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