Question:
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?
Solution:
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.