Match List - I with List - II. ( 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 / _0 (B) Electric field

Correct answer: A. (A)-(III), (B)-(II), (C)-(IV), (D)-(I) Inside uniformly charged spherical shell, E=0 A arrow III For uniformly charged infinite plate E=(σ)/(2 _0) B arrow II Outside of spherical shell E=(Q)/(4 π _0 r_2)=(σ R^2)/( _0 r^2) C arrow IV Between two plates E=(σ)/( _0) D arrow I

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Physics
Chapter: Electrostatics

JEE Main 2025 · 29 January · Morning Shift · Physics · Electrostatics

Match List - I with List - II. ( List - I List - II (A) Electric...

Match List - I with List - II. (

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

) Choose the correct answer from the options given below :

Options

A(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
B(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
C(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
D(A)-(IV), (B)-(I), (C)-(III), (D)-(II)

Correct answer

A. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Step-by-step solution

Inside uniformly charged spherical shell,

E=0

\therefore \mathrm{A} \rightarrow \mathrm{III}

For uniformly charged infinite plate

E=\frac{\sigma}{2 \varepsilon_0}

\therefore \mathrm{B} \rightarrow \mathrm{II}

Outside of spherical shell

E=\frac{Q}{4 \pi \varepsilon_0 r_2}=\frac{\sigma R^2}{\varepsilon_0 r^2}

\therefore \mathrm{C} \rightarrow \mathrm{IV}

Between two plates

E=\frac{\sigma}{\varepsilon_0}

\therefore \mathrm{D} \rightarrow \mathrm{I}

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