QJEEAdv22Ph02c8: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph02c8 Solution: A disk of radius with uniform positive charge density is placed …

Question

Verified Solution • 2025–26 Session moderate

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{σ}{2 ϵ_{0}} (\sqrt{R^{2} + z^{2}} - z) .

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}&gt;0

. In addition to the Coulomb force, the particle experiences a vertical force

\vec{F}=-c \hat{k}

with

c&gt;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.

Accepted Answer

\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

U_{(z) net} = c (4 R) = 4 R c

When,

z=z_0=\frac{25}{7} \mathrm{R}

and;

\mathrm{B}=\frac{1}{4}

U_{(z) net} = C [4 \times \frac{26 R}{7} - 3 \times \frac{25 R}{7}] = \frac{29 R c}{7}

When,

z=z_0=\frac{3 R}{7}

and

\beta=\frac{1}{4}

U_{(z) net} = 3 R c

When,

Z=\frac{R}{\sqrt{3}}

and

\beta=\frac{1}{4}

U_{(z) n t} = 2.887 Rc

As per option (A), particle reaches at origin with the

\mathrm{KE}

\frac{d (U)_{z}}{d z} = 0 at z = \frac{3 R}{\sqrt{7}}

So, at

\beta = \frac{1}{4}

and

z=\frac{3 R}{\sqrt{7}}

U_{z (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}}&gt;\mathrm{U}_{(z) \text { net }}

at

z=\frac{3 \mathrm{R}}{\sqrt{7}}

Particle will return to

z_0

As per option

D,(\beta&gt;1

and

z_0&gt;0)

U_{z(\text { net })}

increases continuously with

\mathrm{Z}

. So, particle reaches the origin.

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