In the following circuit, the equivalent capacitance between terminal A and terminal B is:

The circuit is a bridge. Let the capacitors be $C_1,C_2,C_3,C_4$ in the arms and $C_5$ in the middle.Since $\frac{C_1}{C_2}=\frac{C_3}{C_4}=\frac{2}{2}=1$, the bridge is balanced.No charge flows through the middle capacitor ($2\mu \text{F}$), so it can be removed.Top branch: Two $2\mu \text{F}$ in series $\Rightarrow C_{top}=1\mu \text{F}$.Bottom branch: Two $2\mu \text{F}$ in series $\Rightarrow C_{bot}=1\mu \text{F}$.These branches are in parallel: $C_{eq}=1+1=2\mu \text{F}$.