Question

Two large metal sheets having surface charge density $+\sigma$ and $-\sigma$ are kept parallel to each other at a small separation distance $d$. The electric field at any point in the region between the plates is

• A. $\frac{\sigma }{{ϵ}_0}$
• B. $\frac{\sigma }{2{ϵ}_0}$
• C. $\frac{2\sigma }{{ϵ}_0}$
• D. $\frac{\sigma }{4{ϵ}_0}$

Answer 1

The correct option is A $\frac{\sigma }{{ϵ}_0}$

Electric field due to sheet $A$, is ${\overrightarrow{E}}_1=\frac{\sigma }{2{\epsilon }_0}i$ {in right side of the sheet}

Also, electric field due to sheet $B$, is ${\overrightarrow{E}}_2=\frac{\sigma }{2{\epsilon }_0}i$ {in left side of the sheet}

$\therefore$ Resultant electric field by parallelogram law of vector addition,

$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$ {$\because \theta =0^0$ between ${\overrightarrow{E}}_1\&{\overrightarrow{E}}_2$}

$=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}$

$\overrightarrow{E}=\frac{\sigma }{{\epsilon }_0}$

$\therefore$ Option (A) is the correct answer.