Calculating effective capacitance of a circuit Please refer to the image below.
The question asks: 'The effective capacitance between terminals A and B in the network shown in the adjoining figure is?'  
Here is my working:
(C5 and C6 connected in series.. in parallel to C2 and C4 connected in series).. connected in series with C1 and C3.
Series(C5+C6) = 1 microFarad
Series(C2+C4) = 1.5 microFarads
Parallel (1+1.5) = 2.5 microFarads
Series(C1+C3+2.5) = 15/26 microFarads (Answer)  
The correct answer from the marking scheme is 8/3 microFarads. Could someone please explain how to get to this answer? I am really confused.  
Thanks

 A: The key to understand the issue is that between the upper and lower "corner" of the circuit the voltage is always zero, therefore no current will flow across $\mathrm C_5$ and $\mathrm C_6$.
In "corners" I mean the points common to $\mathrm C_3$-$\mathrm C_4$ and $\mathrm C_1$-$\mathrm C_2$. These pairs of capacitors are effectively voltage dividers. Whatever the voltage is between $A$ and $B$, these dividers divide it by the same 2:1 ratio, yielding the same voltage.
Because of this you can entirely omit $\mathrm C_5$ and $\mathrm C_6$.
What's left is $\mathrm C_1-\mathrm C_2$ series $=2\;\mathrm{\mu F}$ paralleled with $\mathrm C_3-\mathrm C_4$ in series $= 2/3 \:\mathrm{\mu F}$, so  $6/3 + 2/3 = 8/3 \:\mathrm{\mu F}$.
A: 
Using Kirchhoff's First Rule,finding charges in capacitors between D & C (Let it be $Q_m$). 
$$
Q_1-Q_m-Q_2=0\\
Q_3+Q_m-Q_4=0\\
$$
Using Kirchhoff's Second Rule,finding voltages loops ADC & DBC.
$$
\frac{Q_1}{C_1}+\frac{Q_m}{C_m}-\frac{Q_3}{C_3}=0\\
\frac{Q_2}{C_2}-\frac{Q_4}{C_4}-\frac{Q_m}{C_m}=0
$$
When the Bridge is balanced,$Q_m=0$,So the set of equations can be rewritten as,
$$
Q_1=Q_2\\
Q_3=Q_4\\
Q_1.C_3=Q_3.C_1\space--(1)\\
Q_2.C_4=Q_4.C_2\space--(2)
$$
Dividing 1 & 2,We get,
$$
\frac{C_3}{C_4}=\frac{C_1}{C_2}
$$
From this we can see that the bridge is balanced. Since $\frac{3}{6}=\frac{1}{2}$
Therefore Capacitors $C_5$ & $C_6$ can be taken off.
