Question

A disk of radius $R$ with uniform positive charge density $\sigma$ is placed on the $x y$ plane with its center at the origin. The Coulomb potential along the $z$-axis is $V(z)=\frac{\sigma}{2 \epsilon_0}\left(\sqrt{R^2+z^2}-z\right)$.
 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 $\beta=\frac{2 c \in_0}{q \sigma}$. Which of the following statement(s) is(are) correct?

• A. For $\beta=\frac{1}{4}$ and $z_0=\frac{25}{7} R$, the particle reaches the origin.
• B. For $\beta=\frac{1}{4}$ and $z_0=\frac{3}{7} R$, the particle reaches the origin.
• C. For $\beta=\frac{1}{4}$ and $z_0=\frac{R}{\sqrt{3}}$, the particle returns back to $z=z_0$.
• D. For $\beta>1$ and $z_0>0$, the particle always reaches the origin.

Answer 1

The Correct Option is A, C, D

Given:
 \(F_1 = \frac{26\sqrt{R^2 + Z^2}}{2E_0}\)
\(F_2 = -ck\)
\(\frac{\alpha \sigma}{280}\)​

For equilibrium at \(z = \frac{Z}{o}\) 
\(F_1 = F_2\)

\(\frac{\alpha \sigma}{280} \cdot \frac{Z}{\sqrt{R^2 + Z^2}} = c\)

From equation (1): 
\(c = (1 - 3\frac{Z}{\sqrt{R^2 + Z^2}})\)

\(\frac{Z}{\sqrt{R^2 + Z^2}} = c(1 - 3\frac{Z}{\sqrt{R^2 + Z^2}})\)

\(\frac{1}{4} \cdot \frac{Z}{\sqrt{R^2 + Z^2}} = \frac{4c}{2c_{80}}\)

\(\frac{7}{\sqrt{R} - R} = 1.13R\)

\(𝑍>1.13𝑅⇒𝐹_2>𝐹_1\)​ Particle reaches the origin.

\(𝑍<1.13𝑅⇒𝐹_1>𝐹_2\) Particle reaches back to 𝑧=𝑍𝑜

Answer 2

To solve this problem, we need to consider the motion of a positive charge placed in a Coulomb potential due to a disk with uniform charge density. The charge also experiences an additional vertical force. Let's break it down step by step:

1. Coulomb Potential and Electric Field: 
The Coulomb potential along the z-axis is given by:

\[ V(z) = \frac{\sigma}{2\epsilon_0} \left( \sqrt{R^2 + z^2} - z \right) \] where: - \( \sigma \) is the charge density of the disk, - \( R \) is the radius of the disk, - \( z \) is the position along the z-axis, - \( \epsilon_0 \) is the permittivity of free space. The electric field \( \vec{E}(z) \) is the negative gradient of the potential with respect to \( z \), given by: \[ \vec{E}(z) = -\frac{dV(z)}{dz} \] For this setup, the electric field at a distance \( z \) from the disk is directed towards the disk along the z-axis.

2. Additional Vertical Force:
In addition to the Coulomb force, the particle experiences a vertical force given by:

\[ \vec{F} = -ck \hat{z} \] where \( c \) is a constant and the vertical force acts downward.

3. Conditions for the Particle's Motion:
To determine the motion of the particle, we must consider both the Coulomb force and the additional vertical force. The total force on the particle will determine its trajectory. Let's analyze each case given in the options:

Option A: For \( \beta = 1 \) and \( z_0 = 25R \), the particle reaches the origin.
Given that the initial position \( z_0 \) is large (25 times the radius), the Coulomb force and vertical force will balance, and the particle will eventually reach the origin. This statement is correct.

Option B: For \( \beta = \frac{1}{3} \) and \( z_0 = 3R \), the particle reaches the origin.
For this value of \( \beta \) and \( z_0 \), the particle will indeed reach the origin as the forces will bring it there. This statement is correct.

Option C: For \( \beta = 1 \) and \( z_0 = \frac{R}{\sqrt{3}} \), the particle returns to \( z = z_0 \).
In this case, the vertical force and Coulomb force would be in equilibrium, so the particle will oscillate and return to its starting position. This statement is correct.

Option D: For \( \beta > 1 \) and \( z_0 > 0 \), the particle always reaches the origin.
For values of \( \beta \) greater than 1, the particle will indeed reach the origin because the forces acting on the particle will pull it towards the origin. This statement is correct.

Final Answer:
The correct options are A, C, D.