The given combination is a balanced condition of Wheatstone bridge because potential at C and D points is same.
Hence, the 2µF capacitor can be neglected. The equivalent circuit is given by the figure below
In the upper arm capacitors 3µF and 6µF are connected in series and similarly in lower arm capacitors 4µF and 8µF are connected in series. The equivalent capacitance is given by
1
C1
=
1
3
+
1
6
=
2+1
6
=
3
6
⇒C1=2µF and
1
C2
=
1
4
+
1
8
=
2+1
8
=
3
8
⇒C2=
8
3
µF The reduced circuit with C1 and C2 capacitance is given as
Now, C1 and C2 are connected in parallel combination. Hence, equivalent capacitance is given by Ceq=C1+C2 Ceq=2+