Plates A and B constitute an isolated, charged parallel-plate capacitor. The inner surfaces (I and IV) of A and B have charges +Q and –Q respectively. A third plate C with charge +Q is now introduced midway between A and B. Which of the following statements is not correct?

Field in C’=O

$\frac{\mathrm{Q}-\mathrm{x}}{{\mathrm{ϵ}}_0}+\frac{\mathrm{x}}{\mathrm{ϵ}}+\frac{\mathrm{y}}{{\mathrm{E}}_0}-\frac{\mathrm{Q}-\mathrm{y}}{{\mathrm{ϵ}}_0}-\frac{\mathrm{z}}{{\mathrm{E}}_0}+\frac{\mathrm{Q}+\mathrm{z}}{{\mathrm{ϵ}}_0}=0$

Q + y + Q + y + Q = 0 ; $\mathrm{y}=\raisebox{1ex}{$-\mathrm{Q}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Field in A’ = 0

$\frac{\mathrm{Q}-\mathrm{x}}{{\mathrm{ϵ}}_0}-\frac{\mathrm{x}}{{\mathrm{ϵ}}_0}-\frac{\mathrm{y}}{{\mathrm{ϵ}}_0}-\frac{\mathrm{Q}-\mathrm{y}}{{\mathrm{ϵ}}_0}-\frac{\mathrm{z}}{{\mathrm{ϵ}}_0}+\frac{\mathrm{Q}+\mathrm{z}}{{\mathrm{ϵ}}_0}=0$

$\mathrm{Q}-x-x-\mathrm{y}+\mathrm{Q}+\mathrm{y}-\mathrm{z}+\mathrm{Q}+\mathrm{z}=0;\frac{\mathrm{Q}}{2}=x$

Field in B’ = 0

$\begin{array}{l}\frac{\mathrm{Q}}{2{\mathrm{ϵ}}_0}+\frac{\mathrm{Q}}{2{\mathrm{ϵ}}_0}-\frac{\mathrm{Q}}{2{\mathrm{ϵ}}_0}+\frac{3\mathrm{Q}}{2{\mathrm{ϵ}}_0}+\frac{\mathrm{z}}{{\mathrm{ϵ}}_0}+\frac{\mathrm{Q}+\mathrm{z}}{{\mathrm{ϵ}}_0}=0\\ \mathrm{Q}+\mathrm{Q}+\mathrm{z}+\mathrm{Q}+\mathrm{z}=0\Rightarrow \mathrm{z}=\frac{-3\mathrm{Q}}{2}\end{array}$

for AC, Voltage difference ${\mathrm{v}}_{\mathrm{AC}}=\frac{\mathrm{Q}}{2\mathrm{C}}$, for BC ${\mathrm{v}}_{\mathrm{BC}}=\frac{3\mathrm{Q}}{2\mathrm{C}}$. The potential differences are not equal.