JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Work, Power & Energy

A disk of radius

R

with uniform positive charge density

σ

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}} (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} > 0

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

F = - c \hat{k}

with

c > 0

. Let

β = \frac{2 c ϵ _{0}}{q σ}

. Which of the following statement(s) is(are) correct?

(A)For $β = \frac{1}{4}$ and $z_{0} = \frac{25}{7} R$ , the particle reaches the origin.✓ Correct
(B)For $β = \frac{1}{4}$ and $z_{0} = \frac{3}{7} R$ , the particle reaches the origin.
(C)For $β = \frac{1}{4}$ and $z_{0} = \frac{R}{3}$ , the particle returns back to $z = z_{0}$ .✓ Correct
(D)For $β > 1$ and $z_{0} > 0$ , the particle always reaches the origin.✓ Correct

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

Solution

Using work-energy theorem from

z_{0}

z

q [V (z_{0}) - V (z)] + W_{ext} = K_{f}

where

W_{ext} = c z_{0} - cz

(work by downward force).

At

z = 0

K_{f} = \frac{q σ}{2 ϵ _{0}} [R^{2} + z_{0}^{2} - z_{0} - R] + c z_{0}

Substituting

c = \frac{q σ β}{2 ϵ _{0}}

K_{f} = \frac{q σ}{2 ϵ _{0}} [R^{2} + z_{0}^{2} - z_{0} - R + β z_{0}]

The particle reaches origin if

K_{f} \geq 0

.

(A) For

β = \frac{1}{4}

z_{0} = \frac{25 R}{7}

: Substituting gives

K_{f} > 0

. Correct.

(B) For

β = \frac{1}{4}

z_{0} = \frac{3 R}{7}

: Substituting gives

K_{f} < 0

. Incorrect.

(C) For

β = \frac{1}{4}

z_{0} = \frac{R}{3}

: Substituting gives

K_{f} = 0

, so particle just reaches origin with zero velocity and returns. Correct.

(D) For

β > 1

, the external force dominates and

K_{f} > 0

for all

z_{0} > 0

. Correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options