This is a classic problem of an LC (inductor-capacitor) circuit where we need to find the charge on the capacitor as a function of time.

Given:

• Inductance, $L = 2$ Henrys,
• Capacitance, $C = 1$ Farad,
• Initial charge on the capacitor, $Q(0) = 4$ Coulombs,
• Initial current $I(0) = 0$, indicating no initial current flow.

Step 1: Set Up the Differential Equation

In an LC circuit, the charge $Q(t)$ on the capacitor satisfies the differential equation: $L \frac{d^2 Q}{dt^2} + \frac{Q}{C} = 0$

Substituting $L = 2$ and $C = 1$, we get: $2 \frac{d^2 Q}{dt^2} + Q = 0$

Simplifying, this becomes: $\frac{d^2 Q}{dt^2} + \frac{1}{2} Q = 0$

Step 2: Solve the Differential Equation

The general solution for a differential equation of the form: $\frac{d^2 Q}{dt^2} + \omega^2 Q = 0$ is: $Q(t) = A \cos(\omega t) + B \sin(\omega t)$ where $\omega = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Thus, our solution becomes: $Q(t) = A \cos\left(\frac{t}{\sqrt{2}}\right) + B \sin\left(\frac{t}{\sqrt{2}}\right)$

Step 3: Apply Initial Conditions

1. Initial Charge $Q(0) = 4$: $Q(0) = A \cos(0) + B \sin(0) = A$ So, $A = 4$.

2. Initial Current $I(0) = 0$, where current $I(t) = \frac{dQ}{dt}$: $I(t) = -A \frac{1}{\sqrt{2}} \sin\left(\frac{t}{\sqrt{2}}\right) + B \frac{1}{\sqrt{2}} \cos\left(\frac{t}{\sqrt{2}}\right)$ At $t = 0$, $I(0) = 0$, which gives: $0 = -A \frac{1}{\sqrt{2}} \cdot 0 + B \frac{1}{\sqrt{2}} \cdot 1 = B \frac{1}{\sqrt{2}}$ So, $B = 0$.

Final Solution

Thus, the charge on the capacitor for $t > 0$ is: $Q(t) = 4 \cos\left(\frac{t}{\sqrt{2}}\right)$

Would you like a detailed breakdown of each step or have further questions?

Here are some related questions you might find useful to explore:

1. What is the relationship between charge and current in an LC circuit?
2. How does changing the inductance or capacitance affect the oscillation frequency in an LC circuit?
3. How do we interpret the physical meaning of the oscillations in the charge of the capacitor?
4. What happens if we introduce resistance to this LC circuit, creating an RLC circuit?
5. How can we derive the energy stored in the capacitor and inductor over time in an LC circuit?

Tip: In an LC circuit, the total energy remains constant and oscillates between the inductor's magnetic field and the capacitor's electric field.