Let the charge surface density of plate A be σ 1, the left and right charge surface densities of plate C be σ2 and σ3 respectively, and the charge surface density of plate B be σ4.

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. . .