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The maximum charge on a capacitor in an LC circuit plays a crucial role in defining the energy capacity of the system. In this context, the capacitor stores electrical energy as charge. The relationship between charge and voltage in a capacitor is given by the formula:\[ Q_{max} = C \times V \]where \(C\) represents the capacitance and \(V\) is the voltage across the capacitor.
In this exercise, the given capacitor has a capacitance of \(1.0\, \mathrm{nF} = 1.0 \times 10^{-9}\, \mathrm{F}\), and it experiences a maximum voltage of \(3.0\, \mathrm{V}\).
When we substitute these values into the formula, we find the maximum charge:- \(Q_{max} = 1.0 \times 10^{-9}\, \mathrm{F} \times 3.0\, \mathrm{V} = 3.0 \times 10^{-9}\, \mathrm{C}\)This value signifies the peak amount of charge the capacitor can hold in this particular LC circuit configuration.

In an LC circuit, the maximum current is often reached when the energy shifts entirely from the electric field of the capacitor to the magnetic field of the inductor.
This transition is critical for understanding how LC circuits oscillate without energy loss in an ideal scenario.
The maximum current \(I_{max}\) is related to the maximum charge \(Q_{max}\) and the angular frequency \(\omega\), according to the equation:\[ I_{max} = \omega \times Q_{max} \]The angular frequency \(\omega\) is determined by the formula:\[ \omega = \frac{1}{\sqrt{LC}} \]For our problem, with \(L = 3.0\, \mathrm{mH} = 3.0 \times 10^{-3}\, \mathrm{H}\) and \(C = 1.0 \times 10^{-9}\, \mathrm{F}\),- Calculating \(\omega\), we find: \[ \omega = \frac{1}{\sqrt{1.0 \times 10^{-9} \times 3.0 \times 10^{-3}}} \approx \frac{1}{\sqrt{3.0 \times 10^{-12}}}\]- Thus, the maximum current is: \[ I_{max} = \omega \times 3.0 \times 10^{-9}\, \mathrm{C}\]The result offers insights into the current peaks, essential for designing circuits with predictable performance under oscillating conditions.

In LC circuits, energy can be stored in the magnetic field created by the current through the inductor.
The maximum energy stored in this magnetic field is captured when the current reaches its maximum value.
This energy is represented by the formula:\[ E_{max} = \frac{1}{2} L I_{max}^2 \]where \(L\) is the inductance, and \(I_{max}\) is the maximum current found previously.
Given \(L = 3.0 \times 10^{-3}\, \mathrm{H}\), substituting the computed value of \(I_{max}\) from the earlier section helps us determine this energy.- First, calculate \(I_{max}^2\) as needed.- Then plug this into the energy equation: \[ E_{max} = \frac{1}{2} \times 3.0 \times 10^{-3} \times I_{max}^2 \]This quantifies how much energy is temporarily contained within the inductor's magnetic field, illustrating the vital role inductors play in energy storage and management within circuits.