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/ε 0 (B) Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density σ. (II) σ/2ε 0 (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 Choose the correct answer from the options given below :

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

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

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

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

Electrostatics PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q74 with Solution

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/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) $\frac{\sigma}{\epsilon_0 r^2}$

Choose the correct answer from the options given below :

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)

View written solutionFree

Correct answer: C

Find the electric field for each case in List-I

We use standard results from electrostatics.

(A) Electric field inside a uniformly charged spherical shell

For a spherical shell of radius $R$ with uniform surface charge density $\sigma$ , the electric field at any point inside the shell is

$E=0$

for all $r<R$ .

So,

$(A) \to (III)$

(B) Electric field due to an infinite plane sheet

For a uniformly charged infinite plane sheet with surface charge density $\sigma$ , the electric field on either side is

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

So,

$(B) \to (II)$

Outside the shell, it behaves like a point charge placed at the center.

Total charge on shell:

$Q=4\pi R^2\sigma$

Hence at distance $r>R$ ,

=\frac{1}{4\pi\varepsilon_0}\cdot \frac{4\pi R^2\sigma}{r^2} =\frac{\sigma R^2}{\varepsilon_0 r^2}$$ Now compare with the given List-II entries. Entry (IV) is written as $$\frac{\sigma}{\varepsilon_0 r^2}$$ This appears to be missing a factor of $R^2$, but among the provided options, the intended match for the field outside the spherical shell is clearly **(IV)**. So, $$(C) \to (IV)$$ --- 5. **(D) Electric field between two oppositely charged infinite parallel sheets** Field due to one infinite sheet: $$E=\frac{\sigma}{2\varepsilon_0}$$ Between oppositely charged sheets, the fields add up: $$E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}$$ So, $$(D) \to (I)$$ --- 6. **Final matching** We get: - $(A) \to (III)$ - $(B) \to (II)$ - $(C) \to (IV)$ - $(D) \to (I)$ This corresponds to **Option C**. --- 7. **Compare with stored correct answer** Stored correct answer = **C** Our derived answer = **C** So the answer agrees with the stored correct answer.

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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/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) $\frac{\sigma}{\epsilon_0 r^2}$

Choose the correct answer from the options given below :

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)

View written solutionFree

Correct answer: C

Find the electric field for each case in List-I

We use standard results from electrostatics.

(A) Electric field inside a uniformly charged spherical shell

For a spherical shell of radius $R$ with uniform surface charge density $\sigma$ , the electric field at any point inside the shell is

$E=0$

for all $r<R$ .

So,

$(A) \to (III)$

(B) Electric field due to an infinite plane sheet

For a uniformly charged infinite plane sheet with surface charge density $\sigma$ , the electric field on either side is

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

So,

$(B) \to (II)$

Outside the shell, it behaves like a point charge placed at the center.

Total charge on shell:

$Q=4\pi R^2\sigma$

Hence at distance $r>R$ ,

=\frac{1}{4\pi\varepsilon_0}\cdot \frac{4\pi R^2\sigma}{r^2} =\frac{\sigma R^2}{\varepsilon_0 r^2}$$ Now compare with the given List-II entries. Entry (IV) is written as $$\frac{\sigma}{\varepsilon_0 r^2}$$ This appears to be missing a factor of $R^2$, but among the provided options, the intended match for the field outside the spherical shell is clearly **(IV)**. So, $$(C) \to (IV)$$ --- 5. **(D) Electric field between two oppositely charged infinite parallel sheets** Field due to one infinite sheet: $$E=\frac{\sigma}{2\varepsilon_0}$$ Between oppositely charged sheets, the fields add up: $$E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}$$ So, $$(D) \to (I)$$ --- 6. **Final matching** We get: - $(A) \to (III)$ - $(B) \to (II)$ - $(C) \to (IV)$ - $(D) \to (I)$ This corresponds to **Option C**. --- 7. **Compare with stored correct answer** Stored correct answer = **C** Our derived answer = **C** So the answer agrees with the stored correct answer.

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