Simple Capacitors Circuit 
I can't solve this simple circuit, made of 5 capacitors.I know i only have to use parallel and series rules. The solution in the book is $4/3$ C
 A: This is a balanced Wheatstone Bridge. Hence no current will flow through the $3C$ capacitor. Thus you can simply remove the central wire and have a system of the $2$ branches on the top and on the bottom. $$$$The $2$ capacitors in the top branch are in series, and hence have a net capacitance of $\dfrac{1}{C_{net, top}}=\dfrac11+\dfrac12=\dfrac32C\Rightarrow C_{net,top}=\dfrac23C$.$$$$ Similarly the bottom branch has a capacitance of $C_{net,bottom}=\dfrac23 C$. $$$$Now the two branches on the top, and on the bottom, are in parallel, and hence have a net capacitance of $C_{net,circuit}=C_{net,top}+C_{net,bottom}=\dfrac23+\dfrac23=\dfrac43 C$
A: It's similar to working with resistances but the other way round as seen here.
Try and tackle it by "collapsing" the circuit bit by bit replacing every two capacitors with one that has the same capacitance as the configuration of the two.
The capacitor in the middle has no potential difference across it (due to the symmetry of the circuit along the horizontal axis) so it does not contribute anything to the circuit.
