We consider a parallel plate capacitor.Now the electric field due to a plane sheet of charge is Q/2Ae.This can be calculated using gauss law.

Thus the field existing between two plates of a capacitor is Q/2Ae.Here Q is the magnitude of charge existing on each plate. But the charges on the two plates are opposite in sign. Thus if the charge on the plate setting up the charge is Q, then the charge on the other plate is -Q.

As we know, F=Q*E

Thus force acting on the other plate due to the first plate is(-Q)*Q/2Ae.

As we know force between charges of opposite signs is attractive................So here is the answer to your doubt

 If charge on teh plate = Q

then F = QE

here E = charge density / Permitivity in free space