Initially, the switch is open, and the capacitor has a charge q$0$q$0.$ The switch is then closed, and the changes in the system are observed. It turns out that the equation describing the subsequent changes in charge, current, and voltage is very similar to that of simple harmonic motion, studied in mechanics. To obtain this equation, we will use the law of conservation of energy.

Initially, the entire energy of the system is stored in the capacitor. When the circuit is closed, the capacitor begins to discharge through the inductor. As the charge of the capacitor decreases, so does its energy. On the other hand, as the current through the inductor increases, so does the energy stored in the inductor. Assuming no heat loss and no emission of electromagnetic waves, energy is conserved, and at any point in time, the sum of the energy stored in the capacitor UCUC and the energy stored in the inductor ULUL is a constant UU:

U$=$UC$+$UL$=$q$22$C$+$LI$22$U$=$UC$+$UL$=$q$22$C$+$LI$22,$

whereqqis the charge on the capacitor andIIis the current through the inductor $($qqandIIare functions of time, of course$).$ For this problem, take clockwise current to be positive.

Recall that the top plate of the capacitor is positively charged at t$=0$t$=0.$ Once the switch is closed, the current will start to increase. In what direction will this current go$?$