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ExamsJEE AdvancedPhysics

A closed Gaussian surface encloses charge Q1, while charge Q2 sits outside it. Four scenarios are described in List-I based on the signs of Q1 and Q2. List-II lists possible combinations of electric field E at an arbitrary point on the surface and the total electric flux phi through the surface. List-I (P) Q1 = 0 and Q2 is a positive charge (Q) Both Q1 and Q2 are positive charges (R) Q1 is positive and Q2 is negative (S) Both Q1 and Q2 are negative charges List-II (1) E = 0, phi = 0 (2) E != 0, phi = 0 (3) E != 0, phi != 0 (4) E = 0, phi != 0 Match each item in List-I with all applicable items from List-II.

  1. (A) P -> 2; Q -> 3,4; R -> 3,4; S -> 3,4
  2. (B) P -> 1; Q -> 1,3; R -> 3,4; S -> 3,4
  3. (C) P -> 3,4; Q -> 1,2; R -> 3; S -> 4
  4. (D) P -> 1; Q -> 2; R -> 3; S -> 4

Correct answer: (A) P -> 2; Q -> 3,4; R -> 3,4; S -> 3,4

Solution

By Gauss's law, phi = Q1/epsilon₀ regardless of Q2. For P, Q1=0 so phi=0 but Q2 causes nonzero E everywhere on the surface (option 2 only). For Q, R, S: Q1 is nonzero so phi != 0; E is nonzero at most points but can be zero at special points on the surface due to superposition, making both options 3 and 4 possible.

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