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JEE Advanced Physics Electrostatics 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

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 = 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 k ^  with c > 0 . Let β = 2 c ε 0 q σ . Which of the following statement(s) is(are) correct?

Options

  1. A. For β = 1 4  and z 0 = 25 7 R , the particle reaches the origin.
  2. B. For β = 1 4  and z 0 = 3 7 R , the particle reaches the origin.
  3. C. For β = 1 4  and z 0 = R 3 , the particle returns back to  z = z 0
  4. D. For β > 1  and z 0 > 0 , the particle always reaches the origin.

Answer

D. For β > 1  and z 0 > 0 , the particle always reaches the origin.

Step-by-step solution

For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore, c z ≥ σ q 2 ϵ 0 R - R 2 + z 2 - z ⇒ c 2 ϵ 0 σ q ≥ R - R 2 + z 2 - z z ⇒ β ≥ R - R 2 + z 2 - z z For option A R - R 2 + z 2 - z z = R - R 1 + 625 49 - 25 7 R 25 7 ≃ 0 . 242 Since,  0 . 25 > 0 . 242 Option A is correct. For option B R - R 2 + z 2 - z z = R - R 1 + 9 49 - 3 7 R 3 7 ≃ 0 . 79 Since,  0 . 25 < 0 . 79 Option B is incorrect For option C R - R 2 + z 2 - z z = R - R 1 + 1 3 - 1 3 R 1 3 ≃ 0 . 73 Since,  0 . 25 < 0 . 73 Therefore, the particle returns to  z 0 Hence, option C is correct. For option D, For any value of  z > 0 R - R 2 + z 2 - z z < 1 Hence, option D is correct.

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