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

Answer: (A), (C), (D)

$W _{ el }+ W _{ ext }= k _{ f }- k _{ i }$
$qv _{ i }- qv _{ f }+ W _{ ext }= k _{ f }- k _{ i } $
$ \frac{ q \sigma}{2 \epsilon_0}\left[\sqrt{ R ^2+ Z ^2}- Z \right]-\frac{ q \sigma R }{2 \epsilon_0}+ CZ = k _{ f }-0 $
$ C =\frac{ q \sigma B }{2 \epsilon_0}$
Substitute $\beta \& Z$, calculate kinetic energy at $Z =0$
If kinetic energy is positive, then particle will reach at origin
If kinetic energy is negative, then particle will not reach at origin.