91Ó°ÊÓ

The energy stored in the electric field of the capacitor is given by: \( U_C = \frac{1}{2}C(V_C)^2 \) The energy stored in the magnetic field of the inductor is given by: \( U_L = \frac{1}{2}LI^2 \)

Using \(V_C = \frac{Q(t)}{C}\) and \(I(t) = \frac{dQ(t)}{dt}\), we can rewrite the energies as: \( U_C = \frac{1}{2}C\left(\frac{Q(t)}{C}\right)^2 = \frac{1}{2}\frac{(Q(t))^2}{C} \) \( U_L = \frac{1}{2}L\left(\frac{dQ(t)}{dt}\right)^2 = \frac{1}{2}L\left(I(t)\right)^2 \)

According to the problem, the energy stored in the electric field is three times the energy stored in the magnetic field. So, we have: \( U_C = 3 U_L \) Plugging in the expressions for energies, we get: \( \frac{1}{2}\frac{(Q(t))^2}{C} = 3\left(\frac{1}{2}L\left(I(t)\right)^2\right) \) Simplifying the equation, we find the relation between Q(t) and I(t): \( Q(t) = 3 \sqrt{CL} I(t) \)

Using Kirchhoff's voltage law for a basic LC circuit, the governing differential equation is: \( L \frac{d^2Q(t)}{dt^2} + \frac{Q(t)}{C} = 0 \) In our case, since \( L = C \), the equation simplifies to: \( \frac{d^2Q(t)}{dt^2} + Q(t) = 0 \) The general solution to this equation is: \( Q(t) = Q_0\cos(t) + C_1\sin(t) \)

Since \(I(t) = \frac{dQ(t)}{dt}\), we have: \( I(t) = -Q_0\sin(t) + C_1\cos(t) \) Now, using the relation between Q(t) and I(t) found in step 3, we get: \( Q_0\cos(t) + C_1\sin(t) = 3 \sqrt{CL} \left(-Q_0\sin(t) + C_1\cos(t)\right) \) At \(t=0\), we have \(Q(0) = Q_0\), so the equation becomes: \( Q_0 = 3 \sqrt{CL} C_1 \) Notice, however, that we need the time t (in milliseconds) when the condition holds. We can find this by using: 1. Converting the capacitance and inductance values to standard units, i.e., \(C = \frac{18}{\pi} \times 10^{-3}\) F and \(L = \frac{18}{\pi} \times 10^{-3}\) H 2. Using the time period of oscillation formula for an LC circuit: \( T = 2\pi\sqrt{LC} \) Substituting the given values into the formula, we get: \( T = 2\pi\sqrt{\left(\frac{18}{\pi}\times 10^{-3}\right)\left(\frac{18}{\pi}\times 10^{-3}\right)} \equiv 6 \mathrm{ms} \) Due to the symmetry of the situation, the condition is satisfied when the time is three-quarters of the time period of oscillation. Therefore: \(t = \frac{3}{4}T \) Substituting the obtained value for T, we get: \( t = \frac{3}{4}\times 6 \mathrm{ms} = 4.5 \mathrm{ms} \) Solution: The energy stored in the electric field is three times the energy stored in the magnetic field at a time of 4.5 milliseconds.