Match List-I with List-II. (A) Electric field inside (distance $\mathrm{r}>0$ from center) of a uniformly charged spherical shell with surface charge density $\sigma$, and radius R . (B) Electric field at distance $\mathrm{r}>0$ from a uniformly charged infinite plane sheet with surface charge density $\sigma$. (C) Electric field outside (distance $\mathrm{r}>0$ from center) of a uniformly charged spherical shell with surface charge density $\sigma$, and radius R (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density $\sigma$. (I) $\sigma / \varepsilon_{0}$ (II) $\sigma / 2 \varepsilon_{0}$ (III) 0 (IV) $\frac{\sigma \mathrm{R}^{2}}{\varepsilon_{0} \mathrm{r}^{2}}$ Choose the correct answer from the options given below :

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

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

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

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

Electrostatics — JEE Main · 2025 · 29 Jan 2025 · Shift 1 Question with Solution

Match List-I with 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 .
(B) Electric field at distance

r > 0

from a uniformly charged infinite plane sheet with surface charge density

σ

.
(C) Electric field outside (distance

r > 0

from center) of a uniformly charged spherical shell with surface charge density

σ

, and radius R
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density

σ

.

(I)

σ / ε_{0}

(II)

σ /2 ε_{0}

(III) 0
(IV)

\frac{σ R ^{2}}{ε _{0} r ^{2}}

Choose the correct answer from the options given below :

Answer: (D)

For a uniformly charged spherical shell (surface density

σ

, radius

R

), Gauss's law gives zero field everywhere inside (

r < R

), so (A) matches the zero-field entry.

For an infinite charged plane sheet, the field at any distance from it is

E = \frac{σ}{2 ε _{0}}

independent of distance, so (B) matches this entry.

Outside the shell (

r > R

), the entire charge

Q = σ \cdot 4 π R^{2}

acts as if concentrated at the centre:

E = \frac{Q}{4 π ε _{0} r ^{2}} = \frac{σ R ^{2}}{ε _{0} r ^{2}}

an inverse-square field, so (C) matches the

1/ r^{2}

entry.

Between two infinite oppositely charged parallel sheets, the fields from each sheet add in the gap:

E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}}

so (D) matches this entry.

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

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option