Question

Match List - I with List - II:

List - I: 

(A) Electric field inside (distance \( r > 0 \) from center) of a uniformly charged spherical shell with surface charge density \( \sigma \), and radius \( R \). 
(B) Electric field at distance \( r > 0 \) from a uniformly charged infinite plane sheet with surface charge density \( \sigma \). 
(C) Electric field outside (distance \( r > 0 \) from center) of a uniformly charged spherical shell with surface charge density \( \sigma \), and radius \( R \). 
(D) Electric field between two oppositely charged infinite plane parallel sheets with uniform surface charge density \( \sigma \). 

List - II: 

(I) \( \frac{\sigma}{\epsilon_0} \) 
(II) \( \frac{\sigma}{2\epsilon_0} \) 
(III) 0 
(IV) \( \frac{\sigma}{\epsilon_0 r^2} \) Choose the correct answer from the options given below:

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

Hint:

For spherical shells and infinite planes, Gauss's law is often the simplest method to find electric fields.

Answer 1

The Correct Option is A

To solve the given problem, we need to match the scenarios described in List I with their corresponding electric field expressions from List II. Let's analyze each scenario:

List I:

1. (A) Electric field inside a uniformly charged spherical shell: For a uniformly charged spherical shell, the electric field inside (distance \( r > 0 \) from the center) is zero, due to the symmetry of the shell. This matches with List II: (III).
2. (B) Electric field at distance \( r > 0 \) from a uniformly charged infinite plane sheet: The electric field due to an infinite plane sheet with surface charge density \( \sigma \) is given by \( \frac{\sigma}{2\epsilon_0} \), regardless of the distance from the sheet. This matches with List II: (II).
3. (C) Electric field outside a uniformly charged spherical shell: At a point outside a uniformly charged spherical shell, the electric field is the same as if all the charge were concentrated at the center, given by \( \frac{\sigma}{\epsilon_0 r^2} \). 
4. (D) Electric field between two oppositely charged infinite plane parallel sheets: The electric field between two oppositely charged infinite planes each with surface charge density \( \sigma \) results in a field \( \frac{\sigma}{\epsilon_0} \). This matches with List II: (I).

Matching results:

Thus, the matches are:

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

The correct answer from the options is:

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

Answer 2

Step 1: Understand the given physical situations.
We are given four different charge distributions and need to match them with their corresponding expressions for the electric field using Gauss’s law.

Step 2: Recall Gauss’s law.
Gauss’s law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space \( \varepsilon_0 \):
\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}. \] We’ll use this to find the electric field for each configuration.

Step 3: Analyze each case from List - I.
(A) Electric field inside a uniformly charged spherical shell (\( r < R \)):
Inside a spherical shell, the enclosed charge is zero. Hence, the electric field is:
\[ E = 0. \] Corresponds to (III).

(B) Electric field at a distance \( r > 0 \) from an infinite plane sheet with uniform surface charge density \( \sigma \):
From Gauss’s law, for a single infinite sheet of charge, the field is constant and given by:
\[ E = \frac{\sigma}{2\varepsilon_0}. \] Corresponds to (II).

(C) Electric field outside a uniformly charged spherical shell (\( r > R \)):
The field behaves as if all the charge were concentrated at the center of the sphere. Hence,
\[ E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2} = \frac{\sigma}{\varepsilon_0 r^2} \text{ (since } Q = 4\pi R^2 \sigma \text{)}. \] Corresponds to (IV).

(D) Electric field between two oppositely charged infinite parallel sheets each with uniform surface charge density \( \sigma \):
Between the plates, the fields from both sheets add up, giving:
\[ E = \frac{\sigma}{\varepsilon_0}. \] Corresponds to (I).

Step 4: Final matching.
(A) - (III)
(B) - (II)
(C) - (IV)
(D) - (I)

Final Answer:
\[ \boxed{(A)-(III), (B)-(II), (C)-(IV), (D)-(I)} \]