Question

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 $E_I, E_{I I}$ and $E_{I I I}$ are:

• A. $\vec{E}_I=0, \vec{E}_{I I}=\frac{\sigma}{\epsilon_0} \hat{n}, E_{I I I}=0$
• B. $\vec{E}_I=\frac{2 \sigma}{\epsilon_0} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{2 \sigma}{\epsilon_0} \hat{n}$
• C. $\vec{E}_I=-\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{\epsilon_0} \hat{n}$
• D. $\vec{E}_I=\frac{\sigma}{2 \epsilon_0} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_0} \hat{n}$

Answer 1

The Correct Option is C

The correct answer is (C) : $\vec{E}_I=-\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{\epsilon_0} \hat{n}$
Assuming RHS to be $\hat{n}$
${\vec{E}}_I=\frac{\sigma }{2{ϵ}_0}(-\hat{n})+\frac{\sigma }{2{ϵ}_0}(-\hat{n})=-\frac{\sigma }{{ϵ}_0}\hat{n}$
${\vec{E}}_{II}=0$
${\vec{E}}_{III}=\frac{\sigma }{2{ϵ}_0}(\hat{n})+\frac{\sigma }{2{ϵ}_0}(\hat{n})=\frac{\sigma }{{ϵ}_0}(\hat{n})$