin.

Thus, the first option is correct.