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

Let us take a point P right of the plate.
The electric field at P is the vector sum of the
field due to both the surfaces. Hence,
`vecE = vec(E_1) + vec(E_2)` ,brgt `= (Q_1)/(2Aepsilon_0) hatn + (Q_2)/(2Aepsilon_0)hatn`
`=(Q_1+Q_2)/(2Aepsilon_0) hatn = (Q)/(2Aepsilon_0)hatn` .
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