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Electrostatics question

2023 · 1 Feb · Shift 1 · Q44
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  5. /2023 · 1 Feb · Shift 1 · Q44

Electrostatics question

2023 · 1 Feb · Shift 1 · Q44

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Let σ\sigmaσ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region EI,EIIE_{I}, E_{I I}EI​,EII​ and EIIIE_{I I I}EIII​ are: JEE Main 2023 (Online) 1st February Morning Shift Physics - Electrostatics Question 102 English
  1. A
    E⃗I=0,E⃗II=σϵ0n^,EIII=0\vec{E}_{I}=0, \vec{E}_{I I}=\frac{\sigma}{\epsilon_{0}} \hat{n}, E_{I I I}=0EI​=0,EII​=ϵ0​σ​n^,EIII​=0
  2. B
    E⃗I=σ2ϵ0n^,E⃗II=0,E⃗III=σ2ϵ0n^\vec{E}_{I}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}EI​=2ϵ0​σ​n^,EII​=0,EIII​=2ϵ0​σ​n^
  3. C
    E⃗I=−σϵ0n^,E⃗II=0,E⃗III=σϵ0n^\vec{E}_{I}=-\frac{\sigma}{\epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{\epsilon_{0}} \hat{n}EI​=−ϵ0​σ​n^,EII​=0,EIII​=ϵ0​σ​n^
  4. D
    E⃗I=2σϵ0n^,E⃗II=0,E⃗III=2σϵ0n^\vec{E}_{I}=\frac{2 \sigma}{\epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{2 \sigma}{\epsilon_{0}} \hat{n}EI​=ϵ0​2σ​n^,EII​=0,EIII​=ϵ0​2σ​n^
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