Why is $E(t) = 0$ in this electrical circuit?

Question

Word problem

Equation from textbook

Based on the equation above from the textbook, I made $E(t) = 0$, where $E(t)$ represents the sum of the voltage drop throughout the circuit. But why is that true? Because we aren't given the total energy drop throughout the circuit.

Answer 1

$E(t)=0$ since there are no batteries (or any other source) attached to the circuit.

You then get $$Q''(t)=-\frac{1}{LC}Q(t),$$

which has the solution $$Q(t)=A\sin\left(\frac{1}{\sqrt{LC}}t\right)+B\cos\left(\frac{1}{\sqrt{LC}}t\right)$$

The first initial condition, $Q'(0)=0,$ means that $A=0$. The second, $Q(0)=Q_0$, then lands us at $$Q(t)=Q_0\cos\left(\frac{1}{\sqrt{LC}}t\right).$$ The solution is oscillating because of the inductor.