JEE Main20231 Feb 2023Morning ShiftPhysicsElectrostaticsActual
Let σ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region E I , E I I and E I I I
Options
- AE → I = 2 σ ∈ 0 n ^ , E → I I = 0 , E → I I I = 2 σ ∈ 0 n ^
- BE → I = 0 , E → I I = σ ∈ 0 n ^ , E → I I I = 0
- CE → I = σ 2 ∈ 0 n ^ , E → I I = 0 , E → I I I = σ 2 ∈ 0 n ^
- DE → I = - σ ∈ 0 n ^ , E → I I = 0 , E → I I I = σ ∈ 0 n ^
Correct answer
D. E → I = - σ ∈ 0 n ^ , E → I I = 0 , E → I I I = σ ∈ 0 n ^
Step-by-step solution
Electric field due to charged sheet for charge density + σ is E → = + σ 2 ∈ 0 and for - σ is E → = - σ 2 ∈ 0 . Assuming RHS to be n ^ E → I = σ 2 ∈ 0 - n ^ + σ 2 ∈ 0 - n ^ = - σ ∈ 0 n ^ (leftwards) E → I I = σ 2 ∈ 0 - n ^ + σ 2 ∈ 0 n ^ = 0 E → I I I = σ 2 ∈ 0 n ^ + σ 2 ∈ 0 n ^ = σ ∈ 0 n ^ (rightwards)