Question 301: Electrostatics & Capacitance

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:

Accepted Answer

Assuming RHS to be $\hat{n}$
$
\begin{aligned}
& \vec{E}_{\mathrm{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}{\in_0}(\hat{n})
\end{aligned}
$

Answer

$\vec{E}_{I}=0, \vec{E}_{I I}=\frac{\sigma}{\epsilon_{0}} \hat{n}, E_{I I I}=0$

Answer

$\vec{E}_{I}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}$

Answer

$\vec{E}_{I}=-\frac{\sigma}{\epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{\epsilon_{0}} \hat{n}$

Answer

$\vec{E}_{I}=\frac{2 \sigma}{\epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{2 \sigma}{\epsilon_{0}} \hat{n}$

Electrostatics & Capacitance

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:

Choose the correct answer: