Question 283 of 311

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

A${\vec{E}}_{I}=\dfrac{2\sigma }{{\in }_{0}}\hat{n},{\vec{E}}_{II}=0,{\vec{E}}_{III}=\dfrac{2\sigma }{{\in }_{0}}\hat{n}$B${\vec{E}}_{I}=0,{\vec{E}}_{II}=\dfrac{\sigma }{{\in }_{0}}\hat{n},{\vec{E}}_{III}=0$C${\vec{E}}_{I}=\dfrac{\sigma }{2{\in }_{0}}\hat{n},{\vec{E}}_{II}=0,{\vec{E}}_{III}=\dfrac{\sigma }{2{\in }_{0}}\hat{n}$D${\vec{E}}_{I}=\dfrac{-\sigma }{{\in }_{0}}\hat{n},{\vec{E}}_{II}=0,{\vec{E}}_{III}=\dfrac{\sigma }{{\in }_{0}}\hat{n}$

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Answer: (D) ${\vec{E}}_{I}=\dfrac{-\sigma }{{\in }_{0}}\hat{n},{\vec{E}}_{II}=0,{\vec{E}}_{III}=\dfrac{\sigma }{{\in }_{0}}\hat{n}$

Electric field due to charged sheet for charge density $+\sigma$ is $\vec{E}=\dfrac{+\sigma }{2{\in }_{0}}$ and for $-\sigma$ is $\vec{E}=\dfrac{-\sigma }{2{\in }_{0}}$.
Assuming RHS to be $\hat{n}$
${\vec{E}}_{I}=\dfrac{\sigma }{2{\in }_{0}}\left(-\hat{n}\right)+\dfrac{\sigma }{2{\in }_{0}}\left(-\hat{n}\right)=-\dfrac{\sigma }{{\in }_{0}}\hat{n}$ (leftwards)
${\vec{E}}_{II}=\dfrac{\sigma }{2{\in }_{0}}\left(-\hat{n}\right)+\dfrac{\sigma }{2{\in }_{0}}\left(\hat{n}\right)=0$
${\vec{E}}_{III}=\dfrac{\sigma }{2{\in }_{0}}\left(\hat{n}\right)+\dfrac{\sigma }{2{\in }_{0}}\left(\hat{n}\right)=\dfrac{\sigma }{{\in }_{0}}\left(\hat{n}\right)$ (rightwards)