Question

Two infinite, parallel and non-conducting sheets carries equal positive charge density. One is placed in the $y-z$ plane and the other at distance $x=a$. Take potential $V=0$ at $x=0$. Then choose the correct option(s).

• A. For $0\le x\le a$, potential $V_x=0$
• B. For $x\le a$, potential $V_x=-\frac{\sigma }{{ϵ}_0}(x-a)$
• C. For $x\ge a$, potential $V_x=-\frac{\sigma }{{ϵ}_0}(x-a)$
• D. For $x\le 0$, potential $V_x=-\frac{\sigma }{{ϵ}_0}x$

Answer 1

Answer: (A), (C), (D)

For $x\le 0$, potential $V_x=-\frac{\sigma }{{ϵ}_0}x$


We know that for an infinite, parallel and non-conducting sheets, electric field is given by $E=\frac{\sigma }{2{ϵ}_o}$, where $\sigma$ is charge density.

For $0\le x\le a$ :

${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=0$ because charge density is same.

So,

$\Delta V=-{\overrightarrow{E}}_{\text{net}}.\overrightarrow{dx}=0\Rightarrow V_x-V_{x=0}=0$

$\Rightarrow V_x=0$

For $x\ge a$ :

${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\frac{\sigma }{{ϵ}_o}\hat{i}$

So,

$\Delta V=-{\overrightarrow{E}}_{\text{net}}.\overrightarrow{dx}=-\frac{\sigma }{{ϵ}_o}\hat{i}.(x-a)\hat{i}$

$\Rightarrow V_x-V_{x=0}=-\frac{\sigma }{{ϵ}_o}(x-a)$

$\Rightarrow V_x=-\frac{\sigma }{{ϵ}_o}(x-a)$

For $x\le 0$ :

${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=-\frac{\sigma }{{ϵ}_o}\hat{i}$

So,

$\Delta V=-{\overrightarrow{E}}_{\text{net}}.\overrightarrow{dx}=-(-\frac{\sigma }{{ϵ}_o}\hat{i}).(-x)\hat{i}$

$\Rightarrow V_x-V_{x=0}=-\frac{\sigma }{{ϵ}_o}x$

$\Rightarrow V_x=-\frac{\sigma }{{ϵ}_o}x$