LC Circuit

Suppose we have a capacitor with capacitance $C$ and an inductor with inductance $L$ connected on both ends. This is called an LC circuit. Let $Q$ denote the charge on the capacitor, and let $I=\frac{dQ}{dt}$ denote the current. Suppose that, at time $t=0$, the capacitor has a charge $Q_0$, meaning that $Q=Q_0$ at $t=0$. At time $t=0$, the switch is moved from open to closed.

To analyze the RC circuit mathematically, we use Faraday's Law, which states that, over a closed loop $X$: $$\int_{X} \vec{E}\cdot\vec{dl}=-\frac{d\phi}{dt}$$ From the above, we get the following: $$\frac{Q}{C}=-L\frac{dI}{dt}$$ $$\frac{d^2Q}{dt^2}=-\frac{Q}{LC}$$ Note that this is of the form: $$\frac{d^2Q}{dt^2}=-\omega^2 Q$$ where $\omega=\frac{1}{\sqrt{LC}}$. This means that $Q$ follows a simple harmonic oscillation!

The general solution to this differential equation is of the form: $$Q=Q_{max}\cos(\omega t+\phi)$$ where $Q_{max}$ is the maximum charge of the capacitor and $\phi$ is the phase angle. From this, we can easily compute the current: $$I=\frac{dQ}{dt}=-Q_{max}\omega\sin(\omega t+\phi)$$

In this particular case, we know that $Q=Q_0$ at time $t=0$, and since the inductor will fight the change in current, we know that $I=0$ when the switch is closed, which is also $t=0$. Therefore, we get the following: $$Q_{max}\cos\phi=Q_0$$ $$Q_{max}\omega\sin\phi=0$$ Therefore, $\phi=0$ and $Q_{max}=Q_0$, so the solution in our specific case is: $$Q=Q_0\cos\frac{t}{\sqrt{LC}}$$ The period and frequency of this oscillation are, respectively: $$T=2\pi\sqrt{LC}$$ $$f=\frac{1}{2\pi\sqrt{LC}}$$