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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2023 · 1 February · Morning Shift · Physics · Electrostatics

Let σ be the uniform surface charge density of two infinite thin...

Let σ\sigma be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region EI,EII{E}_{I}, {E}_{II} and EIII{E}_{III}

Let σ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields

Options

  1. A\overset{\to}{E}I = \dfrac{2\sigma}{{\in}_{0}}\overset{^}{n}, \overset{\to}{E}II = 0, \overset{\to}{E}III = \dfrac{2\sigma}{{\in}_{0}}\overset{^}{n}
  2. B\overset{\to}{E}I = 0, \overset{\to}{E}II = \dfrac{\sigma}{{\in}_{0}}\overset{^}{n}, \overset{\to}{E}III = 0
  3. C\overset{\to}{E}I = \dfrac{\sigma}{2{\in}_{0}}\overset{^}{n}, \overset{\to}{E}II = 0, \overset{\to}{E}III = \dfrac{\sigma}{2{\in}_{0}}\overset{^}{n}
  4. D\overset{\to}{E}I = \dfrac{- \sigma}{{\in}_{0}}\overset{^}{n}, \overset{\to}{E}II = 0, \overset{\to}{E}III = \dfrac{\sigma}{{\in}_{0}}\overset{^}{n}

Correct answer

D. \overset{\to}{E}I = \dfrac{- \sigma}{{\in}_{0}}\overset{^}{n}, \overset{\to}{E}II = 0, \overset{\to}{E}III = \dfrac{\sigma}{{\in}_{0}}\overset{^}{n}

Step-by-step solution

Let σ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields

Electric field due to charged sheet for charge density +σ+ \sigma is E→=+σ2∈0\overset{\to}{E} = \dfrac{+ \sigma}{2{\in}_{0}} and for −σ- \sigma is E→=−σ2∈0\overset{\to}{E} = \dfrac{- \sigma}{2{\in}_{0}}.

Assuming RHS to be \overset{^}{n}

\overset{\to}{E}I = \dfrac{\sigma}{2{\in}_{0}}\left(- \overset{^}{n}\right) + \dfrac{\sigma}{2{\in}_{0}}\left(- \overset{^}{n}\right) = - \dfrac{\sigma}{{\in}_{0}}\overset{^}{n} (leftwards)

\overset{\to}{E}II = \dfrac{\sigma}{2{\in}_{0}}\left(- \overset{^}{n}\right) + \dfrac{\sigma}{2{\in}_{0}}\left(\overset{^}{n}\right) = 0

\overset{\to}{E}III = \dfrac{\sigma}{2{\in}_{0}}\left(\overset{^}{n}\right) + \dfrac{\sigma}{2{\in}_{0}}\left(\overset{^}{n}\right) = \dfrac{\sigma}{{\in}_{0}}\left(\overset{^}{n}\right) (rightwards)

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