To find the effective capacitance between points E and F, we need to analyze the circuit and determine which capacitors are in **series** and which are in **parallel**. Capacitors in series add reciprocally, while capacitors in parallel add directly.

Starting from point E, the two capacitors connected to it are in **series**. Their combined capacitance is:

C_{1} + C_{2} = 2µF + 2µF = 4µF

This 4µF equivalent capacitor is in **parallel** with the third 2µF capacitor, so the total capacitance between E and F is:

C_{total} = C_{1} + C_{2} || C_{3}

C_{total} = C_{1} + C_{2} + C_{3} / (C_{1} + C_{2}) || C_{3}

C_{total} = (C_{1} + C_{2}) * C_{3} / (C_{1} + C_{2} + C_{3})

C_{total} = (4µF) * (2µF) / (4µF + 2µF)

C_{total} = 8µF / 6µF

C_{total} = 1.33µF

Therefore, the effective capacitance between E and F is **1.33µF**.