When B is not grounded, the charge density on both sides of A is Q/(2S), and the electric field intensity in the center of A plate is zero. Due to electrostatic induction, the inner charge density of plate B is -Q/(2S) and the outer charge density is Q/(2S), which can be obtained by Gauss theorem. The electric field intensity between AB E 1 = Q/(2εS), U 1 = E 1d = Qd/(2εS). Because E 1 has nothing to do with d, the electric field intensity outside plate A and plate B is E 1, and the electric field energy w has nothing to do with d, so the force between AB is 0 (moving AB plate, the electric field energy is constant, and the electric field force does not do work, so the electric field force between the two plates is neither attractive nor repulsive, and this force can only be 0).
For the case of grounding B, due to electrostatic induction, the inner charge density of B plate is -Q/S, and the outer charge density is 0; For plate A, the inner charge density is Q/S and the outer charge density is 0, which can be obtained by Gauss theorem. The electric field intensity between AB E2 = Q/(εS), U2 = E2d = Qd/(εS). At this time, there is only one electric field between the plates, and the energy of aW/ad electric field is the energy stored in the capacitor w = u2q/2 = (q 2) d/(2 ε s), and F2 =-а w/а d =-. F2-LT0 indicates that there is gravity between the two plates.
To sum up, the answer is: q/(2 ε s); qd/(2εS)； 0; q/(εS)； qd/(εS)； Q 2/(2 ε s)。 (Note: For simplicity, I will abbreviate ε0 as ε)