Question

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
$V(z)=\frac{\sigma }{2{ϵ}_0}\left(\sqrt{R^2+z^2}-z\right)$
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 $\beta =\frac{2c{\in }_0}{q\sigma }$. 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)

$W_{el}+W_{ext}=k_f-k_i$
$q{v}_i-q{v}_f+W_{ext}=k_f-k_i$
$\frac{q\sigma }{2{ϵ}_0}\left[\sqrt{R^2+Z^2}-Z\right]-\frac{q\sigma R}{2{ϵ}_0}+CZ=k_f-0$
$C=\frac{q\sigma B}{2{ϵ}_0}$
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.