JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 1)
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:
_1st_February_Morning_Shift_en_1_1.png)
$$\vec{E}_{I}=0, \vec{E}_{I I}=\frac{\sigma}{\epsilon_{0}} \hat{n}, E_{I I I}=0$$
$$\vec{E}_{I}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}$$
$$\vec{E}_{I}=-\frac{\sigma}{\epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{\sigma}{\epsilon_{0}} \hat{n}$$
$$\vec{E}_{I}=\frac{2 \sigma}{\epsilon_{0}} \hat{n}, \vec{E}_{I I}=0, \vec{E}_{I I I}=\frac{2 \sigma}{\epsilon_{0}} \hat{n}$$
Explanation
Assuming RHS to be $\hat{n}$
$$ \begin{aligned} & \vec{E}_{\mathrm{I}}=\frac{\sigma}{2 \epsilon_0}(-\hat{n})+\frac{\sigma}{2 \epsilon_0}(-\hat{n})=-\frac{\sigma}{\epsilon_0} \hat{n} \\\\ & \vec{E}_{I I}=0, \\\\ & \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_0}(\hat{n})+\frac{\sigma}{2 \epsilon_0}(\hat{n})=\frac{\sigma}{\in_0}(\hat{n}) \end{aligned} $$
$$ \begin{aligned} & \vec{E}_{\mathrm{I}}=\frac{\sigma}{2 \epsilon_0}(-\hat{n})+\frac{\sigma}{2 \epsilon_0}(-\hat{n})=-\frac{\sigma}{\epsilon_0} \hat{n} \\\\ & \vec{E}_{I I}=0, \\\\ & \vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_0}(\hat{n})+\frac{\sigma}{2 \epsilon_0}(\hat{n})=\frac{\sigma}{\in_0}(\hat{n}) \end{aligned} $$
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