Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Capacitance of the capacitor, C = 30μF = 30 x 10⁻⁶F

Inductance of the inductor, L = 27 mH = 27 x 10⁻³ H

Charge on the capacitor, Q = 6 mC = 6 x 10⁻³ C

Total energy stored in the capacitor can be calculated as:

E = 1/2   x Q²/C  = 1/2 x (6 x 10⁻³)²/(30 x 10⁻⁶) = 6/10 = 0.6 J

Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.