Each capacitor is of 2μF
Max energy stored in capacitor between A and B [Single capacitor between A and B] in μJ is .
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Solution
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. ∴Q=CV=7C/18×9=7C/2
Voltage of 7C/4 is Q7C/4=7C/27C/4=2V ∴ The voltage difference between A and C is 2V, and so the voltage difference between A and B is 1V by symmetry. ⇒ the energy in the given capacitor is E=12CV2=12×2μ×12=1μJ