If an isolated infinite plate contains a charge `Q_1` on one of its surfaces and a charge `Q_2` on its other surface, then prove that electric field intensity at a point in front of the plate will be `Q//2Aepsilon_0`, where `Q = Q_1+Q_2`

Explanation

Electric field at point P : `vec(E)=vec(E)_(Q_(1))+vec(E)_(Q_(2))=Q_(1)/(2A epsilon_(0)) hat(n)+Q_(2)/(2 A epsilon_(0))hat(n)=(Q_(1)+Q_(2))/(2A epsilon_(0))hat(n)=Q/(2A epsilon_(0))hat(n)`

[This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces].