The electric field on two sides of a large charged plate is shown in figure. The charge density on the plate in SI units is given by (ε0  is  the  permittivity  of  free  space  in  SI  units)

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  E0 is theexternal electric field. Therefore, E0+E=12 Vm−1.............(i)E0−E=8 Vm−1.............(ii)Solving Eqs. (i) and (ii), E=2  Vm−1  and  E0=10  Vm−1Now, electric field due to plate is σ/2ε0=2.Therefore, σ=4ε0