Calculate the quivalent capacitance of the combination between P and Q as shown in figure below.

Explanation

The equivalent circuit is as shown in figure below

Above circuit behave as Wheatstone bridge
`because" "(2muF)/(2muF)=(4muF)/(4muF)`
Therefore, the current through `6muF` will be zero
So, the equivalent circuit will be

`C_("eq")=((C_(1)C_(2))/(C_(1)+C_(2)))+(C_(3)C_(4))/(C_(3)+C_(4))`
`=(2xx4)/(2+4)+(2xx4)/(2+4)=(2xx4xx2)/(6)=(8)/(3)muF`