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