JEE Advance - Physics (2022 - Paper 2 Online - No. 12)
A disk of radius $\mathrm{R}$ 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?
$$ V(z)=\frac{\sigma}{2 \epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right) . $$
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?
For $\beta=\frac{1}{4}$ and $z_{0}=\frac{25}{7} R$, the particle reaches the origin.
For $\beta=\frac{1}{4}$ and $z_{0}=\frac{3}{7} R$, the particle reaches the origin.
For $\beta=\frac{1}{4}$ and $z_{0}=\frac{R}{\sqrt{3}}$, the particle returns back to $z=z_{0}$.
For $\beta>1$ and $z_{0}>0$, the particle always reaches the origin.
Explanation
$ \mathrm{V}(z) =\frac{\sigma}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right) $
$ \mathrm{U}(z)_{\mathrm{net}} =\frac{\sigma q}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+\mathrm{cz} $
$ =c\left[\frac{\sigma q}{2 c \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+z\right]$
When, $z=0$ and $\beta=1 / 4$
$$ \mathrm{U}_{(z) \text { net }}=c(4 \mathrm{R})=4 R c $$
When, $z=z_0=\frac{25}{7} \mathrm{R}$ and; $\mathrm{B}=\frac{1}{4}$
$$ U_{(z) \text { net }}=C\left[4 \times \frac{26 R}{7}-3 \times \frac{25 R}{7}\right]=\frac{29 R c}{7} $$
When, $z=z_0=\frac{3 R}{7}$ and $\beta=\frac{1}{4}$
$$ \mathrm{U}_{(z) \text { net }}=3 R c $$
When, $Z=\frac{R}{\sqrt{3}}$ and $\beta=\frac{1}{4}$
$$ \mathrm{U}_{(z) n \mathrm{t}}=2.887 \mathrm{Rc} $$
As per option (A), particle reaches at origin with the $\mathrm{KE}$
$$ \frac{d(U)_z}{d z}=0 \text { at } z=\frac{3 R}{\sqrt{7}} $$
So, at $\beta = \frac{1}{4}$ and $z=\frac{3 R}{\sqrt{7}}$
$$ \mathrm{U}_{z(\text { net })}=\sqrt{7} R c=2.645 R c $$
Similarly, for option (B),
$\mathrm{U}_{(z) \text { net }}$ at $z=\frac{\mathrm{R}}{\sqrt{3}}$ the kinetic energy at origin
will become negative and mearly equal to $3 R c$
Also, for option (C)
$\mathrm{U}_{(z) \text { net }}$ at $\mathrm{Z}=\frac{\mathrm{R}}{\sqrt{3}}>\mathrm{U}_{(z) \text { net }}$ at $z=\frac{3 \mathrm{R}}{\sqrt{7}}$
Particle will return to $z_0$
As per option $D,(\beta>1 $ and $z_0>0) $ $U_{z(\text { net })}$ increases continuously with $\mathrm{Z}$. So, particle reaches the origin.
$ \mathrm{U}(z)_{\mathrm{net}} =\frac{\sigma q}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+\mathrm{cz} $
$ =c\left[\frac{\sigma q}{2 c \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+z\right]$
When, $z=0$ and $\beta=1 / 4$
$$ \mathrm{U}_{(z) \text { net }}=c(4 \mathrm{R})=4 R c $$
When, $z=z_0=\frac{25}{7} \mathrm{R}$ and; $\mathrm{B}=\frac{1}{4}$
$$ U_{(z) \text { net }}=C\left[4 \times \frac{26 R}{7}-3 \times \frac{25 R}{7}\right]=\frac{29 R c}{7} $$
When, $z=z_0=\frac{3 R}{7}$ and $\beta=\frac{1}{4}$
$$ \mathrm{U}_{(z) \text { net }}=3 R c $$
When, $Z=\frac{R}{\sqrt{3}}$ and $\beta=\frac{1}{4}$
$$ \mathrm{U}_{(z) n \mathrm{t}}=2.887 \mathrm{Rc} $$
As per option (A), particle reaches at origin with the $\mathrm{KE}$
$$ \frac{d(U)_z}{d z}=0 \text { at } z=\frac{3 R}{\sqrt{7}} $$
So, at $\beta = \frac{1}{4}$ and $z=\frac{3 R}{\sqrt{7}}$
$$ \mathrm{U}_{z(\text { net })}=\sqrt{7} R c=2.645 R c $$
Similarly, for option (B),
$\mathrm{U}_{(z) \text { net }}$ at $z=\frac{\mathrm{R}}{\sqrt{3}}$ the kinetic energy at origin
will become negative and mearly equal to $3 R c$
Also, for option (C)
$\mathrm{U}_{(z) \text { net }}$ at $\mathrm{Z}=\frac{\mathrm{R}}{\sqrt{3}}>\mathrm{U}_{(z) \text { net }}$ at $z=\frac{3 \mathrm{R}}{\sqrt{7}}$
Particle will return to $z_0$
As per option $D,(\beta>1 $ and $z_0>0) $ $U_{z(\text { net })}$ increases continuously with $\mathrm{Z}$. So, particle reaches the origin.
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