JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
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) are:
- (a) E⃗_I=0, E⃗_(I I)=σ/ε₀ n, E_(I I I)=0
- (b) E⃗_I=σ/(2 ε₀) n, E⃗_(I I)=0, E⃗_(I I I)=σ/(2 ε₀) n
- (c) E⃗_I=-σ/ε₀ n, E⃗_(I I)=0, E⃗_(I I I)=σ/ε₀ n
- (d) E⃗_I=(2 σ)/ε₀ n, E⃗_(I I)=0, E⃗_(I I I)=(2 σ)/ε₀ n
Answer and solution
Answer (c) E⃗_I=-σ/ε₀ n, E⃗_(I I)=0, E⃗_(I I I)=σ/ε₀ n
Solution
Physics sir · 2 steps
Each sheet gives σ/2ε₀ away from itself; add the two arrows region by region.
- One sheet alone. Take a Gauss pillbox that cuts through an infinite sheet. Flux leaves through both flat faces, so 2EA = (σ A)/ε₀. This field points away from a positive sheet on both sides and does not depend on the distance from the sheet.
Eₒₙₑₛₕₑₑₜ = σ/2ε₀
Region Field of left sheet Field of right sheet Net field I (left of both) σ/2ε₀, to the left σ/2ε₀, to the left -σ/ε₀n II (between) σ/2ε₀, to the right σ/2ε₀, to the left 0 III (right of both) σ/2ε₀, to the right σ/2ε₀, to the right +σ/ε₀n - So. With n pointing to the right, E⃗_I = -σ/ε₀n, E⃗_(II) = 0 and E⃗_(III) = σ/ε₀n. That is option (c).
The idea to keep. Give each sheet its own σ/2ε₀ arrow pointing away from positive charge, then add region by region. Like charges cancel between the sheets; opposite charges cancel outside.
Answer from the NTA final answer key. Solution by PrepWiser.