Question

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`

Answer 1

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

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