The electric field on two sides of a large charged plat

SOLUTION

From the figure, it is clear that the  plate is placed in an external electric field. Let the electric field due 
to plate be E, and $E_0$ is the external electric field.

$\therefore E_0+E=12V{m}^{-1}$

$E_0-E=8V{m}^{-1}$
 
Solving equations, $E=2V{m}^{-1}$  and $E_0=10V{m}^{-1}$

Now, electric field due to plate =$\frac{\sigma }{2{ϵ}_0}=2\therefore \sigma =4{ϵ}_0$