Question

Each capacitor is of $2\mu F$
Max energy stored in capacitor between A and B [Single capacitor between A and B] in $\mu J$ is .

Answer 1

By using wheatstone bridge and symmetry, the circuit can be altered into the following.
By combining the series and parallel capacitors, we get
The voltage of 9 V is distributed among C, 7C/4 and C.
Their net capacitance is $7C/18$.
$\therefore Q=CV=7C/18\times 9=7C/2$
Voltage of $7C/4$ is $\frac{Q}{7C/4}=\frac{7C/2}{7C/4}=2V$
$\therefore$ The voltage difference between A and C is 2V, and so the voltage difference between A and B is 1V by symmetry.
$\Rightarrow$ the energy in the given capacitor is
$E=\frac{1}{2}C{V}^2=\frac{1}{2}\times 2\mu \times 1^2=1\mu J$