Because the electric field intensity in the conductor plate in electrostatic balance is zero everywhere, it is determined by Gauss theorem:
(σ 1+σ2)S=0 (σ3+σ4)S=0
So σ 1=-σ2, σ3=-σ4.
According to gauss theorem, the electric field intensity between AC E 1=σ2/ε0 BC E2=σ3/ε0.
Then the potential difference between AC: U 1=(σ2/ε0)(d/2)
Bc: U2 = (σ 3/ε 0) D.
Meaning: U 1=U2
So: (σ 2/ε 0) (d/2) = (σ 3/ε 0) d ............. (1).
According to the law of conservation of charge: (σ2+σ3)S=Q............(2)
At the same time (1)(2) available: σ2= σ3=
Therefore, the charge quantity QA of plate QA =-σ2s = the charge quantity QB of plate QB =-σ3s =
You do the math yourself. . .