Final answer:
The initial charge Q0 on the capacitor in an LC circuit can be found by equating the energy stored in the capacitor to the energy in the inductor at maximum current, using the given values for capacitance, inductance, and maximum current.
Step-by-step explanation:
The student's question concerns an LC (inductor-capacitor) circuit that includes a switch, a capacitor, and an inductor. Initially, the capacitor is fully charged with a charge Q0 on its positive plate. The question states that after the switch is closed, the maximum current in the circuit is 14 mA and asks for the amount of initial charge Q0 on the capacitor.
In such an LC circuit, the energy stored in the capacitor with charge Q0 is given by the equation Uc = 1/2 * C * V^2, where V is the voltage across the capacitor. Since energy is conserved in the LC circuit, this energy is then transferred to the inductor at the moment when the current is maximum. The energy in the inductor at maximum current (Imax) is given by Ul = 1/2 * L * I^2.
Setting the energy of the capacitor equal to the energy of the inductor, we get:
1/2 * C * V^2 = 1/2 * L * I^2
We are given that the maximum current I is 14 mA (or 0.014 A), the inductance L is 27 H, and the capacitance C is 30 nF (or 30 * 10^-9 F). Rearranging the equation to solve for V, we get:
V^2 = (L * I^2) / C
Once V is determined, we can use the relationship Q0 = C * V to find the initial charge on the capacitor. Plugging in the given values:
V^2 = (27 * (0.014)^2) / (30 * 10^-9)
V = sqrt((27 * (0.014)^2) / (30 * 10^-9))
With V calculated, we then find Q0 as:
Q0 = 30 * 10^-9 * V
The correct answer for Q0 will match one of the options provided, which requires the calculation to be completed.