Determine the equivalent capacitance between the terminals A and B...

Starting from the left end of the given capacitor network, the leftmost vertical capacitor ($2\ \mu\text{F}$) and the adjacent horizontal capacitor ($2\ \mu\text{F}$) are connected in series.

Their equivalent capacitance is:

$C_1 = \dfrac{2 \times 2}{2 + 2} = 1\ \mu\text{F}$

This equivalent capacitor $C_1$ is in parallel with the first vertical $1\ \mu\text{F}$ capacitor.

Their equivalent capacitance is:

$C_2 = 1 + 1 = 2\ \mu\text{F}$

Now, $C_2$ is in series with the next horizontal $2\ \mu\text{F}$ capacitor.

Their equivalent capacitance is:

$C_3 = \dfrac{2 \times 2}{2 + 2} = 1\ \mu\text{F}$

This equivalent capacitor $C_3$ is in parallel with the second vertical $1\ \mu\text{F}$ capacitor.

Their equivalent capacitance is:

$C_4 = 1 + 1 = 2\ \mu\text{F}$

Finally, $C_4$ is in series with the rightmost horizontal $2\ \mu\text{F}$ capacitor connected to terminal A.

The equivalent capacitance between terminals A and B is:

$C_{AB} = \dfrac{2 \times 2}{2 + 2} = 1\ \mu\text{F}$

Answer: $1\ \mu\text{F}$