Solution

Let $i$ be the total current flowing from the battery and through the first inductor $L$. At the junction, this current splits. Let $q$ be the charge on the capacitor plate at time $t$. The current flowing through the capacitor branch is $i_c = \frac{dq}{dt}$.

By Kirchhoff’s Current Law (KCL), the current flowing through the second inductor (which is in parallel with the capacitor) is $(i – \frac{dq}{dt})$.

Since the second inductor and the capacitor are in parallel, the potential difference across them must be equal: $$ V_L = V_C $$ $$ L \frac{d}{dt}\left(i – \frac{dq}{dt}\right) = \frac{q}{C} $$ $$ L \frac{di}{dt} – L \frac{d^2q}{dt^2} = \frac{q}{C} $$ Rearranging for $L \frac{di}{dt}$: $$ L \frac{di}{dt} = \frac{q}{C} + L \frac{d^2q}{dt^2} \quad \dots (1) $$

Now, apply Kirchhoff’s Voltage Law (KVL) to the outer loop containing the battery, the first inductor, and the capacitor: $$ \mathcal{E} – L \frac{di}{dt} – \frac{q}{C} = 0 $$ Substitute the value of $L \frac{di}{dt}$ from equation (1) into this KVL equation: $$ \mathcal{E} – \left( \frac{q}{C} + L \frac{d^2q}{dt^2} \right) – \frac{q}{C} = 0 $$ $$ \mathcal{E} – \frac{2q}{C} – L \frac{d^2q}{dt^2} = 0 $$ $$ L \frac{d^2q}{dt^2} + \frac{2q}{C} = \mathcal{E} $$ Dividing by $L$, we get the standard differential equation for a harmonic oscillator with a constant forcing term: $$ \frac{d^2q}{dt^2} + \frac{2}{LC}q = \frac{\mathcal{E}}{L} $$

The angular frequency of oscillation is given by $\omega^2 = \frac{2}{LC}$, so $\omega = \sqrt{\frac{2}{LC}}$. The general solution for the charge $q(t)$ is the sum of the complementary function (transient) and the particular integral (steady state): $$ q(t) = q_{steady} + A \cos(\omega t) + B \sin(\omega t) $$ For the steady state (equilibrium), $\frac{d^2q}{dt^2} = 0$, so: $$ \frac{2}{LC}q_{steady} = \frac{\mathcal{E}}{L} \implies q_{steady} = \frac{C\mathcal{E}}{2} $$ Thus: $$ q(t) = \frac{C\mathcal{E}}{2} + A \cos(\omega t) + B \sin(\omega t) $$

Initial Conditions at $t=0$:

1. The capacitor is initially uncharged, so $q(0) = 0$.
2. Due to the inductors, current cannot change instantaneously from zero. Thus $i(0) = 0$ and the current through the second inductor is also zero. Consequently, $\dot{q}(0) = 0$.

Using $q(0) = 0$: $$ 0 = \frac{C\mathcal{E}}{2} + A(1) + 0 \implies A = -\frac{C\mathcal{E}}{2} $$ Using $\dot{q}(0) = 0$: $$ \dot{q}(t) = -A\omega \sin(\omega t) + B\omega \cos(\omega t) $$ $$ 0 = 0 + B\omega \implies B = 0 $$

Substituting $A$ and $B$ back into the charge equation: $$ q(t) = \frac{C\mathcal{E}}{2} – \frac{C\mathcal{E}}{2} \cos(\omega t) = \frac{C\mathcal{E}}{2} [1 – \cos(\omega t)] $$

Finally, the voltage across the capacitor $V_c(t)$ is given by $q(t)/C$: $$ V_c(t) = \frac{\mathcal{E}}{2} [1 – \cos(\omega t)] $$ Substituting $\omega = \sqrt{\frac{2}{LC}}$: