Calculate the potential difference and the energy stored in the capacitor C2 in the circuit shown in the figure. Given potential at A is 90 V, C1 = 20µF, C2 = 30µF and C3 = 15µF

Equivalent capacitance of the circuit
1/Ceq = 1/C1 + 1/C2 + 1/C3
∴1/Ceq = 1/20 + 1/30 + 1/15
⇒Ceq = 6.67µF
Thus charge flowing through the circuit
Q = CeqV = 6.67µ × 90 = 600µC
Potential difference across C2
V2 = Q/C2 = 600/30 = 20 volts
Energy stored in C2
E2 = 1/2C2V2² = 1/2 × 30µ × (20)² = 6 × 10⁻³J