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

To find the equivalent capacitance between terminal A and terminal B, we can analyze the given circuit diagram. The circuit consists of five capacitors, each with a capacitance of 2 μF, arranged in a bridge network.

Let us label the internal junctions of the bridge:
- Let C be the junction between the top-left 2 μF capacitor and the top-right 2 μF capacitor.
- Let D be the junction between the bottom-left 2 μF capacitor and the bottom-right 2 μF capacitor.

The arrangement forms a Wheatstone bridge where:
- The top-left arm is $C_{\mathrm{AC}}=2\text{μF}$
- The top-right arm is $C_{\mathrm{CB}}=2\text{μF}$
- The bottom-left arm is $C_{\mathrm{AD}}=2\text{μF}$
- The bottom-right arm is $C_{\mathrm{DB}}=2\text{μF}$
- The central bridge arm is $C_{\mathrm{CD}}=2\text{μF}$

We check the balance condition of the Wheatstone bridge by calculating the ratio of the capacitances in the arms:
$\frac{C_{\mathrm{AC}}}{C_{\mathrm{CB}}}=\frac{2\text{μF}}{2\text{μF}}=1$
$\frac{C_{\mathrm{AD}}}{C_{\mathrm{DB}}}=\frac{2\text{μF}}{2\text{μF}}=1$

Since the ratio of the arms is equal:
$\frac{C_{\mathrm{AC}}}{C_{\mathrm{CB}}}=\frac{C_{\mathrm{AD}}}{C_{\mathrm{DB}}}$
The bridge is balanced. This means that the electric potential at node C is equal to the electric potential at node D ($V_C=V_D$). As a result, no charge flows through or accumulates on the central capacitor ($C_{\mathrm{CD}}$), and it can be removed from the circuit analysis.

Once the central capacitor is removed, the circuit simplifies into two parallel branches:
1. Upper Branch: The capacitors $C_{\mathrm{AC}}$ and $C_{\mathrm{CB}}$ are connected in series.
2. Lower Branch: The capacitors $C_{\mathrm{AD}}$ and $C_{\mathrm{DB}}$ are connected in series.

Calculating the equivalent capacitance of the upper branch ($C_{\mathrm{upper}}$):
$C_{\mathrm{upper}}=\frac{C_{\mathrm{AC}}·C_{\mathrm{CB}}}{C_{\mathrm{AC}}+C_{\mathrm{CB}}}=\frac{2·2}{2+2}=1\text{μF}$

Calculating the equivalent capacitance of the lower branch ($C_{\mathrm{lower}}$):
$C_{\mathrm{lower}}=\frac{C_{\mathrm{AD}}·C_{\mathrm{DB}}}{C_{\mathrm{AD}}+C_{\mathrm{DB}}}=\frac{2·2}{2+2}=1\text{μF}$

Now, the equivalent capacitance of the entire circuit between A and B ($C_{\mathrm{AB}}$) is the parallel combination of the upper and lower branches:
$C_{\mathrm{AB}}=C_{\mathrm{upper}}+C_{\mathrm{lower}}=1\text{μF}+1\text{μF}=2\text{μF}$

Therefore, the equivalent capacitance between terminal A and terminal B is 2 μF.