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

Answer: (C)

Assuming RHS to be $\hat{n}$
$ \vec{E}_{ I }=\frac{\sigma}{2 \epsilon_0}(-\hat{n})+\frac{\sigma}{2 \epsilon_0}(-\hat{n})=-\frac{\sigma}{\epsilon_0} \hat{n} $
$ \vec{E}_{I I}=0 $
$ \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_0}(\hat{n})+\frac{\sigma}{2 \epsilon_0}(\hat{n})=\frac{\sigma}{\epsilon_0}(\hat{n})$