JEE Advanced 2024 · Paper 1

JEE Advanced 2024 Physics question on Electrostatics

Four identical thin, square metal sheets, S1S_1, S2S_2, S3S_3 and S4S_4, each of side aa are kept parallel to each other with equal distance dd (d≪ad \ll a) between them, as shown in the figure. Let C0=ε0a2/dC_0 = \varepsilon_0 a^2/d, where ε0\varepsilon_0 is the permittivity of free space. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
  1. A.
    P→3;Q→2;R→4;S→5P \to 3; Q \to 2; R \to 4; S \to 5
  2. B.
    P→2;Q→3;R→3;S→1P \to 2; Q \to 3; R \to 3; S \to 1
  3. C.
    P→3;Q→2;R→4;S→1P \to 3; Q \to 2; R \to 4; S \to 1
  4. D.
    P→3;Q→2;R→2;S→5P \to 3; Q \to 2; R \to 2; S \to 5
Show answer

Correct option: C

[JEE Advanced 2024 · Paper 1]

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