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← JEE Advanced 2022 · Paper 2
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JEE Advanced 2022 · Paper 2, Physics question 12

A disk of radius RR with uniform positive charge density σ\sigma is placed on the xyxy plane with its center at the origin. The Coulomb potential along the zz-axis is
V(z)=σ2ϵ0(R2+z2−z).V(z) = \frac{\sigma}{2\epsilon_0}\left(\sqrt{R^2 + z^2} - z\right).
A particle of positive charge qq is placed initially at rest at a point on the zz axis with z=z0z = z_0 and z0>0z_0 > 0. In addition to the Coulomb force, the particle experiences a vertical force F⃗=−c k^\vec F = -c\,\hat k with c>0c > 0. Let β=2cϵ0qσ\beta = \dfrac{2c\epsilon_0}{q\sigma}. Which of the following statement(s) is(are) correct?

Choose one or more options
Answer and worked solution

Official answer: A, C, D

Potential energy. In units of qσ2ϵ0\frac{q\sigma}{2\epsilon_0}:
u(z)=R2+z2−z+βz,u′(z)=zR2+z2−1+β.u(z)=\sqrt{R^2+z^2}-z+\beta z,\qquad u'(z)=\frac z{\sqrt{R^2+z^2}}-1+\beta.
Starting from rest at z0z_0, the particle moves toward lower uu.

Case β=14\beta=\frac14. u′=0u'=0 at zR2+z2=34\frac z{\sqrt{R^2+z^2}}=\frac34, i.e. z∗=3R7≈1.13Rz^*=\frac{3R}{\sqrt7}\approx1.13R. This is a minimum of uu.

  • Starting above z∗z^*, the particle moves down. It reaches the origin if u(z0)≥u(0)=Ru(z_0)\ge u(0)=R.
  • Starting below z∗z^*, it is pushed up and oscillates about z∗z^*, returning to z0z_0.

Check the options.

  • (A) z0=257Rz_0=\frac{25}7R: u=R(6747−2528)≈1.03R≥Ru=R\left(\frac{\sqrt{674}}7-\frac{25}{28}\right)\approx1.03R\ge R. It reaches the origin. True.
  • (B) z0=37R<z∗z_0=\frac37R<z^*: it moves away from the disk. False.
  • (C) z0=R3<z∗z_0=\frac R{\sqrt3}<z^*: it oscillates and returns to z0z_0. True.
  • (D) β>1\beta>1: u′>0u'>0 for every zz, so the net force always points toward the disk, and the particle reaches the origin. True.

(A), (C), (D).