Compute the equivalent capacitance between terminals A and B for...

The given circuit represents a Wheatstone bridge.

The ratio of capacitances in the arms connected to terminal A is $\dfrac{C_{AX}}{C_{AY}} = \dfrac{1}{2}$.

The ratio of capacitances in the arms connected to terminal B is $\dfrac{C_{XB}}{C_{YB}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Since $\dfrac{C_{AX}}{C_{AY}} = \dfrac{C_{XB}}{C_{YB}}$, the Wheatstone bridge is balanced.

Therefore, the potential at junction X is equal to the potential at junction Y, and no charge flows through the $5\ \mu\text{F}$ capacitor. It can be removed from the circuit.

The equivalent capacitance of the upper branch ($1\ \mu\text{F}$ and $3\ \mu\text{F}$ in series) is:
$C_{upper} = \dfrac{1 \times 3}{1 + 3} = \dfrac{3}{4} = 0.75\ \mu\text{F}$

The equivalent capacitance of the lower branch ($2\ \mu\text{F}$ and $6\ \mu\text{F}$ in series) is:
$C_{lower} = \dfrac{2 \times 6}{2 + 6} = \dfrac{12}{8} = 1.5\ \mu\text{F}$

The total equivalent capacitance between A and B is the parallel combination of $C_{upper}$ and $C_{lower}$:
$C_{eq} = C_{upper} + C_{lower} = 0.75 + 1.5 = 2.25\ \mu\text{F}$