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

JEE Advanced Physics Question (2022) — Solution

Question

In the figure, the inner (shaded) region  A represents a sphere of radius r A = 1 , within which the electrostatic charge density varies with the radial distance r from the center as ρ A = k r , where k is positive. In the spherical shell B of outer radius r B , the electrostatic charge density varies as ρ B = 2 k r . Assume that dimensions are taken care of. All physical quantities are in their SI units. ​​​​​​​ Which of the following statement(s) is/(are) correct?

Options

  1. A. If  r B = 3 2 , then the electric field is zero everywhere outside B . 
  2. B. If  r B = 3 2 , then the electric potential just outside  B is  k ε 0 .
  3. C. If  r B = 2 , then the total charge of the configuration is  15 π k .
  4. D. If  r B = 5 2 , then the magnitude of the electric field just outside  B  is  13 π k ε 0 .

Answer

B. If  r B = 3 2 , then the electric potential just outside  B is  k ε 0 .

Step-by-step solution

Since both the densities are positive, the total charge can not be zero. Hence, option A is incorrect. For total charge, Q Total = ∫ 0 r A k r 4 π r 2 d r + ∫ r A r B 2 k r 4 π r 2 d r = 4 π k 4 r A 4 + 8 π k 2 r B 2 - r A 2 = π k + 4 π k r B 2 - r A 2 If  r B = 3 2 Q Total   = π k + 4 π k 9 4 - 1 = π k + 4 π k 5 4 = 6 π k The potential just outside  B  will be, V = 1 4 π ε 0 Q total r B = 1 4 π ε 0 6 π k r B = 3 k 2 2 3 ε 0 = k ε 0 Hence, option B is correct. If  r B = 2 Q Total   = π k + 4 π k 4 - 1 = 13 π k Hence, option C is incorrect If  r B = 5 2 Q Total   = π k + 4 π k 25 4 - 1 = π k + π k 21 = 22 π k Therefore, the electric field just outside B will be, E = 1 4 π ε 0 Q Total r B 2 1 4 π ε 0 22 π k 25 × 4 = 22 k 25 ε 0 Hence, option D is incorrect.

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