Question

Two infinite parallel, non conducting sheets carry equal positive charge density $\sigma$. One of them is placed in the $yz$ plane at $x=0$ and the other at $x=a$.
Take potential $V=0$ at $x=0.$ Then,

• A. For $0\le x\le a$, potential $V_x=0$
• B. For $x\ge a$, 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), (B), (D)

The correct options are
A For $0\le x\le a$, potential $V_x=0$
B For $x\ge a$, potential $V_x=-\frac{\sigma }{{ϵ}_0}(x-a)$
D For $x\le 0$, potential $V_x=\frac{-\sigma }{{ϵ}_0}x$

For $0\le x\le a,$ Electric field $E_x=0$
For $x\ge a,$ Electric field $E_x=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$
For $x\le 0,$ Electric field $E_x=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$


Given that potential at $x=0,V=0$

For $0\le x\le a:V_x-V_0=[-{\int }_0^x{E}_{x}dx]$
As $E_x=0\Rightarrow V_x=0$

For $x\ge a;V_x=-{\int }_a^x{E}_{x}dx+V_{\left(a\right)}$
$V_x=[-{\int }_a^x\frac{\sigma }{{ϵ}_0}dx]+V_{\left(a\right)}=-\frac{\sigma }{{ϵ}_0}(x-a)$ ( As $V_a=0$)

For $x\le 0;V_x=-{\int }_0^x{E}_{x}dx+V_{\left(0\right)}=[-\frac{\sigma }{{ϵ}_0}.x]+V_{\left(0\right)}$
$V_x=[-\frac{\sigma }{{ϵ}_0}.x]$