Question

Two large metal sheets having surface charge density $+\sigma$ and $-\sigma$ are kept perpendicular to each other at a small separation distance $d$. The electric field at any point is 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 B $\frac{\sigma }{2{ϵ}_0}$

Electric field in between those perpendicular plate is given by the,

$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

Now, Electric field due plate (1), is given by,

${\overrightarrow{E}}_1=\frac{-\sigma }{2{\epsilon }_0}$

But in case of $E_2$ w know that

$\varphi =E_{0}da$

$=EAcos\theta$

and we can see from the figure that $\overrightarrow{E}$ and area for $(+\sigma )$ charge sheet are $\perp$, hence there $\varphi =0$, because $\cos \theta =0$

$\Rightarrow {\overrightarrow{E}}_2=0$

$\therefore \overrightarrow{E}=\frac{\sigma }{2{\epsilon }_0}or\frac{-\sigma }{2{\epsilon }_0}$ {Option (B) is the correct }