What happens to the inductor voltage if the inductor is disconnected from the LC circuit exactly when its energy is zero

Question

In this thought experiment we consider the tank circuit, where the energy is exchanged between the capacitor and inductor. The energy when stored in the capacitor is $E_C = \frac{1}{2} \cdot C \cdot V_C^2$ and while stored in the inductor is $E_L = \frac{1}{2} \cdot L \cdot I_L^2$.

Question 1:

Imagine that the inductor is disconnected exactly when its energy is zero, i.e when its current is zero. The current through inductor has a sine like shape. The moment it crosses zero, its slope is not zero hence since the voltage on the inductor is $V_L = L \cdot\frac{d I_L}{dt}$ this will not be zero at that instance. What will happen ? Will the voltage "create an arc" ... why? Actually the voltage on the inductor (being in parallel to cthe capacitor) is exactly the voltage on the capacitor ... i.e "not that big" to create an arc, is it? I tend to think that the voltage on the inductor, if disconnected as specified will simply drop to zero!

Question 2:

What if the now disconnected capacitor will be connected to another inductor (which does not have a potential difference or current flowing through it, i.e "fresh"), what will happen? Will the oscillations continue with the same phase as before?

Answer 1

Sorry for my poor english. My native language is french.

In the first case: the current being zero, there is no induced emf linked to a sudden variation in current. The capacitor retains its charge (which was the maximum charge during the oscillations).

In the second case: you have a simple circuit $(L, C)$ with the initial conditions $q (0)$ known and $i (0) = 0$. The solution is $q(t)=q(0)\cos(\omega t)$ With these initial conditions, there is no transient regime.