Question

Two infinite, parallel, non-conducting sheets carry equal positive charge density $\sigma$. One is placed in the yz plane at $x=0$ and the other at distance $x=a$. 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), (C), (D)

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

Now , Since both are infinite plates and carry same charge density therefore, electric field between them will be equal to zero.
Now, potential will be constant between them and at $x=0;V=0$ and V = constant between the plates.
Therefore, V=0 between the plates means $0\le x\le a$
Now electric field beyond $x=a$ is $2\times \sigma /2{ϵ}_o=\sigma /{ϵ}_o$
We know that,
$V=-{\int }_a^x\overrightarrow{E}.\overrightarrow{dx}$
$V=-E(x-a)$
$V=-\sigma (x-a)/{ϵ}_o$
and for $x<0$
$V=-{\int }_x^0\overrightarrow{E}.\overrightarrow{dx}$
$V=Ex$
$V=\sigma x/{ϵ}_o$
option (A)(B)(D) are correct.