Electrostatics Question 19
Question 19 - 01 February - Shift 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 $E_I, E _{II}$ and Surface Charge
$E _{\text{III }}$ are:
(1) $ \vec{E} _I=\frac{2 \sigma}{\epsilon_0} \hat{n}, \vec{E} _{II}=0, \vec{E} _{III}=\frac{2 \sigma}{\epsilon_0} \hat{n}$
(2) $ \vec{E} _I=0, \vec{E} _{II}=\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E} _{III}=0$
(3) $ \vec{E} _I=\frac{\sigma}{2 \epsilon_0} \hat{n}, \vec{E} _{II}=0, \vec{E} _{III}=\frac{\sigma}{2 \epsilon_0} \hat{n}$
(4) $ \vec{E} _I=-\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E} _{II}=0, \vec{E} _{III}=\frac{\sigma}{\epsilon_0} \hat{n}$
Show Answer
Answer: (4)
Solution:
Assuming RHS to be $\hat{n}$
$ \begin{aligned} & \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}) \end{aligned} $
