Answer:
The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.
Explanation:
In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).
At any time, the sum of both energies can be expressed as follows:
E = 1/2 Q² / C + 1/2 L I²
In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.
When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.
The total energy, when I is maximum, can be written as follows:
E = 1/2 L I² (1)
In our case, when I= 2.4A, E= 56 mJ.
So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.
Solving for L in (1):
L = 2 *.56 mJ / (2.4)² A² = 20 mH
The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:
Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ
As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:
Ec = 56 mJ - 14.4 mJ = 41.6 mJ
is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)