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The maximum voltage across the capacitor V is found when the charge is maximum on the plates of the capacitor. The voltage is related to the capacitance and charge of the capacitor in the form${\mathrm{V}}_{\max }=\frac{\mathrm{Q}}{\mathrm{C}}$

Substitute the values we have,

${\mathrm{V}}_{\max }=\frac{6×{10}^{-6}\mathrm{C}}{2.5×{10}^{-4}\mathrm{F}}=0.024\mathrm{V}$

Therefore, the maximum voltage across the capacitor is$0.024\mathrm{V}$

To obey the conservation law of energy, the maximum current at the inductor flows when the energy is maximum in the coil. This energy comes from the fully discharged process of the capacitor

$$\begin{gathered} {\mathrm{U}}_{\mathrm{L}}={\mathrm{U}}_{\mathrm{C}} \\ \frac{1}{2}{{\mathrm{Li}}^2}_{\max }=\frac{{\mathrm{Q}}^2}{2\mathrm{C}} \\ {\mathrm{i}}_{\max }=\frac{\mathrm{Q}}{\sqrt{\mathrm{LC}}} \end{gathered}$$

Substitute the the values for Q, L and $\mathit{C}$we have,

$$\begin{gathered} {\mathrm{i}}_{\max }=\frac{6×{10}^{-6}\mathrm{C}}{\sqrt{\left(0.06\mathrm{H}\right)\left(2.5×{10}^{-4}\mathrm{F}\right)}} \\ =1.55×{10}^{-3}\mathrm{A} \end{gathered}$$

Therefore, the maximum current in the inductor is$1.55×{10}^{-3}\mathrm{A}$

The maximum energy is stored when the maximum current flows through the inductor and this energy is given by${\mathrm{U}}_{\mathrm{Lmax}}=\frac{1}{2}{{\mathrm{Li}}_{\max }}^2$Substitute the values we have

$$\begin{gathered} {\mathrm{U}}_{\mathrm{Lmax}}=\frac{1}{2}\left(0.06\mathrm{H}\right){\left(1.55×{10}^{-3}\mathrm{A}\right)}^2 \\ =7.21×{10}^{-8}\mathrm{J} \end{gathered}$$

Therefore, the maximum energy stored in the inductor is$7.21×{10}^{-8}\mathrm{J}$

When the current is half its value in the inductor, therefore, the energy in the inductor is one four the maximum energy is${\mathrm{U}}_{\mathrm{L}}=\frac{1}{4}{\mathrm{U}}_{\max }=1.80×{10}^{-8}\mathrm{J}$

The remaining energy will be stored in the capacitor. So, we can get the charge of the capacitor by

$$\begin{gathered} {\mathrm{U}}_{\mathrm{C}}=\frac{3}{4}{\mathrm{U}}_{\max }=\frac{(\sqrt{3/4){{\mathrm{Q}}_{\max }}^2}}{2\mathrm{C}}=\frac{{\mathrm{q}}^2}{2\mathrm{C}} \\ \mathrm{q}=\sqrt{\frac{3}{4}{\mathrm{Q}}_{\max }}=5.20×{10}^{-6}\mathrm{C} \end{gathered}$$

Therefore${\mathrm{U}}_{\mathrm{L}}=1.80×{10}^{-8}\mathrm{J},{\mathrm{U}}_{\mathrm{C}}=\frac{{\mathrm{q}}^2}{2\mathrm{C}},\mathrm{q}=5.20×{10}^{-6}\mathrm{C}$