Will energy in an LC circuit become 0 if its is disconnected when the capacitor is fully discharged? In a recent test I had a question in which there was an LC circuit with an inductor a capacitor and a switch. According to the answer key

If switch is opened when capacitor is fully charged energy of LC system remains same.


If switch is opened when capacitor is fully discharged energy of LC system becomes 0.

I can understand the first one but not the second one. The answer keys to this particular exam do tend to be wrong once in a while so I thought I'd get a second opinion.!
 A: When the capacitor has no charge, the energy is in the magnetic field of the inductor, which is associated with a current flow.  If the switch is open, the current cannot flow. The magnetic field collapses, leaving no energy. (The collapse of the field will cause a large voltage spike, and probably an arc across the opening switch. The energy is dissipated there.)
A: 
If switch is opened when capacitor is fully charged energy of LC
system remains same.

If it's a series LC circuit, then the energy will be $E=\frac{CV^2}{2}$ all residing in the electric field of the capacitor. If it is a parallel LC circuit of an ideal inductor and capacitor, the energy will remain the same but it will be exchanged between the electric field of the capacitor and magnetic field of the inductor at the resonant frequency.

If switch is opened when capacitor is fully discharged energy of LC
system becomes 0.

The only place where energy, if any, can be stored is in the magnetic field of the inductor, $E=\frac{LI^2}{2}$ and that would require current in the circuit prior to the switch opening. Once the switch is opened, current cannot flow. The energy stored in the inductor will be dissipated in an arc between the switch contacts as they open, caused by the high voltage induced in the inductor in response to the attempt to disrupt the current by the switch.
Hope this helps.
