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

Hint:

When dealing with electric fields from different charge distributions, remember to apply Gauss’s Law to find the electric field for symmetric configurations like spherical shells and infinite sheets.

Answer 1

The Correct Option is A

- The electric field inside a uniformly charged spherical shell is 0 (Coulomb’s Law), hence \( {(A)-(III)} \). 
- The electric field due to a uniformly charged infinite plane sheet is \( \frac{\sigma}{2\epsilon_0} \), hence \( {(B)-(II)} \). 
- The electric field outside a uniformly charged spherical shell behaves like that of a point charge and is \( \frac{\sigma}{\epsilon_0 r^2} \), hence \( {(C)-(IV)} \). 
- The electric field between two oppositely charged infinite plane sheets is \( \frac{\sigma}{\epsilon_0} \), hence \( {(D)-(I)} \). Thus, the correct answer is \( {(1)} \).