Question

In an LCcircuit, L= 85.0 mH and C= 3.20 mF. During the oscillations the maximum current in the inductor is 0.850 mA. What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA?

Short answer

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

Step-by-step solution

Energy conservation in LC circuit

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.

Calculation of maximum charge on the capacitor in LC circuit

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

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

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

Calculation of the magnitude of the charge on the capacitor

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

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

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