Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC.

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

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

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

Total energy stored in the capacitor can be calculated by the relation,

$E=\frac{1}{2} \frac{Q^{2}}{C}$

$=\frac{1}{2} \times \frac{\left(6 \times 10^{-3}\right)^{2}}{30 \times 10^{-6}}$

$=\frac{6}{10}=0.6 \mathrm{~J}$

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