Physics - Elecrostatics Question with Solution | TestHub

Electric field due to charged sheet for charge density $+\sigma$ is $\overrightarrow{E}=\frac{+\sigma }{2{\in }_0}$ and for $-\sigma$ is $\overrightarrow{E}=\frac{-\sigma }{2{\in }_0}$.

Assuming RHS to be $\hat{n}$

${\overrightarrow{E}}_I=\frac{\sigma }{2{\in }_0}\left(-\hat{n}\right)+\frac{\sigma }{2{\in }_0}\left(-\hat{n}\right)=-\frac{\sigma }{{\in }_0}\hat{n}$  (leftwards)

${\overrightarrow{E}}_{II}=\frac{\sigma }{2{\in }_0}\left(-\hat{n}\right)+\frac{\sigma }{2{\in }_0}\left(\hat{n}\right)=0$

${\overrightarrow{E}}_{III}=\frac{\sigma }{2{\in }_0}\left(\hat{n}\right)+\frac{\sigma }{2{\in }_0}\left(\hat{n}\right)=\frac{\sigma }{{\in }_0}\left(\hat{n}\right)$    (rightwards)