A capacitor and a coil trade energy forever

Take a capacitor holding charge $Q$ and an inductor carrying current $I = dQ/dt$, and go once around the loop:

$$-L\frac{dI}{dt} - \frac{Q}{C} = 0 \qquad\Longrightarrow\qquad L\frac{d^2Q}{dt^2} + \frac{Q}{C} = 0.$$

Compare with the mass on a spring, $m\,\ddot{x} + kx = 0$. Term for term, $L$ is the mass and $1/C$ is the spring constant. That is not a metaphor; it is the same differential equation, so it has the same solution:

$$Q(t) = Q_0\cos(\omega t + \phi), \qquad \omega = \frac{1}{\sqrt{LC}}, \qquad T = 2\pi\sqrt{LC}.$$

Notice what is not in the equation: no first-derivative term, because there is no resistor. An RC circuit obeys $R\,dQ/dt + Q/C = 0$, whose solution is the exponential $Q_0 e^{-t/RC}$. A first-order equation relaxes; a second-order equation with a restoring term oscillates. Read the order of the equation before you sketch anything.

The current follows by differentiating: $I = dQ/dt = -\omega Q_0 \sin(\omega t + \phi)$, so its amplitude is $I_{\max} = \omega Q_0 = Q_0/\sqrt{LC}$.

The energy lives in two places and their sum never changes:

$$U = \frac{Q^2}{2C} + \frac{1}{2}LI^2 = \frac{Q_0^2}{2C} = \frac{1}{2}LI_{\max}^2.$$

That single line settles the phase question. When $Q = \pm Q_0$ the capacitor term is already the whole budget, so the inductor term is zero and $I = 0$. A quarter period later the capacitor is empty, the inductor holds everything, and $I = I_{\max}$. Charge and current cannot peak together: if they did, the total energy would be $Q_0^2/2C + LI_{\max}^2/2$, twice what the circuit was given.

1. $t = 0$: $Q = Q_0$, $I = 0$, all energy in the capacitor.
2. $t = T/4$: $Q = 0$, $I = I_{\max}$, all energy in the inductor.
3. $t = T/2$: $Q = -Q_0$, $I = 0$, capacitor fully charged with the opposite polarity.
4. $t = 3T/4$: $Q = 0$, $I = -I_{\max}$, current running the other way.

At any instant in between, the split is fixed by the same conservation line. Given $Q$, the current is $I = \omega\sqrt{Q_0^2 - Q^2}$: it depends on how much charge is left to lose, not on the charge that is present, which is why $I$ and $Q$ run ninety degrees out of step rather than in proportion.

$\omega = 1/\sqrt{LC}$ is the most-mangled line in the topic, and every mangling has a name. Inverting it, $\omega = \sqrt{LC}$, makes larger components ring faster and gives units of seconds for a frequency. Dropping the root, $\omega = 1/LC$, scales too hard. Reading $T$ as proportional to $C$ scales the right way by the wrong power.

The check that catches all three is the oscillator you already trust: $T = 2\pi\sqrt{m/k}$. Heavier ($L$ up) is slower. Softer ($1/C$ down, so $C$ up) is slower. Both enter under a square root, so

$$C \to 4C \quad\Longrightarrow\quad T \to 2T, \qquad L \to L/4 \quad\Longrightarrow\quad T \to T/2.$$

Quadrupling the capacitance does not quadruple the frequency; it halves it. And $T$ never depends on $Q_0$: like a mass on a spring, the LC circuit takes the same time per cycle whether it was started with a large charge or a small one. Only the amplitude, and with it $I_{\max} = \omega Q_0$, remembers how hard it was pushed.

Units referee any doubt. One henry is one volt-second per ampere; one farad is one coulomb per volt. Their product is $(\text{V}\cdot\text{s}/\text{A})(\text{C}/\text{V}) = \text{s}\cdot\text{C}/\text{A} = \text{s}^2$, so $\sqrt{LC}$ is a time and $1/\sqrt{LC}$ is a rate. $\sqrt{LC}$ on its own is not.

The commonest wrong sketch has $Q(t)$ sliding smoothly down to zero the way it does through a resistor. Ask what the inductor is doing at the instant the capacitor empties. It is carrying $I_{\max}$, and an inductor does not let its current stop: $L\,dI/dt$ would have to be infinite. So the current keeps flowing, and the only place it can put charge is back on the capacitor, with the plates reversed. The capacitor is refilled to $-Q_0$, the current drops to zero, and the whole thing runs again the other way. That is $Q_0\cos(\omega t)$, forever, with no resistance to take a share.

Three pictures, and the equations behind them:

1. RC: first order, $Q = Q_0 e^{-t/RC}$. Monotonic, never negative, never returns.
2. Ideal LC: second order, no damping, $Q = Q_0\cos(\omega t)$. Passes through zero every half period, constant amplitude.
3. RLC: both terms present. Oscillates and decays, because $I^2R$ withdraws energy every cycle.

The inductor is a flywheel, not a drain. A drain would leave $I = 0$ when $Q = 0$; the flywheel is spinning fastest exactly then. That one distinction separates the second picture from the first, and it is the distinction the loop rule made in the first section: no $R$, no first-derivative term, no decay.