JEE Advanced 2022 — Paper 2 — PHY-II solution

JEE Advanced2022Paper 2PHY-II

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.
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 $z = z_{0}$ .
D
For $β > 1$ and $z_{0} > 0$ , the particle always reaches the origin.

Solution

The total potential energy (Coulomb plus the extra force) is $U (z) = q V (z) + cz = \frac{q σ}{2 ε _{0}} (R^{2} + z^{2} - z) + cz$ .

Using $β = \frac{2 c ε _{0}}{q σ}$ , factor: $U (z) = \frac{q σ}{2 ε _{0}} [R^{2} + z^{2} - z + β z]$ . For $β = 1/4$ this becomes proportional to $4 R^{2} + z^{2} - 3 z$ (up to a positive constant).

Setting $\frac{d U}{d z} = 0$ for $β = 1/4$ : $\frac{4 z}{R ^{2} + z ^{2}} = 3 \Rightarrow z^{*} = \frac{3 R}{7}$ , a maximum of $- U$ (i.e. a barrier).

Energy at barrier: $U (z^{*}) \propto 4 R^{2} + 9 R^{2} /7 - 3 \cdot \frac{3 R}{7} = 7 R$ . Energy at origin: $U (0) \propto 4 R$ . So the barrier lies above the origin.

(A) $z_{0} = 25 R /7$ : $U (z_{0}) \propto 4 \cdot 26 R /7 - 3 \cdot 25 R /7 = 29 R /7 > 4 R$ . Particle has enough energy to overcome the barrier and reach $z = 0$ with positive KE. Correct.

(B) $z_{0} = 3 R /7$ : $U (z_{0}) \propto 4 49 + 9 R /7 - 9 R /7 = (458 - 9) R /7 \approx 3 R < 4 R$ . Energy insufficient to reach origin. Incorrect.

(C) $z_{0} = R / 3$ : $U (z_{0}) \propto 4 \cdot 2 R / 3 - 3 R / 3 = 5 R / 3 \approx 2.89 R$ . Since $U (z_{0}) < U (0) = 4 R$ and the barrier at $3 R / 7$ exceeds $U (z_{0})$ as well (barrier value $7 R \approx 2.646 R$ ... need to compare: $5/ 3 \approx 2.887 > 7 \approx 2.646$ , so the particle has enough energy to cross the barrier away from origin and oscillates back). The particle turns around and returns to $z_{0}$ since it cannot reach origin (energy $< U (0)$ ). Correct.

(D) For $β > 1$ , $U (z) = \frac{q σ}{2 ε _{0}} [R^{2} + z^{2} + (β - 1) z]$ has $\frac{d U}{d z} = \frac{q σ}{2 ε _{0}} [\frac{z}{R ^{2} + z ^{2}} + (β - 1)] > 0$ for all $z > 0$ . So $U$ is monotonically increasing, force pushes the particle toward origin regardless of $z_{0}$ . Correct.

$∴$ Correct options: (A), (C), (D).

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