The electric field on two sides of a large charged plate is shown...

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.



$∴ {E}_{0} + \text{E}=1 2 {V m}^{- 1}$

${E}_{0} - \text{E}=8 {V m}^{- 1}$

Solving equations, E=2 V m^{-1} and E_{0}=10 V m^{-1}

Now, electric field due to plate $=\dfrac{\sigma}{2 {ε}_{0}}=2$

$∴ \sigma=4 {ε}_{0}$