Match List - I with List - II. Choose the correct answer from the options given below: List - I List - II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) σ/ε0 (B) Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density σ. (II) σ/2ε0 (C) Electric field outside (distance r>0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (III) 0 (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ. (IV) σε₀ r² (a) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) (b) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (c) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (d) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

Answer: (c). (A) Inside a charged shell there is no enclosed charge, so E = 0 → (III). (B) Infinite sheet: E = (σ)/(2ε₀) → (II). (C) Outside a shell: E = (σ 4π R²)/(4πε₀ r²) = (σ R²)/(ε₀ r²), the 1/r² form → (IV). (D) Between oppositely charged sheets the fields add: E = (σ)/(ε₀) → (I). So the answer is (c) A-III, B-II, C-IV, D-I.

Match List - I with List - II. Choose the correct answer from the… | JEE Main 2025 (29 Jan, Shift 1)

Q34JEE Main 2025 · 29 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

Match List - I with List - II.

Choose the correct answer from the options given below:

List - I	List - II
(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.	(I) σ/ε0
(B) Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density σ.	(II) σ/2ε0
(C) Electric field outside (distance r>0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.	(III) 0
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.	(IV) $\frac{\sigma}{\epsilon_0 r^{2}}$

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Answer: (c)

(A) Inside a charged shell there is no enclosed charge, so $E = 0$ → (III).
(B) Infinite sheet: $E = \frac{\sigma}{2\varepsilon_0}$ → (II).
(C) Outside a shell: $E = \frac{\sigma 4\pi R^2}{4\pi\varepsilon_0 r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}$ , the $\frac{1}{r^2}$ form → (IV).
(D) Between oppositely charged sheets the fields add: $E = \frac{\sigma}{\varepsilon_0}$ → (I).
So the answer is (c) A-III, B-II, C-IV, D-I.

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Electric Flux & Gauss's Law question from JEE Main 2025 (29 Jan, Shift 1)

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