91Ó°ÊÓ

The stored energy in capacitor in addition to the stored energy in inductor at any instant equals to the maximum stored energy in the capacitor.

Mathematically it can be given by

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

Where , $\mathrm{U}{}_{\mathrm{cmax}}$is the maximum energy stored in capacitor and ${\mathrm{U}}_{\mathrm{L}}$and ${\mathrm{U}}_{\mathrm{C}}$are the energy stored in inductor and capacitor respectively.

Using

$\frac{1}{2}\mathrm{Li}+\frac{{\mathrm{q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$

For maximum current in LC circuit stored charge on capacitor must be zero .

So,

$$\begin{gathered} \frac{1}{2}\mathrm{Li}{\left({\mathrm{i}}_{\max }\right)}^2=\frac{{\mathrm{Q}}^2}{2\mathrm{C}} \\ \mathrm{Q}={\mathrm{i}}_{\max }\sqrt{\mathrm{LC}} \end{gathered}$$

Put the values of constants in above equation

$$\begin{gathered} \mathrm{Q}=\left(0.850×{10}^{-3}\mathrm{A}\right)\sqrt{0.085\mathrm{H}×3.20×{10}^{-6}\mathrm{F}} \\ \mathrm{Q}=4.43×{10}^{-7}\mathrm{C} \end{gathered}$$

Thus, the maximum charge on the capacitor is $4.43×{10}^{-7}\mathrm{C}$

Using,

$\frac{1}{2}\mathrm{Li}+\frac{{\mathrm{q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$

$\mathrm{q}=\sqrt{{\mathrm{Q}}^2-{\mathrm{LCi}}^2}$

Put the values of constants in above equation

$$\begin{gathered} \mathrm{q}=\sqrt{{\left(4.43×{10}^{-7}\mathrm{C}\right)}^2-0.085\mathrm{H}×3.20×{10}^{-6}×{\left(5.00×{10}^{-4}\right)}^2} \\ \mathrm{q}=3.58×{10}^{-7}\mathrm{C} \end{gathered}$$

Thus, the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA is $3.58×{10}^{-7}\mathrm{C}$.