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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 $\beta=\frac{2 c \epsilon_0}{q \sigma}$. 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.
$A,B,C$
$A,B$
$A,C$
$A,C,D$
Solution
$W _{ el }+ W _{ est }= k _{ t }- k _{ i }$
$qv _{ i }- qv _{ t }+ W _{ est }= k _{ t }- k _{ i }$
$\frac{ q \sigma }{2 \epsilon_0}\left[\sqrt{ R ^2+ Z ^2}- Z \right]-\frac{ q \sigma R }{2 \epsilon_0}+ CZ = k _{ t }-0$
$C =\frac{ q \sigma B }{2 \epsilon_0}$
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.